SATHEE: Limits and Derivatives

Limits and Derivatives

Short Answer Type Questions

1. Evaluate $lim_{x \to 3} \frac{x^{2} - 9}{x - 3}$ .

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Solution

Given,

$\begin{aligned} lim_{x \to 3} \frac{x^{2} - 9}{x - 3} & = lim_{x \to 3} \frac{x^{2} - (3)^{2}}{x - 3} \\ = lim_{x \to 3} \frac{(x + 3) (x - 3)}{(x - 3)} = lim_{x \to 3} (x + 3) \\ = 3 + 3 = 6 \end{aligned}$

2. Evaluate $lim_{x \to 1 / 2} \frac{4 x^{2} - 1}{2 x - 1}$ .

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Solution

Given,

$\begin{aligned} lim_{x \to 1 / 2} \frac{4 x^{2} - 1}{2 x - 1} & = lim_{x \to 1 / 2} \frac{(2 x)^{2} - (1)^{2}}{2 x - 1} \\ = lim_{x \to 1 / 2} \frac{(2 x + 1) (2 x - 1)}{(2 x - 1)} = lim_{x \to 1 / 2} (2 x + 1) \\ = 2 \times \frac{1}{2} + 1 = 1 + 1 = 2 \end{aligned}$

3. Evaluate $lim_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h}$ .

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Solution

Given, $lim_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h} = lim_{h \to 0} \frac{(x + h)^{1 / 2} - (x)^{1 / 2}}{x + h - x}$

$= lim_{h \to 0} \frac{(x + h)^{1 / 2} - (x)^{1 / 2}}{(x + h) - x}$ $[∵ lim_{x \to a} \frac{x^{n} - a^{n}}{x - a} = n a^{n - 1}]$

$= \frac{1}{2} x^{\frac{1}{2} - 1} = \frac{1}{2} x^{- 1 / 2}$ $[∵ h \to 0 \Rightarrow x + h \to x]$

$= \frac{1}{2 \sqrt{x}}$

4. Evaluate $lim_{x \to 0} \frac{(x + 2)^{1 / 3} - 2^{1 / 3}}{x}$ .

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Solution

Given, $lim_{x \to 0} \frac{(x + 2)^{1 / 3} - 2^{1 / 3}}{x} = lim_{x \to 0} \frac{(x + 2)^{1 / 3} - 2^{1 / 3}}{(x + 2) - 2}$

$= \frac{1}{3} \times 2^{\frac{1}{3} - 1}$

$= \frac{1}{3} \times (2)^{- 2 / 3} [∵ lim_{x \to a} \frac{x^{n} - a^{n}}{x - a} = n a^{n - 1}]$

$= \frac{1}{3 (2)^{2 / 3}} [∵ x \to 0 \Rightarrow x + 2 \to 2]$

5. Evaluate $lim_{x \to 0} \frac{(1 + x)^{6} - 1}{(1 + x)^{2} - 1}$ .

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Solution

Given, $lim_{x \to 0} \frac{(1 + x)^{6} - 1}{(1 + x)^{2} - 1} = lim_{x \to 0} \frac{\frac{(1 + x)^{6} - 1}{x}}{\frac{(1 + x)^{2} - 1}{x}}$ [dividing numerator and denominator by $x$ ]

$= lim_{x \to 0} \frac{\frac{(1 + x)^{6} - 1}{(1 + x) - 1}}{\frac{(1 + x)^{2} - 1}{(1 + x) - 1}}$ $[∵ x \to 0 \Rightarrow 1 + x \to 1]$

$= \frac{lim_{x \to 0} \frac{(1 + x)^{6} - (1)^{6}}{(1 + x) - 1}}{lim_{x \to 0} \frac{(1 + x)^{2} - (1)^{2}}{(1 + x) - 1}} [∵ lim_{x \to a} \frac{f (x)}{g (x)} = \frac{lim_{x \to a} f (x)}{lim_{x \to a} g (x)}]$

$= \frac{6 (1)^{6 - 1}}{2 (1)^{2 - 1}} [∴ lim_{x \to a} \frac{x^{n} - a^{n}}{x - a} = n a^{n - 1}]$

$= \frac{6 \times 1}{2 \times 1} = \frac{6}{2} = 3$

6. Evaluate $lim_{x \to a} \frac{(2 + x)^{5 / 2} - (a + 2)^{5 / 2}}{x - a}$ .

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Solution

Given, $lim_{x \to a} \frac{(2 + x)^{5 / 2} - (a + 2)^{5 / 2}}{x - a} = lim_{x \to a} \frac{(2 + x)^{5 / 2} - (a + 2)^{5 / 2}}{(2 + x) - (a + 2)}$

$= \frac{5}{2} (a + 2)^{\frac{5}{2} - 1} [∵ lim_{x \to a} \frac{x^{n} - a^{n}}{x - a} = n a^{n - 1}]$

$= \frac{5}{2} (a + 2)^{3 / 2} [∵ x \to a \Rightarrow x + 2 \to a + 2]$

7. Evaluate $lim_{x \to 1} \frac{x^{4} - \sqrt{x}}{\sqrt{x} - 1}$ .

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Solution

Given, $lim_{x \to 1} \frac{x^{4} - \sqrt{x}}{\sqrt{x} - 1} = lim_{x \to 1} \frac{\sqrt{x} [(x)^{7 / 2} - 1]}{\sqrt{x} - 1}$

$\begin{matrix} = lim_{x \to 1} \frac{(x)^{7 / 2} - 1}{\sqrt{x} - 1} \cdot lim_{x \to 1} \sqrt{x} & [∵ lim_{x \to a} f (x) \cdot g (x) = lim_{x \to a} f (x) \cdot lim_{x \to a} g (x)] \\ = lim_{x \to 1} \frac{\frac{x^{7 / 2} - 1}{x - 1}}{\frac{(x)^{1 / 2} - 1}{x - 1}} \cdot 1 \\ = \frac{lim_{x \to 1} \frac{x^{7 / 2} - 1}{x - 1}}{lim_{x \to 1} \frac{(x)^{1 / 2} - 1}{x - 1}} \\ = \frac{\frac{7}{2} (1)^{\frac{7}{2} - 1} - \frac{7}{2}}{\frac{1}{2}} = 7 \\ \frac{1}{2} (1)^{\frac{1}{2} - 1} \frac{1}{2} \end{matrix}$

8. Evaluate $lim_{x \to 2} \frac{x^{2} - 4}{\sqrt{3 x - 2} - \sqrt{x + 2}}$ .

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Solution

Given, $lim_{x \to 2} \frac{x^{2} - 4}{\sqrt{3 x - 2} - \sqrt{x + 2}} = lim_{x \to 2} \frac{(x^{2} - 4) \sqrt{3 x - 2} + \sqrt{x + 2})}{(\sqrt{3 x - 2} - \sqrt{x + 2}) (\sqrt{3 x - 2} - \sqrt{x + 2})}$

$= lim_{x \to 2} \frac{(x^{2} - 4) (\sqrt{3 x - 2} + \sqrt{x + 2})}{(\sqrt{3 x - 2})^{2} - (\sqrt{x + 2})^{2}}$

$\begin{aligned} [∵ (a + b) (a - b) = a^{2} - b^{2}] \\ = lim_{x \to 2} \frac{(x^{2} - 4) (\sqrt{3 x - 2} + \sqrt{x + 2})}{(3 x - 2) - (x + 2)} \\ = lim_{x \to 2} \frac{(x^{2} - 4) (\sqrt{3 x - 2} + \sqrt{x + 2})}{3 x - 2 - x - 2} \\ = lim_{x \to 2} \frac{(x^{2} - 4) (\sqrt{3 x - 2} + \sqrt{x + 2})}{2 x - 4} \\ = lim_{x \to 2} \frac{(x + 2) (x - 2) (\sqrt{3 x - 2} + \sqrt{x + 2})}{2 (x - 2)} \\ = lim_{x \to 2} \frac{(x + 2) (\sqrt{3 x - 2} + \sqrt{x + 2})}{2} \\ = \frac{(2 + 2) (\sqrt{6 - 2} + \sqrt{2 + 2})}{2} \\ = \frac{4 (2 + 2)}{2} = 8 \end{aligned}$

9. Evaluate $lim_{x \to \sqrt{2}} \frac{x^{4} - 4}{x^{2} + 3 \sqrt{2} x - 8}$ .

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Solution

Given, $lim_{x \to \sqrt{2}} \frac{x^{4} - 4}{x^{2} + 3 \sqrt{2} x - 8} = lim_{x \to \sqrt{2}} \frac{(x^{2})^{2} - (2)^{2}}{x^{2} + 3 \sqrt{2} x - 8}$

$\begin{aligned} = lim_{x \to \sqrt{2}} \frac{(x^{2} - 2) (x^{2} + 2)}{x^{2} + 4 \sqrt{2} x - \sqrt{2} x - 8} \\ = lim_{x \to \sqrt{2}} \frac{(x - \sqrt{2}) (x + \sqrt{2}) (x^{2} + 2)}{x (x + 4 \sqrt{2)} - \sqrt{2} (x + 4 \sqrt{2})} \\ = lim_{x \to \sqrt{2}} \frac{(x - \sqrt{2}) (x + \sqrt{2}) (x^{2} + 2)}{(x - \sqrt{2}) (x + 4 \sqrt{2})} \\ = lim_{x \to \sqrt{2}} \frac{(x + \sqrt{2}) (x^{2} + 2)}{(x + 4 \sqrt{2})} \\ = \frac{(\sqrt{2} + \sqrt{2}) [(\sqrt{2})^{2} + 2]}{(\sqrt{2} + 4 \sqrt{2})} \\ = \frac{2 \sqrt{2} (2 + 2)}{5 \sqrt{2}} = \frac{8}{5} \end{aligned}$

10. Evaluate $lim_{x \to 1} \frac{x^{7} - 2 x^{5} + 1}{x^{3} - 3 x^{2} + 2}$ .

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Solution

Given,

$\begin{aligned} lim_{x \to 1} \frac{x^{7} - 2 x^{5} + 1}{x^{3} - 3 x^{2} + 2} \\ = & lim_{x \to 1} \frac{x^{7} - x^{5} - x^{5} + 1}{x^{3} - x^{2} - 2 x^{2} + 2} \\ = & lim_{x \to 1} \frac{x^{5} (x^{2} - 1) - 1 (x^{5} - 1)}{x^{2} (x - 1) - 2 (x^{2} - 1)} \end{aligned}$

On dividing numerator and denominator by $(x - 1)$ , then

$\begin{aligned} = lim_{x \to 1} \frac{\frac{x^{5} (x^{2} - 1)}{(x - 1)} - \frac{1 (x^{2} - 1)}{(x - 1)}}{\frac{x^{2} (x - 1)}{(x - 1)} - \frac{2 (x^{2} - 1)}{(x - 1)}} \\ = \frac{lim_{x \to 1} x^{5} (x + 1) - lim_{x \to 1} \frac{x^{5} - 1}{x - 1}}{lim_{x \to 1} x^{2} - lim_{x \to 1} (x + 1)} \\ = \frac{1 \times 2 - 5 \times (1)^{4}}{1 - 2 \times 2} = \frac{2 - 5}{1 - 4} \\ = \frac{- 3}{- 3} = 1 \end{aligned}$

11. Evaluate $lim_{x \to 0} \frac{\sqrt{1 + x^{3}} - \sqrt{1 - x^{3}}}{x^{2}}$ .

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Solution

Given, $lim_{x \to 0} \frac{\sqrt{1 + x^{3}} - \sqrt{1 - x^{3}}}{x^{2}} = lim_{x \to 0} \frac{\sqrt{1 + x^{3}} - \sqrt{1 - x^{3}}}{x^{2}} \cdot \frac{\sqrt{1 + x^{3}} + \sqrt{1 - x^{3}}}{\sqrt{1 + x^{3}} + \sqrt{1 - x^{3}}}$

$= lim_{x \to 0} \frac{(1 + x^{3}) - (1 - x^{3})}{x^{2} (\sqrt{1 + x^{3}} + \sqrt{1 - x^{3}})}$ $= lim_{x \to 0} \frac{1 + x^{3} - 1 + x^{3}}{x^{2} (\sqrt{1 + x^{3}} + \sqrt{1 - x^{3}})}$

$= lim_{x \to 0} \frac{2 x^{3}}{x^{2} (\sqrt{1 + x^{3}} + \sqrt{1 - x^{3}})}$

$= lim_{x \to 0} \frac{2 x}{(\sqrt{1 + x^{3}} + \sqrt{1 - x^{3}})}$

$= 0$

12. Evaluate $lim_{x \to - 3} \frac{x^{3} + 27}{x^{5} + 243}$ .

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Solution

Given, $lim_{x \to - 3} \frac{x^{3} + 27}{x^{5} + 243} = lim_{x \to - 3} \frac{\frac{x^{3} + 27}{x + 3}}{\frac{x^{5} + 243}{x + 3}}$

$\begin{matrix} = lim_{x \to - 3} \frac{\frac{x^{3} - (- 3)^{3}}{x - (- 3)}}{\frac{x^{5} - (- 3)^{5}}{x - (- 3)}} = \frac{lim_{x \to - 3} \frac{x^{3} - (- 3)^{3}}{x - (- 3)}}{lim_{x \to - 3} \frac{x^{5} - (- 3)^{5}}{x - (- 3)}} & [∵ lim_{x \to a} \frac{f (x)}{g (x)} = \frac{lim_{x \to a} f (x)}{lim_{x \to a} g (x)}] \\ = \frac{3 (- 3)^{3 - 1}}{5 (- 3)^{5 - 1}} = \frac{3}{5} \frac{(- 3)^{2}}{(- 3)^{4}} & [∵ lim_{x \to a} \frac{x^{n} - a^{n}}{x - a} = n a^{n - 1}] \\ = \frac{3}{5 (- 3)^{2}} = \frac{3}{45} = \frac{1}{15} \end{matrix}$

13. Evaluate $lim_{x \to 1 / 2} (\frac{8 x - 3}{2 x - 1} - \frac{4 x^{2} + 1}{4 x^{2} - 1})$ .

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Solution

Given, $lim_{x \to 1 / 2} (\frac{8 x - 3}{2 x - 1} - \frac{4 x^{2} + 1}{4 x^{2} - 1}) = lim_{x \to 1 / 2} [\frac{(8 x - 3) (2 x + 1) - (4 x^{2} + 1)}{(4 x^{2} - 1)}]$

$\begin{aligned} = lim_{x \to 1 / 2} [\frac{16 x^{2} + 8 x - 6 x - 3 - 4 x^{2} - 1}{4 x^{2} - 1}] \\ = lim_{x \to 1 / 2} [\frac{12 x^{2} + 2 x - 4}{4 x^{2} - 1}] \\ = lim_{x \to 1 / 2} [\frac{2 (6 x^{2} + x - 2)}{4 x^{2} - 1}] \end{aligned}$

$\begin{aligned} = lim_{x \to 1 / 2} \frac{2 (6 x^{2} + 4 x - 3 x - 2)}{4 x^{2} - 1} \\ = lim_{x \to 1 / 2} \frac{2 [2 x (3 x + 2) - 1 (3 x + 2)]}{4 x^{2} - 1} \\ = lim_{x \to 1 / 2} \frac{2 [(3 x + 2) (2 x - 1)]}{(2 x)^{2} - (1)^{2}} \\ = lim_{x \to 1 / 2} \frac{2 (3 x + 2) (2 x - 1)}{(2 x - 1) (2 x + 1)} \\ = lim_{x \to 1 / 2} \frac{2 (3 x + 2)}{2 x - 1} = \frac{2 (3 \times \frac{1}{2} + 2)}{2 \times \frac{1}{2} + 1} \\ = \frac{3}{2} + 2 = \frac{7}{2} \end{aligned}$

14. Find the value of $n$ , if $lim_{x \to 2} \frac{x^{n} - 2^{n}}{x - 2} = 80, n \in N$ .

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Solution

Given,

$lim_{x \to 2} \frac{x^{n} - 2^{n}}{x - 2} = 80$

$\begin{aligned} \Rightarrow & n (2)^{n - 1} = 80 & [∵ lim_{x \to a} \frac{x^{n} - a^{n}}{x - a} = n a^{n - 1}] \\ \Rightarrow & n (2)^{n - 1} = 5 \times 16 \\ \Rightarrow & n \times 2^{n - 1} = 5 \times (2)^{4} \\ \Rightarrow & n \times 2^{n - 1} = 5 \times (2)^{5 - 1} \\ ∴ & n = 5 \end{aligned}$

15. Evaluate $lim_{x \to 0} \frac{\sin 3 x}{\sin 7 x}$ .

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Solution

Given, $lim_{x \to 0} \frac{\frac{\sin 3 x}{3 x} \cdot 3 x}{\frac{\sin 7 x}{7 x} \cdot 7 x} = \frac{lim_{x \to 0} \frac{\sin 3 x}{3 x}}{lim_{x \to 0} \frac{\sin 7 x}{7 x}} \cdot \frac{3 x}{7 x}$

$= \frac{3}{7} \cdot \frac{lim_{x \to 0} \frac{\sin 3 x}{3 x}}{lim_{x \to 0} \frac{\sin 7 x}{7 x}}$

$= \frac{3}{7} [∵ x \to 0 \Rightarrow (k x \to 0), here k is real number]$

16. Eavaluate $lim_{x \to 0} \frac{\sin^{2} 2 x}{\sin^{2} 4 x}$ .

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Solution

Given, $lim_{x \to 0} \frac{\sin^{2} 2 x}{\sin^{2} 4 x} = lim_{x \to 0} \frac{\sin^{2} 2 x}{[\sin 2 (2 x)]^{2}}$

$\begin{aligned} = lim_{x \to 0} \frac{\sin^{2} 2 x}{(2 \sin 2 x \cos 2 x)^{2}} \\ = lim_{x \to 0} \frac{\sin^{2} 2 x}{4 \sin^{2} 2 x \cos^{2} 2 x} & [∵ \sin 2 θ = 2 \sin θ \cos θ] \\ = lim_{x \to 0} \frac{1}{4 \cos^{2} 2 x} = \frac{1}{4} & [∵ \cos 0 = 1] \end{aligned}$

17. Evaluate $lim_{x \to 0} \frac{1 - \cos 2 x}{x^{2}}$ .

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Solution

Given, $lim_{x \to 0} \frac{1 - \cos 2 x}{x^{2}} = lim_{x \to 0} \frac{1 - 1 + 2 \sin^{2} x}{x^{2}} [∵ \cos 2 x = 1 - 2 \sin^{2} x]$

$\begin{matrix} = lim_{x \to 0} \frac{2 \sin^{2} x}{x^{2}} = 2 lim_{x \to 0} \frac{\sin^{2} x}{x^{2}} \\ = 2 lim_{x \to 0} (\frac{\sin x}{x})^{2} & [∵ lim_{x \to 0} (\frac{\sin x}{x})^{2} = 1] \\ = 2 \times 1 = 2 \end{matrix}$

18. Evaluate $lim_{x \to 0} \frac{2 \sin x - \sin 2 x}{x^{3}}$ .

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Solution

Given, $lim_{x \to 0} \frac{2 \sin x - \sin 2 x}{x^{3}} = lim_{x \to 0} \frac{2 \sin x - 2 \sin x \cos x}{x^{3}} [∵ \sin 2 x = 2 \sin x \cos x]$

$\begin{aligned} = lim_{x \to 0} \frac{2 \sin x (1 - \cos x)}{x^{3}} \\ = 2 lim_{x \to 0} \frac{\sin x}{x} \cdot lim_{x \to 0} (\frac{1 - \cos x}{x^{2}}) \end{aligned}$

$\begin{matrix} = 2 \cdot 1 lim_{x \to 0} \frac{1 - \cos x}{x^{2}} & [∵ lim_{x \to 0} \frac{\sin x}{x} = 1] \\ = 2 lim_{x \to 0} \frac{1 - 1 + 2 \sin^{2} \frac{x}{2}}{x^{2}} = 2 lim_{x \to 0} \frac{2 \sin^{2} \frac{x}{2}}{4 \times \frac{x^{2}}{4}} \\ = \frac{2 \cdot 2}{4} lim_{x \to 0} (\frac{\sin \frac{x}{2}}{\frac{x}{2}})^{2} = lim_{x \to 0} (\frac{\sin \frac{x^{2}}{2}}{\frac{x}{2}})^{2} = 1 \end{matrix}$

19. Evaluate $lim_{x \to 0} \frac{1 - \cos m x}{1 - \cos n x}$ .

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Solution

Given, $lim_{x \to 0} \frac{1 - \cos m x}{1 - \cos n x} = lim_{x \to 0} \frac{1 - 1 + 2 \sin^{2} \frac{m x}{2}}{1 - 1 + 2 \sin^{2} \frac{n x}{2}}$

$[∵ \cos m x = 1 - 2 \sin^{2} \frac{m x}{2}$

and $\sin n x = 1 - 2 \sin^{2} \frac{n x}{2}]$

$= lim_{x \to 0} \frac{\sin^{2} \frac{m x}{2}}{\sin^{2} \frac{n x}{2}} = lim_{x \to 0} \frac{\frac{\sin^{2} \frac{m x}{2}}{(\frac{m x}{2})^{2}} \cdot (\frac{m x}{2})^{2}}{\frac{\sin^{2} \frac{n x}{2}}{(\frac{n x}{2})^{2}} \cdot (\frac{n x}{2})^{2}} \= \frac{lim_{x \to 0} (\frac{\sin \frac{m x}{2}}{\frac{m x}{2}})^{2}}{lim_{x \to 0} (\frac{\sin \frac{n x}{2}}{\frac{n x}{2}})^{2}} \cdot \frac{m^{2}}{n^{2}}$

$[∵ lim_{x \to 0} \frac{\sin x}{x} = 1]$

$[∵ x \to 0 \Rightarrow k x \to 0]$

20. Evaluate $lim_{x \to π / 3} \frac{\sqrt{1 - \cos 6 x}}{\sqrt{2} \frac{π}{3} - x}$ .

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Solution

Given, $lim_{x \to π / 3} \frac{\sqrt{1 - \cos 6 x}}{\sqrt{2} (\frac{π}{3} - x)} = lim_{x \to π / 3} \frac{\sqrt{1 - 1 + 2 \sin^{2} 3 x}}{\sqrt{2} (\frac{π}{3} - x)}$

$[∵ \cos 2 x = 1 - 2 \sin^{2} x]$

$= lim_{x \to π / 3} \frac{\sqrt{2} \sin 3 x}{\sqrt{2} (\frac{π}{3} - x)} = lim_{x \to π / 3} \frac{\sin 3 x}{\frac{π}{3} - x}$

$= lim_{x \to π / 3} \frac{\sin (π - 3 x)}{\frac{π - 3 x}{3}} [∵ \sin (π - θ) = \sin θ]$

$= 3 lim_{x \to π / 3} \frac{\sin (π - 3 x)}{(π - 3 x)} = 3 \times 1 [∵ lim_{x \to 0} \frac{\sin x}{x} = 1]$

$= 3 [∵ x \to \frac{π}{3} \Rightarrow x - \frac{π}{3} \to 0]$

21. Evaluate $lim_{x \to \frac{π}{4}} \frac{\sin x - \cos x}{x - \frac{π}{4}}$ .

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Solution

Given, $lim_{x \to π / 4} \frac{\sqrt{2} (\sin x \cdot \frac{1}{\sqrt{2}} - \cos x \cdot \frac{1}{\sqrt{2}})}{(x - \frac{π}{4})} = lim_{x \to π / 4} \frac{\sqrt{2} (\sin x \cos \frac{π}{4} - \cos x \cdot \sin \frac{π}{4})}{(x - \frac{π}{4})}$

$\begin{matrix} = lim_{x \to π / 4} \frac{\sqrt{2} \sin x - \frac{π}{4}}{x - \frac{π}{4}} \\ = \sqrt{2} lim_{x \to π / 4} \frac{\sin x - \frac{π}{4}}{x - \frac{π}{4}} = \sqrt{2} & [∵ lim_{x \to 0} \frac{\sin x}{x} = 1] \end{matrix}$

$[∵ x \to \frac{π}{4} \Rightarrow x - \frac{π}{4} \to 0]$

22. Evaluate $lim_{x \to π / 6} \frac{\sqrt{3} \sin x - \cos x}{x - \frac{π}{6}}$ .

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Solution

Given, $lim_{x \to π / 6} \frac{\sqrt{3} \sin x - \cos x}{x - \frac{π}{6}} = lim_{x \to π / 6} \frac{2 \frac{\sqrt{3}}{2} \sin x - \frac{1}{2} \cos x}{x - \frac{π}{6}}$

$= lim_{x \to π / 6} \frac{2 (\sin x \cos \frac{π}{6} - \cos x \sin \frac{π}{6})}{(x - \frac{π}{6})}$ $= 2 lim_{x \to π / 6} \frac{\sin (x - \frac{π}{6})}{(x - \frac{π}{6})}$

$= 2 [∵ \sin A \cos B - \cos A \sin B = \sin (A - B)]$

$[∵ lim_{x \to 0} \frac{\sin x}{x} = 1]$

$[∵ x \to \frac{π}{6} \Rightarrow (x - \frac{π}{6}) \to 0]$

23. Evaluate $lim_{x \to 0} \frac{\sin 2 x + 3 x}{2 x + \tan 3 x}$ .

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Solution

Given, $lim_{x \to 0} \frac{\sin 2 x + 3 x}{2 x + \tan 3 x} = lim_{x \to 0} \frac{\frac{\sin 2 x + 3 x}{2 x} \cdot 2 x}{\frac{2 x + \tan 3 x}{3 x} \cdot 3 x}$

$= lim_{x \to 0} \frac{(\frac{\sin 2 x}{2 x} + \frac{3 x}{2 x}) 2 x}{(\frac{2 x}{3 x} + \frac{\tan 3 x}{3 x}) 3 x} = lim_{x \to 0} \frac{\frac{\sin 2 x}{2 x} + \frac{3}{2}}{\frac{2}{3} + \frac{\tan 3 x}{3 x}} \cdot \frac{2}{3}$

$\begin{aligned} = \frac{2}{3} lim_{x \to 0} \frac{\frac{\sin 2 x}{2 x} + \frac{3}{2}}{\frac{2}{3} + lim_{x \to 0} \frac{\tan 3 x}{3 x}} \\ = \frac{2}{3} (\frac{1 + \frac{3}{2}}{\frac{2}{3} + 1}) [∵ lim_{x \to 0} \frac{\sin x}{x} = 1 and lim_{x \to 0} \frac{\tan x}{x} = 1] \\ = \frac{2}{3} \times \frac{\frac{2}{5}}{\frac{5}{3}} = \frac{2}{3} \times \frac{5}{2} \times \frac{3}{5} = 1 \end{aligned}$

24. Evaluate $lim_{x \to a} \frac{\sin x - \sin a}{\sqrt{x} - \sqrt{a}}$ .

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Solution

Given, $lim_{x \to a} \frac{\sin x - \sin a}{\sqrt{x} - \sqrt{a}} = lim_{x \to 0} \frac{2 \cos (\frac{x + a}{2}) \sin (\frac{x - a}{2})}{\sqrt{x} - \sqrt{a}}$

$\begin{aligned} = lim_{x \to a} \frac{2 \cos (\frac{x + a}{2}) \sin (\frac{x - a}{2}) (\sqrt{x +} \sqrt{a})}{(\sqrt{x -} \sqrt{a}) (\sqrt{x} + \sqrt{a})} \\ = lim_{x \to 0} \frac{2 \cos (\frac{x + a}{2}) \sin (\frac{x - a}{2}) (\sqrt{x} + \sqrt{a})}{x - a} \\ = 2 lim_{x \to a} \cos (\frac{x + a}{2}) (\sqrt{x} + \sqrt{a}) lim_{x \to 0} \frac{\sin (\frac{x - a}{2})}{\frac{x - a}{2}} \\ = 2 lim_{x \to 0} \cos (\frac{x + a}{2}) (\sqrt{x} + \sqrt{a}) \cdot \frac{1}{2} lim_{x \to 0} \frac{\sin (\frac{x - a}{2})}{\frac{x - a}{2}} \\ = 2 \cdot \cos \frac{a}{2} \cdot \sqrt{a} \cdot \frac{1}{2} \\ = \sqrt{a} \cos \frac{a}{2} \end{aligned}$

25. Evaluate $lim_{x \to π / 6} \frac{\cot^{2} x - 3}{c o s e c x - 2}$ .

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Solution

Given $lim_{x \to π / 6} \frac{\cot^{2} x - 3}{c o s e c x - 2} = lim_{x \to π / 6} \frac{c o s e c^{2} x - 1 - 3}{c o s e c x - 2} [∵ c o s e c^{2} x = 1 + \cot^{2} x]$

$\begin{aligned} = lim_{x \to π / 6} \frac{c o s e c^{2} x - 4}{c o s e c x - 2} = lim_{x \to π / 6} \frac{(c o s e c x)^{2} - (2)^{2}}{c o s e c x - 2} \\ = lim_{x \to π / 6} \frac{(c o s e c x + 2) (c o s e c x - 2)}{(c o s e c x - 2)} = lim_{x \to π / 6} (c o s e c x + 2) \\ = c o s e c \frac{π}{6} + 2 = 2 + 2 = 4 \end{aligned}$

26. Evaluate $lim_{x \to 0} \frac{\sqrt{2} - \sqrt{1 + \cos x}}{\sin^{2} x}$ .

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Solution

Given, $lim_{x \to 0} \frac{\sqrt{2} - \sqrt{1 + \cos x}}{\sin^{2} x} = lim_{x \to 0} \frac{\sqrt{2} - \sqrt{1 + 2 \cos^{2} \frac{x}{2} - 1}}{\sin^{2} x} [∵ \cos x = 2 \cos^{2} \frac{x}{2} - 1]$

$\begin{aligned} = lim_{x \to 0} \frac{\sqrt{2} - \sqrt{2 \cos^{2} \frac{x}{2}}}{\sin^{2} x} \\ = lim_{x \to 0} \frac{\sqrt{2} (1 - \cos \frac{x}{2})}{\sin^{2} x} = lim_{x \to 0} \frac{\sqrt{2} (1 - 1 + 2 \sin^{2} \frac{x}{4})}{\sin^{2} x} \\ = lim_{x \to 0} \frac{\sqrt{2} (2 \sin^{2} \frac{x}{4})}{\sin^{2} x} = lim_{x \to 0} 2 \sqrt{2} \frac{\sin^{2} \frac{x}{4}}{(\frac{x}{4})^{2}} \cdot \frac{(\frac{x}{4})^{2}}{\sin^{2} x} \\ = 2 \sqrt{2} lim_{x \to 0} (\frac{\sin^{\frac{x}{4}}}{\frac{x}{4}})^{2} \cdot lim_{x \to 0} (\frac{x}{\sin x})^{2} \cdot \frac{1}{16} \\ = 2 \sqrt{2} \cdot 1 \cdot 1 \cdot \frac{1}{16} = \frac{1}{4 \sqrt{2}} \end{aligned}$

27. Evaluate $lim_{x \to 0} \frac{\sin x - 2 \sin 3 x + \sin 5 x}{x}$ .

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Solution

Given,

$\begin{aligned} lim_{x \to 0} \frac{\sin x - 2 \sin 3 x + \sin 5 x}{x} & = lim_{x \to 0} \frac{\sin 5 x + \sin x - 2 \sin 3 x}{x} \\ = lim_{x \to 0} \frac{2 \sin 3 x \cos 2 x - 2 \sin 3 x}{x} = lim_{x \to 0} \frac{2 \sin 3 x (\cos 2 x - 1)}{x} \\ = lim_{x \to 0} \frac{2 \sin 3 x}{\frac{1}{3} \times 3 x} (\cos 2 x - 1) = 6 lim_{x \to 0} \frac{\sin 3 x}{3 x} (\cos 2 x - 1) \\ = 6 lim_{x \to 0} \frac{\sin 3 x}{3 x} \cdot lim_{x \to 0} (\cos 2 x - 1) = 6 \times 1 \times 0 = 0 \end{aligned}$

28. If $lim_{x \to 1} \frac{x^{4} - 1}{x - 1} = lim_{x \to k} \frac{x^{3} - k^{3}}{x^{2} - k^{2}}$ , then find the value of $k$ .

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Solution

Given,

$lim_{x \to 1} \frac{x^{4} - 1}{x - 1} = lim_{x \to k} \frac{x^{3} - k^{3}}{x^{2} - k^{2}}$

$\Rightarrow 4 (1)^{4 - 1} = lim_{x \to k} \frac{\frac{x^{3} - k^{3}}{x - k}}{\frac{x^{2} - k^{2}}{x - k}}$

$\begin{matrix} \end{matrix}$

$\begin{matrix} \Rightarrow & 4 = \frac{lim_{x \to k} \frac{x^{3} - k^{3}}{x - k}}{lim_{x \to k} \frac{x^{2} - k^{2}}{x - k}} 1 \\ \Rightarrow & 4 = \frac{3 k^{2}}{2 k} \Rightarrow 4 = \frac{3}{2} k \\ ∴ & k = \frac{4 \times 2}{3} = \frac{8}{3} \end{matrix}$

Differentiate each of the functions w.r.t. $x$ in following questions

29. $\frac{x^{4} + x^{3} + x^{2} + 1}{x}$

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Solution

$\frac{d}{d x} (\frac{x^{4} + x^{3} + x^{2} + 1}{x}) = \frac{d}{d x} (x^{3} + x^{2} + x + \frac{1}{x})$

$\begin{aligned} = \frac{d}{d x} x^{3} + \frac{d}{d x} x^{2} + \frac{d}{d x} x + \frac{d}{d x} (\frac{1}{x}) \\ = 3 x^{2} + 2 x + 1 + (- \frac{1}{x^{2}}) \\ = 3 x^{2} + 2 x + 1 - \frac{1}{x^{2}} \\ = \frac{3 x^{4} + 2 x^{3} + x^{2} - 1}{x^{2}} \end{aligned}$

30. $(x + \frac{1}{x})^{3}$

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Solution

Let

$y = (x + \frac{1}{x})^{3}$

$\begin{aligned} ∴ \frac{d y}{d x} = \frac{d}{d x} (x + \frac{1}{x})^{3} & = 3 (x + \frac{1}{x})^{3 - 1} \frac{d}{d x} (x + \frac{1}{x}) \\ = 3 (x + \frac{1}{x})^{2} (1 - \frac{1}{x^{2}}) \\ = 3 x^{2} - \frac{3}{x^{2}} - \frac{3}{x^{4}} + 3 \end{aligned}$

31. $(3 x + 5) (1 + \tan x)$

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Solution

Let

$\begin{aligned} y & = (3 x + 5) (1 + \tan x) \\ \frac{d y}{d x} & = \frac{d}{d x} [(3 x + 5) (1 + \tan x)] \\ = (3 x + 5) \frac{d}{d x} (1 + \tan x) + (1 + \tan x) \frac{d}{d x} (3 x + 5) [by product rule] \\ = (3 x + 5) (\sec^{2} x) + (1 + \tan x) \cdot 3 \\ = (3 x + 5) \sec^{2} x + 3 (1 + \tan x) \\ = 3 x \sec^{2} x + 5 \sec^{2} x + 3 \tan x + 3 \end{aligned}$

$∴ \frac{d y}{d x} = \frac{d}{d x} [(3 x + 5) (1 + \tan x)]$

32. $(\sec x - 1) (\sec x + 1)$

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Solution

Let

$\begin{matrix} y = (\sec x - 1) (\sec x + 1) \\ y = (\sec^{2} - 1) & [∵ (a + b) (a - b) = a^{2} - b^{2}] \\ = \tan^{2} x \\ ∴ \frac{d y}{d x} = 2 \tan x \cdot \frac{d}{d x} \tan x \\ = 2 \tan x \cdot \sec^{2} x [by chain rule] \end{matrix}$

33. $\frac{3 x + 4}{5 x^{2} - 7 x + 9}$

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Solution

Let $y = \frac{3 x + 4}{5 x^{2} - 7 x + 9}$

$\begin{aligned} ∴ \frac{d y}{d x} & = \frac{(5 x^{2} - 7 x + 9) \frac{d}{d x} (3 x + 4) - (3 x + 4) \frac{d}{d x} (5 x^{2} - 7 x + 9)}{(5 x^{2} - 7 x + 9)^{2}} [by quotient rule] \\ = \frac{(5 x^{2} - 7 x + 9) \cdot 3 - (3 x + 4) (10 x - 7)}{(5 x^{2} - 7 x + 9)^{2}} \\ = \frac{15 x^{2} - 21 x + 27 - 30 x^{2} + 21 x - 40 x + 28}{(5 x^{2} - 7 x + 9)^{2}} \\ = \frac{- 15 x^{2} - 40 x + 55}{(5 x^{2} - 7 x + 9)^{2}} \\ = \frac{55 - 15 x^{2} - 40 x}{(5 x^{2} - 7 x + 9)^{2}} \end{aligned}$

34. $\frac{x^{5} - \cos x}{\sin x}$

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Solution

Let

$\begin{aligned} y & = \frac{x^{5} - \cos x}{\sin x} \\ \frac{d y}{d x} & = \frac{\sin x \frac{d}{d x} (x^{5} - \cos x) - (x^{5} - \cos x) \frac{d}{d x} \sin x}{(\sin x)^{2}} [by quotient rule] \\ = \frac{\sin x (5 x^{4} + \sin x) - (x^{5} - \cos x) \cos x}{\sin^{2} x} \\ = \frac{5 x^{4} \sin x + \sin^{2} x - x^{5} \cos x + \cos^{2} x}{\sin^{2} x} \\ = \frac{5 x^{4} \sin x - x^{5} \cos x + \sin^{2} x + \cos^{2} x}{\sin^{2} x} \\ = \frac{5 x^{4} \sin x - x^{5} \cos x + 1}{\sin^{2} x} \end{aligned}$

35. $\frac{x^{2} \cos \frac{π}{4}}{\sin x}$

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Solution

Let

$\begin{aligned} y & = \frac{x^{2} \cos \frac{π}{4}}{\sin x} = \frac{\frac{x^{2}}{\sqrt{2}}}{\sin x} \\ y & = \frac{1}{\sqrt{2}} \cdot \frac{x^{2}}{\sin x} \\ \frac{d y}{d x} & = \frac{1}{\sqrt{2}} [\frac{\sin x \cdot \frac{d}{d x} x^{2} - x^{2} \frac{d}{d x} \sin x}{\sin^{2} x}] \\ = \frac{1}{\sqrt{2}} [\frac{\sin x \cdot 2 x - x^{2} \cdot \cos x}{\sin^{2} x}] \\ = \frac{1}{\sqrt{2}} \cdot \frac{2 x \sin x - x^{2} \cos x}{\sin^{2} x} \\ = \frac{x}{\sqrt{2}} [2 c o s e c x - x \cot x c o s e c x] \\ = \frac{x}{\sqrt{2}} c o s e c [2 - x \cot x] \end{aligned}$ $[∴ \frac{d y}{d x} = \frac{1}{\sqrt{2}} \frac{\sin x \cdot \frac{d}{d x} x^{2} - x^{2} \frac{d}{d x} \sin x}{\sin^{2} x}]$

36. $(a x^{2} + \cot x) (p + q \cos x)$

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Solution

Let $y = (a x^{2} + \cot x) (p + q \cos x)$

$\begin{aligned} ∴ \frac{d y}{d x} & = (a x^{2} + \cot x) \frac{d}{d x} (p + q \cos x) + (p + q \cos x) \frac{d}{d x} (a x^{2} + \cot x) [by product rule] \\ = (a x^{2} + \cot x) (- q \sin x) + (p + q \cos x) (2 a x - c o s e c^{2} x) \\ = - q \sin x (a x^{2} + \cot x) + (p + q \cos x) (2 a x - c o s e c^{2} x) \end{aligned}$

37. $\frac{a + b \sin x}{c + d \cos x}$

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Solution

Let $y = \frac{a + b \sin x}{c + d \cos x}$

$\begin{aligned} ∴ \frac{d y}{d x} & = \frac{(c + d \cos x) \frac{d}{d x} (a + b \sin x) - (a + b \sin x) \frac{d}{d x} (c + d \cos x)}{(c + d \cos x)^{2}} [by quotinet rule] \\ = \frac{(c + d \cos x) (b \cos x) - (a + b \sin x) (- d \sin x)}{(c + d \cos x)^{2}} \\ = \frac{b \cos x + b d \cos^{2} x + a d \sin x + b d \sin^{2} x}{(c + d \cos x)^{2}} \\ = \frac{b \cos x + a d \sin x + b d (\cos^{2} x + \sin^{2} x)}{(c + d \cos x)^{2}} \\ = \frac{b c \cos x + a d \sin x + b d}{(c + d \cos x)^{2}} \end{aligned}$

38. $(\sin x + \cos x)^{2}$

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Solution

Let

$\begin{aligned} y = (\sin x + \cos x)^{2} \\ ∴ \frac{d y}{d x} = 2 (\sin x + \cos x) (\cos x - \sin x) \\ = 2 (\cos^{2} x - \sin^{2} x) & [by chain rule] \\ = 2 \cos 2 x & [∵ \cos 2 x = \cos^{2} x - \sin^{2} x] \end{aligned}$

39. $(2 x - 7)^{2} (3 x + 5)^{3}$

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Solution

Let

$\begin{matrix} y = (2 x - 7)^{2} (3 x + 5)^{3} \\ \frac{d y}{d x} = (2 x - 7)^{2} \frac{d}{d x} (3 x + 5)^{3} + (3 x + 5)^{3} \frac{d}{d x} (2 x - 7)^{2} & [by product rule] \\ = (2 x - 7)^{2} (3) (3 x + 5)^{2} (3) + (3 x + 5)^{3} 2 (2 x - 7) (2) [by chain rule] \\ = 9 (2 x - 7)^{2} (3 x + 5)^{2} + 4 (3 x + 5)^{3} (2 x - 7) \\ = (2 x - 7) (3 x + 5)^{2} [9 (2 x - 7) + 4 (3 x + 5)] \\ = (2 x - 7) (3 x + 5)^{2} (18 x - 63 + 12 x + 20) \\ = (2 x - 7) (3 x + 5)^{2} (30 x - 43) \end{matrix}$

40. $x^{2} \sin x + \cos 2 x$

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Solution

Let

$\begin{matrix} y = x^{2} \sin x + \cos 2 x \\ \frac{d y}{d x} = \frac{d}{d x} (x^{2} \sin x) + \frac{d}{d x} \cos 2 x \\ = x^{2} \cdot \cos x + \sin x 2 x + (- \sin 2 x) \cdot 2 [by product rule] \\ = x^{2} \cos x + 2 x \sin x - 2 \sin 2 x [by chain urle] \end{matrix}$

41. $\sin^{3} x \cos^{3} x$

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Solution

Let

$\begin{matrix} y = \sin^{3} x \cos^{3} x \\ \frac{d y}{d x} = \sin^{3} x \cdot \frac{d}{d x} \cos^{3} x + \cos^{3} x \frac{d}{d x} \sin^{3} x [by product rule] \\ = \sin^{3} x \cdot 3 \cos^{2} x (- \sin x) + \cos^{3} x \cdot 3 \sin^{2} x \cos x [by chain rule] \\ = - 3 \cos^{2} x \sin^{4} x + 3 \sin^{2} x \cos^{4} x \\ = 3 \sin^{2} x \cos^{2} x (\cos^{2} x - \sin^{2} x) \\ = 3 \sin^{2} x \cos^{2} x \cos 2 x \\ = \frac{3}{4} (2 \sin x \cos x)^{2} \cos 2 x \\ = \frac{3}{4} \sin^{2} 2 x \cos 2 x \end{matrix}$

42. $\frac{1}{a x^{2} + b x + c}$

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Solution

Let $y = \frac{1}{a x^{2} + b x + c} = (a x^{2} + b x + c)^{- 1}$

$\begin{aligned} ∴ \frac{d y}{d x} & = - (a x^{2} + b x + c)^{- 2} (2 a x + b) [by chain rule] \\ = \frac{- (2 a x + b)}{(a x^{2} + b x + c)^{2}} \end{aligned}$

Long Answer Type Questions

Differentiate each of the functions with respect to $x$ in following questions using first principle.

43. $\cos (x^{2} + 1)$

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Solution

Let

$\begin{aligned} f (x) & = \cos (x^{2} + 1) and f (x + h) = \cos {(x + h)^{2} + 1} \\ \frac{d}{d x} f (x) & = lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \\ = lim_{h \to 0} \frac{\cos {(x + h)^{2} + 1} - \cos (x^{2} + 1)}{h} \\ = lim_{h \to 0} \frac{- 2 \sin {\frac{(x + h)^{2} + 1 + x^{2} + 1}{2}} \sin {\frac{(x + h)^{2} + 1 - x^{2} - 1}{2}}}{h} \\ [∵ \cos C - \cos D = - 2 \sin {\frac{C + D}{2}} \cdot \sin {\frac{C - D}{2}}] \\ = lim_{h \to 0} \frac{1}{h} [- 2 \sin {\frac{(x + h)^{2} + x^{2} + 2}{2}} \sin {\frac{(x + h)^{2} - x^{2}}{2}}] \\ = lim_{h \to 0} \frac{1}{h} [- 2 \sin {\frac{(x + h)^{2} + x^{2} + 2}{2}} \sin {\frac{x^{2} + h^{2} + 2 x h - x^{2}}{2}}] \\ = lim_{h \to 0} \frac{1}{h} [- 2 \sin {\frac{(x + h)^{2} + x^{2} + 2}{2}} \sin {\frac{h^{2} + 2 h x}{2}}] \\ = - 2 lim_{h \to 0} \sin {\frac{(x + h)^{2} + x^{2} + 2}{2}} lim_{h \to 0} {\frac{\sin h \frac{h + 2 x}{2}}{h \frac{h + 2 x}{2}}} \times (\frac{h + 2 x}{2}) \\ = - 2 lim_{h \to 0} \sin {\frac{(x + h)^{2} + x^{2} + 2}{2}} lim_{h \to 0} (\frac{h + 2 x}{2}) [∵ lim_{x \to 0} \frac{\sin x}{x} = 1] \\ = - 2 x \sin (x^{2} + 1) \end{aligned}$

$∴ \frac{d}{d x} f (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$

44. $\frac{a x + b}{c x + d}$

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Solution

Let $f (x) = \frac{a x + b}{c x + d}$

$f (x + h) = \frac{a (x + h) + b}{c (x + h) + d}$

$\frac{d}{d x} f (x) = lim_{h \to 0} \frac{1}{h} [f (x + h) - f (x)]$

$= lim_{h \to 0} \frac{1}{h} [\frac{a (x + h) + b}{c (x + h) + d} - \frac{a x + b}{c x + d}]$

$= lim_{h \to 0} \frac{1}{h} [\frac{a x + b + a h}{c (x + h) + d} - \frac{a x + b}{c x + d}]$

$= lim_{h \to 0} \frac{1}{h} [\frac{(a x + a h + b) (c x + d) - (a x + b c (x + h) + d}{c (x + h) + d (c x + d)}]$

$= lim_{h \to 0} \frac{1}{h} [\frac{(a x + a h + b) (c x + d) - (a x + b) (c x + c h + d)}{{c (x + h) + d)} (c x + d)}]$

$a c x^{2} + a c h x + b c x + a d x + a d h + b d$

$= lim_{h \to 0} \frac{1}{h} [\frac{- a c x^{2} + a c h x + a d x + b c x + b c h + b d}{{c (x + h) + d} (c x + d)}]$

$a c x^{2} + a c h x + b c x + a d x + a d h + b d$

$= lim_{h \to 0} \frac{1}{h} [\frac{- a c x^{2} - a c h x - a d x - b c x - b c h - b d}{{c (x + h) + d} (c x + d)}]$

$= lim_{h \to 0} \frac{1}{h} [\frac{a d h - b c h}{{c (x + h) + d} (c x + d)}]$

$= lim_{h \to 0} \frac{a c - b d}{c (x + h) + d (c x + d)}$

$= \frac{a c - b d}{(c x + d)^{2}}$

45. $x^{2 / 3}$

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Solution

Let

Now, $\begin{aligned} f (x) & = x^{2 / 3} \\ f (x + h) & = (x + h)^{2 / 3} \\ \frac{d}{d x} f (x) & = lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \\ = lim_{h \to 0} \frac{1}{h} [(x + h)^{2 / 3} - x^{2 / 3}] \\ = lim_{h \to 0} \frac{1}{h} [x^{2 / 3} (1 + \frac{h}{x})^{2 / 3} - x^{2 / 3}] \end{aligned}$

$\begin{aligned} = lim_{h \to 0} \frac{1}{h} [x^{2 / 3} (1 + \frac{h}{x} \cdot \frac{2}{3} + \frac{2}{3} (\frac{2}{3} - 1) \frac{h^{2}}{x^{2}} + \dots) - 1] \\ [∵ (1 + x)^{h} = 1 + n x + \frac{h (n - 1)}{2!} x^{2} + \dots] \\ = lim_{h \to 0} \frac{1}{h} [x^{2 / 3} (\frac{2}{3} \cdot \frac{h}{x} - \frac{2}{9} \cdot \frac{h^{2}}{x^{2}} + \dots)] \\ = lim_{h \to 0} \frac{x^{2 / 3}}{h} \cdot \frac{2}{3} \frac{h}{x} (1 - \frac{1}{3} \cdot \frac{h}{x} + \dots) \\ = \frac{2}{3} x^{2 / 3 - 1} = \frac{2}{3} x^{- 1 / 3} \end{aligned}$

Alternate Method

Let $\begin{matrix} f (x) = x^{2 / 3} \\ f (x + h) = (x + h)^{2 / 3} \\ \frac{d}{d x} f (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \\ = lim_{(h \to 0)} [\frac{(x + h) 2 / 3 - x^{2 / 3}}{h}] = lim_{(x + h) \to x} [\frac{(x + h) 2 / 3 - x^{2 / 3}}{(x + h) - x}] \\ = \frac{2}{3} (x)^{2 / 3 - 1} & [∵ lim_{x \to a} \frac{x^{x} - a^{n}}{x - a} = n a^{n - 1}] \\ = \frac{2}{3} x^{- 1 / 3} \end{matrix}$

46. $x \cos x$

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Solution

Let

$\begin{aligned} f (x) & = x \cos x \\ ∴ f (x + h) & = (x + h) \cos (x + h) \\ ∴ \frac{d}{d x} f (x) & = lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \\ = lim_{h \to 0} \frac{1}{h} [(x + h) \cos (x + h) - x \cos x] \\ = lim_{h \to 0} \frac{1}{h} [x \cos (x + h) + h \cos (x + h) - x \cos x] \\ = lim_{h \to 0} \frac{1}{h} [x {\cos (x + h) - \cos x} + h \cos (x + h)] \\ = lim_{h \to 0} \frac{1}{h} [x {- 2 \sin \frac{2 x + h}{2} \sin \frac{h}{2}} + h \cos (x + h)] \\ = lim_{h \to 0} [- 2 x (\sin x + \frac{h}{2}) \frac{\sin \frac{h}{2}}{h} + \cos (x + h)] \\ = - 2 lim_{h \to 0} (x \sin x + \frac{h}{2}) lim_{h \to 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}} \cdot \frac{1}{2} + lim_{h \to 0} \cos (x + h) \\ = - 2 \cdot \frac{1}{2} x \sin x + \cos x \\ = \cos x - x \sin x \end{aligned}$

Evaluate each of the following limits in following questions

47. $lim_{y \to 0} \frac{(x + y) \sec (x + y) - x \sec x}{y}$

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Solution

Given, $lim_{y \to 0} \frac{(x + y) \sec (x + y) - x \sec x}{y}$

$\begin{aligned} = lim_{y \to 0} \frac{\frac{x + y}{\cos (x + y)} - \frac{x}{\cos x}}{y} \\ = lim_{y \to 0} \frac{(x + y) \cos x - x \cos (x + y)}{y \cos x \cos (x + y)} \\ = lim_{y \to 0} \frac{x \cos x + y \cos x - x \cos (x + y)}{y \cos x \cos (x + y)} \\ = lim_{y \to 0} \frac{x \cos x - x \cos (x + y) + y \cos x}{y \cos x \cos (x + y)} \\ = lim_{y \to 0} \frac{x {\cos x - \cos (x + y)} + y \cos x}{y \cos x \cos (x + y)} \\ = lim_{y \to 0} \frac{x [- 2 \sin (x + \frac{y}{2}) \sin (\frac{- y}{2})] + y \cos x}{y \cos x \cos (x + y)} \\ [∵ \cos C - \cos D = - 2 \sin \frac{C + D}{2} \cdot \sin \frac{C - D}{2}] \end{aligned}$

$\begin{aligned} = lim_{y \to 0} [\frac{x {2 \sin (x + \frac{y}{2}) \sin \frac{y}{2}} + y \cos x}{y \cos x \cos (x + y)}] \\ = lim_{y \to 0} \frac{2 x \sin (x + \frac{y}{2})}{\cos x \cos (x + y)} \cdot lim_{y \to 0} \frac{\sin \frac{y}{2}}{\frac{y}{2}} \cdot \frac{1}{2} + lim_{y \to 0} \sec (x + y) \\ [∵ lim_{x \to 0} \frac{\sin x}{x} = 1 and x \to 0 \Rightarrow k x \to 0] \end{aligned}$

$\begin{aligned} = lim_{y \to 0} \frac{2 x \sin (x + \frac{y}{2})}{\cos x \cos (x + y)} \cdot \frac{1}{2} + lim_{y \to 0} \sec (x + y) \\ = \frac{2 x \sin x}{\cos x \cos x} \cdot \frac{1}{2} + \sec x \\ = x \tan x \sec x + \sec x \\ = \sec x (x \tan x + 1) \end{aligned}$

48. $lim_{x \to 0} \frac{\sin (α + β) x + \sin (α - β) x + \sin 2 α x}{\cos 2 β x - \cos 2 α x} \cdot x$

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Solution

Given, $lim_{x \to 0} \frac{[\sin (α + β) x + \sin (α - β) x + \sin 2 α x]}{\cos 2 β x - \cos 2 α x} \cdot x$

$= lim_{x \to 0} \frac{[2 \sin α x \cdot \cos β x + \sin 2 α x]}{\cos 2 β x - \sin + \cos 2 α x} = 2 \sin \frac{C + D}{2} \cos \frac{C - D}{2}$

$\begin{aligned} = lim_{x \to 0} \frac{[2 \sin α x \cos β x + \sin 2 α x] x}{2 \sin (α + β) x \sin (α - β) x} [∵ \cos C - \cos D = 2 \sin \frac{C + D}{2} \cdot \sin \frac{D - C}{2}] \\ = lim_{x \to 0} \frac{[2 \sin α x \cos β x + 2 \sin α x \cos α x] x}{2 \sin (α + β) x \sin (α - β) x} \\ = lim_{x \to 0} \frac{2 \sin α x [\cos β x + \cos α x] x}{2 \sin (α + β) x \sin (α - β) x} \\ \begin{array}{c} = lim_{x \to 0} \frac{\sin α x [2 \cos \frac{α + β}{2} x \cos \frac{α - β}{2}] \times x}{2 \sin \frac{α + β}{2} x \cos \frac{α + β}{2} x \cdot 2 \sin \frac{α - β}{2} x \cos \frac{α - β}{2} x} \\ [∵ \cos C + \cos D = 2 \cos \frac{C + D}{2} \cos \frac{C - D}{2} and \sin 2 θ = 2 \sin θ \cos θ] \end{array} \\ = lim_{x \to 0} \frac{\sin α x \cdot x}{2 \sin \frac{α + β}{2} x \sin \frac{α - β}{2} x} \\ = \frac{1}{2} lim_{x \to 0} \frac{\frac{\sin α x}{α x} \cdot x \cdot (α x)}{2 \sin \frac{\frac{α + β}{2} x}{\frac{α + β}{2} x} \cdot \sin \frac{\frac{α - β}{2} x}{\frac{α - β}{2} x} \cdot \frac{α + β}{2} x \cdot \frac{α - β}{2} x} \\ = \frac{\frac{1}{2} lim_{x \to 0} \frac{\sin α x}{α x} \cdot α x^{2}}{lim_{x \to 0} \sin \frac{\frac{α + β}{2} x}{\frac{α + β}{2} lim_{x \to 0} \frac{\frac{α - β}{2} x}{\frac{α - β}{2} x} \cdot \frac{α^{2} - β^{2}}{4} x^{2}}} \end{aligned}$

$∵ lim_{x \to 0} \frac{\sin x}{x} = 1$ and $x \to 0 \Rightarrow k x \to 0$

$\begin{aligned} = \frac{1}{2} \cdot \frac{α \cdot 4}{α^{2} - β^{2}} [\frac{lim_{x \to 0} \frac{\sin α x}{α x}}{lim_{x \to 0} \sin \frac{\frac{α + β}{2} x}{\frac{α + β}{2} x} lim_{x \to 0} \sin \frac{\frac{α - β}{2} x}{\frac{α - β}{2} x}}] \\ = \frac{1}{2} \cdot \frac{4 α}{α^{2} - β^{2}} = \frac{2 α}{α^{2} - β^{2}} \end{aligned}$

49. $lim_{x \to π / 4} \frac{\tan^{3} x - \tan x}{\cos (x + \frac{π}{4})}$

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Solution

Given, $lim_{x \to π / 4} \frac{\tan^{3} x - \tan x}{\cos (x + \frac{π}{4})}$

$\begin{matrix} = lim_{x \to π / 4} \frac{\tan x (\tan^{2} x - 1)}{\cos (x + \frac{π}{4})} = lim_{x \to π / 4} \tan x \cdot lim_{x \to π / 4} (\frac{1 - \tan^{2} x}{\cos (x + \frac{g π}{4})}) \\ = - 1 \times lim_{x \to π / 4} \frac{(1 + \tan x) (1 - \tan x)}{\cos (x + \frac{π}{4})} & [∵ a^{2} - b^{2} = (a + b) (a - b)] \end{matrix}$

$= - lim_{x \to π / 4} (1 + \tan x) lim_{x \to π / 4} [\frac{\cos x - \sin x}{\cos x \cdot \cos (x + \frac{π}{4})}]$

$\begin{aligned} = - (1 + 1) \times lim_{x \to π / 4} \frac{\sqrt{2} [\frac{1}{\sqrt{2}} \cdot \cos x - \frac{1}{\sqrt{2}} \cdot \sin x]}{\cos x \cdot \cos x + \frac{π}{4}} = - 2 \sqrt{2} lim_{x \to π / 4} [\frac{\cos \frac{π}{4} \cdot \cos x - \sin \frac{π}{4} \cdot \sin x}{\cos x \cdot \cos x + \frac{π}{4}}] \\ [∵ \cos A \cdot \cos B - \sin A \sin B = \cos (A + B)] \\ = - 2 \sqrt{2} lim_{x \to π / 4} \frac{\cos (x + \frac{π}{4})}{\cos x \cdot \cos (x + \frac{π}{4})} = - 2 \sqrt{2} \times \frac{1}{\frac{1}{\sqrt{2}}} = - 2 \sqrt{2} \times \sqrt{2} = - 4 \end{aligned}$

50. $lim_{x \to π} \frac{1 - \sin \frac{x}{2}}{\cos \frac{x}{2} (\cos \frac{x}{4} - \sin \frac{x}{4})}$

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Solution

Given, $lim_{x \to π} \frac{1 - \sin \frac{x}{2}}{\cos \frac{x}{2} (\cos \frac{x}{4} - \sin \frac{x}{4})}$

$\begin{aligned} = lim_{x \to π} \frac{\cos^{2} \frac{x}{4} + \sin^{2} \frac{x}{4} - 2 \cdot \sin \frac{x}{4} \cdot \cos \frac{x}{4}}{\cos \frac{x}{2} \cdot (\cos \frac{x}{4} - \sin \frac{x}{4})} [∵ \sin^{2} θ + \cos^{2} θ = 1 \sin 2 θ = 2 \sin θ \cos θ] \\ = lim_{x \to π} \frac{(\cos \frac{x}{4} - \sin \frac{x}{4})^{2}}{(\cos^{2} \frac{x}{4} - \sin^{2} \frac{x}{4}) (\cos \frac{x}{4} - \sin \frac{x}{4})} [∵ \cos^{2} 2 θ = \cos^{2} θ - \sin^{2} θ] \end{aligned}$

$\begin{aligned} = lim_{x \to π} \frac{\cos \frac{x}{4} - \sin \frac{x}{4}}{(\cos \frac{x}{4} + \sin \frac{x}{4}) (\cos \frac{x}{4} - \sin \frac{x}{4})} \\ lim_{x \to π} \frac{1}{\cos \frac{x}{4} + \sin \frac{x}{4}} = \frac{1}{\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \end{aligned}$

51. Show that $lim_{x \to π / 4} \frac{| x - 4 |}{x - 4}$ does not exist,

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Solution

Given,

$lim_{x \to π / 4} \frac{| x - 4 |}{x - 4}$

$L H L = lim_{x \to \frac{π^{-}}{4}} \frac{- (x - 4)}{x - 4} [∵ | x - 4 | = - (x - 4), x < 4]$

$= - 1$

$R H L = lim_{x \to \frac{π^{+}}{4}} \frac{(x - 4)}{x - 4} = 1 [∵ | x - 4 | = (x - 4), x > 4]$

$∴$ LHL $\neq$ RHL

So, limit does not exist.

52. If $f (x) = {\begin{cases} \frac{k \cos x}{π - 2 x}, & when x \neq \frac{π}{2} \\ 3, & when x = \frac{π}{2} \end{cases}$ and $lim_{x \to π / 2} f (x) = f \frac{π}{2}$ , then find the

value of $k$ .

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Solution

Given,

$f (x) = {\begin{cases} \frac{k \cos x}{π - 2 x}, x \neq \frac{π}{2} \\ 3, x = \frac{π}{2} \end{cases}$

$L H L = lim_{x \to \frac{π^{-}}{2}} \frac{k \cos x}{π - 2 x} = lim_{h \to 0} \frac{k \cos \frac{π}{2} - h}{π - 2 \frac{π}{2} - h}$

$= lim_{h \to 0} \frac{k \sinh}{π - π + 2 h} = lim_{h \to 0} \frac{k \sinh}{2 h}$

$= \frac{k}{2} lim_{h \to 0} \frac{\sin h}{h} = \frac{k}{2} \cdot 1 = \frac{k}{2} ∵ lim_{h \to 0} \frac{\sin x}{x} = 1$

$R H L = lim_{x \to \frac{π^{+}}{2}} \frac{k \cos x}{π - 2 x} = lim_{x \to \frac{π}{2}} + \frac{k \cos \frac{π}{2} + h}{π - 2 \frac{π}{2} + h}$

$= lim_{h \to 0} \frac{- k \sin h}{π - π - 2 h} = lim_{h \to 0} \frac{k \sinh}{2 h} = \frac{k}{2} lim_{h \to 0} \frac{\sin h}{2 h} = \frac{k}{2} and f \frac{π}{2} = 3$

$It is given that, lim_{x \to π / 2} f (x) = f \frac{π}{2} \Rightarrow \frac{k}{2} = 3$

$∴ k = 6$

53. If $f (x) = {\begin{cases} x + 2, & x \leq - 1 \\ c x^{2}, & x > - 1^{'} \end{cases}$ then find $c$ when $lim_{x \to - 1} f (x)$ exists.

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Solution

Given,

$\begin{aligned} f (x) = {\begin{cases} x + 2, & x \leq - 1 \\ c x^{2}, & x > - 1 \end{cases} \\ L H L = lim_{x \to - 1^{-}} f (x) = lim_{x \to -^{-}} (x + 2) \\ = lim_{h \to 0} (- 1 - h + 2) = lim_{h \to 0} (1 - h) = 1 \\ R H L = lim_{x \to - 1^{+}} f (x) = lim_{x \to - 1^{+}} c x^{2} = lim_{h \to 0} c (- 1 + h)^{2} \end{aligned}$

$\begin{aligned} ∴ \\ If lim_{x \to - 1} f (x) exist, then L H L = R H L = c \\ ∴ c = 1 \end{aligned}$

Objective Type Questions

54. $lim_{x \to π} \frac{\sin x}{x - π}$ is equal to

(a) 1

(b) 2

(d) -2

Show Answer

Solution

$[∵ \sin θ = \sin (π - θ)]$

$= - lim_{x \to π} \frac{\sin (π - x)}{(π - x)} = - 1 [∵ lim_{x \to 0} \frac{\sin x}{x} = 1 and π - x \to 0 \Rightarrow x \to π]$

55. $lim_{x \to 0} \frac{x^{2} \cos x}{1 - \cos x}$ is equal to

(a) 2

(b) $\frac{3}{2}$

(d) 1

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Solution

(a) Given, $lim_{x \to 0} \frac{x^{2} \cos x}{1 - \cos x} = lim_{x \to 0} \frac{x^{2} \cos x}{2 \sin^{2} \frac{x}{2}}$

$[∵ 1 - \cos x = 2 \sin^{2} \frac{x}{2}]$

$= 2 lim_{x \to 0} \frac{\frac{x}{2}}{\sin^{2} \frac{x}{2}} \cdot lim_{x \to 0} \cos x = 2 \cdot 1 = 2$

56. $lim_{x \to 0} \frac{(1 + x)^{n} - 1}{x}$ is equal to

(a) $n$

(b) 1

(d) 0

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Solution

(a) Given, $lim_{x \to 0} \frac{(1 + x)^{n} - 1}{x} = lim_{x \to 0} \frac{(1 + x)^{n} - 1}{(1 + x) - 1} = lim_{x \to 0} \frac{(1 + x)^{n} - 1}{(1 + x) - 1}$

$\begin{aligned} = lim_{x \to 0} \frac{(1 + x)^{n} - 1^{n}}{(1 + x) - 1} = lim_{(1 + x) \to 1} \frac{(1 + x)^{n} - 1^{n}}{(1 + x) - 1} \\ = n \cdot (1)^{n - 1} = n ∵ lim_{x \to a} \frac{x^{n} - a^{n}}{x - a} = n a^{n - 1} \end{aligned}$

57. $lim_{x \to 1} \frac{x^{m} - 1}{x^{n} - 1}$ is equal to

(a) 1

(b) $\frac{m}{n}$

(d) $\frac{m^{2}}{n^{2}}$

Show Answer

Solution

(b) Given, $lim_{x \to 1} \frac{x^{m} - 1}{x^{n} - 1} = lim_{x \to 1} \frac{\frac{x^{m} - 1}{x - 1}}{\frac{x^{n} - 1}{x - 1}} = \frac{lim_{x \to 1} \frac{x^{m} - 1^{m}}{x - 1}}{lim_{x \to 1} \frac{x^{n} - 1^{n}}{x - 1}}$

$= \frac{m (1)^{m - 1}}{n (1)^{n - 1}} = \frac{m}{n} [∵ lim_{x \to a} \frac{x^{n} - a^{n}}{x - a} = n a^{n - 1}]$

58. $lim_{θ \to 0} \frac{1 - \cos 4 θ}{1 - \cos 6 θ}$ is equal to

(a) $\frac{4}{9}$

(b) $\frac{1}{2}$

(d) -1

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Solution

(a) Given, $lim_{θ \to 0} \frac{1 - \cos 4 θ}{1 - \cos 6 θ} = lim_{θ \to 0} \frac{2 \sin^{2} 2 θ}{2 \sin^{2} 3 θ}$

$[∵ 1 - \cos 2 θ = 2 \sin^{2} θ]$

$\begin{aligned} = \frac{lim_{θ \to 0} \frac{\sin^{2} 2 θ}{(2 θ)^{2}} \cdot (2 θ)^{2}}{lim_{θ \to 0} \frac{\sin^{2} 3 θ}{(3 θ)^{2}} \cdot (3 θ)^{2}} = \frac{4}{9} \cdot \frac{lim_{θ \to 0} (\frac{\sin 2 θ}{2 θ})^{2}}{lim_{θ \to 0} (\frac{\sin 3 θ}{3 θ})^{2}} [∵ lim_{x \to 0} \frac{\sin x}{x} = 1 and x \to 0 \Rightarrow k x \to 0] \\ = \frac{4}{9} \end{aligned}$

59. $lim_{x \to 0} \frac{c o s e c x - \cot x}{x}$ is equal to

(a) $\frac{- 1}{2}$

(b) 1

(d) 1

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Solution

$\begin{aligned} lim_{x \to 0} \frac{c o s e c x - \cot x}{x} \\ = lim_{x \to 0} \frac{\frac{1}{\sin x} - \frac{\cos x}{\sin x}}{x} = lim_{x \to 0} \frac{1 - \cos x}{x \cdot \sin x} \\ = lim_{x \to 0} \frac{2 \sin^{2} \frac{x}{2}}{x \cdot 2 \sin \frac{x}{2} \cos \frac{x}{2}} = lim_{x \to 0} \frac{\tan \frac{x}{2}}{x} \\ = lim_{x \to 0} \frac{\tan \frac{x}{2}}{\frac{x}{2}} \cdot \frac{1}{2} = \frac{1}{2} \end{aligned}$

$[∵ lim_{θ \to 0} \frac{\tan θ}{θ} = 1]$

60. $lim_{x \to 0} \frac{\sin x}{\sqrt{x + 1} - \sqrt{1 - x}}$ is equal to

(a) 2

(b) 0

(d) -1

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Solution

$\begin{aligned} = lim_{x \to 0} \frac{\sin x}{\sqrt{x + 1} - \sqrt{1 - x}} \cdot \frac{\sqrt{x + 1} + \sqrt{1 - x}}{\sqrt{x + 1} + \sqrt{1 - x}} \\ = lim_{x \to 0} \frac{\sin x (\sqrt{x + 1} + \sqrt{1 - x})}{(x + 1) - (1 - x)} \\ = lim_{x \to 0} \frac{\sin x (\sqrt{x + 1} + \sqrt{1 - x})}{x + 1 - 1 + x} \\ = \frac{1}{2} lim_{x \to 0} \frac{\sin x}{x} lim_{x \to 0} (\sqrt{x + 1} + \sqrt{1 - x}) \\ = \frac{1}{2} \cdot 1 \cdot 2 = 1 \end{aligned}$

61. $lim_{x \to π / 4} \frac{\sec^{2} x - 2}{\tan x - 1}$ is

(a) 3

(b) 1

(d) 2

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Solution

(d) Given,

$\begin{aligned} lim_{x \to π / 4} \frac{\sec^{2} x - 2}{\tan x - 1} \\ = & lim_{x \to π / 4} \frac{1 + \tan^{2} x - 2}{\tan x - 1} = lim_{x \to π / 4} \frac{\tan^{2} x - 1}{\tan x - 1} \\ = & lim_{x \to π / 4} \frac{(\tan x + 1) (\tan x - 1)}{(\tan x - 1)} = lim_{x \to π / 4} (\tan x + 1) \\ = & 2 \end{aligned}$

62. $lim_{x \to 1} \frac{(\sqrt{x} - 1) (2 x - 3)}{2 x^{2} + x - 3}$ is equal to

(a) $\frac{1}{10}$

(b) $\frac{- 1}{10}$

(d) None of these

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Solution

(b) Given, $lim_{x \to 1} \frac{(\sqrt{x} - 1) (2 x - 3)}{2 x^{2} + x - 3} = lim_{x \to 1} \frac{(\sqrt{x} - 1) (2 x - 3)}{(2 x + 3) (x - 1)}$

$\begin{aligned} = lim_{x \to 1} \frac{(\sqrt{x} - 1) (2 x - 3)}{(2 x + 3) (\sqrt{x} - 1) (\sqrt{x} + 1)} \\ = lim_{x \to 1} \frac{2 x - 3}{(2 x + 3) (\sqrt{x} + 1)} = \frac{- 1}{5 \times 2} = \frac{- 1}{10} \end{aligned}$

63. If f(x)= ${\begin{cases} \frac{\sin [x]}{[x]}, & [x] \neq 0, where [\cdot] denotes the greatest integer \\ 0, & [x] = 0 \end{cases}$

function, then $lim_{x \to 0} f (x)$ is equal to

(a) 1

(b) 0

(d) Does not exist

Show Answer

Solution

(d) Given,

f(x)= ${\begin{cases} \frac{\sin [x]}{[x]}, & [x] \neq 0 \\ 0, & [x] = 0 \end{cases}$

$∴$ $L H L = lim_{x \to 0^{-}} f (x)$

$\begin{aligned} = lim_{x \to 0^{-}} \frac{\sin [x]}{[x]} = lim_{h \to 0} \frac{\sin [0 - h]}{[0 - h]} \\ = lim_{h \to 0} \frac{- \sin [- h]}{[- h]} = - 1 \\ R H L & = lim_{x \to 0^{+}} f (x) = lim_{x \to 0^{+}} \frac{\sin [x]}{[x]} \\ = lim_{x \to 0^{+}} \frac{\sin [0 + h]}{[0 + h]} = lim_{h \to 0} \frac{\sin [h]}{[h]} = 1 \end{aligned}$

$∵$ $L H L \neq R H L$

So, limit does not exist.

64. $lim_{x \to 0} \frac{| \sin x |}{x}$ is equal to

(a) 1

(b) $= - 1$

(d) None of these

Show Answer

Solution

$\begin{aligned} limit & = lim_{x \to 0} \frac{| \sin x |}{x} \\ ∴ L H L = lim_{x \to 0^{-}} \frac{- \sin x}{x} = - lim_{x \to 0^{-}} \frac{\sin x}{x} = - 1 \\ R H L & = lim_{x \to 0^{+}} \frac{\sin x}{x} = 1 \end{aligned}$

$∵ L H L \neq R H L$

So, limit does not exist.

65. If $f (x) = {\begin{cases} x^{2} - 1, & 0 < x < 2 \\ 2 x + 3, & 2 \leq x < 3^{3} \end{cases}$ then the quadratic equation whose roots are $lim_{x \to 2^{-}} f (x)$ and $lim \underset{x \to 2^{+}}{f (x)}$ is

(a) $x^{2} - 6 x + 9 = 0$

(b) $x^{2} - 7 x + 8 = 0$

(d) $x^{2} - 10 x + 21 = 0$

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Solution

(d) Given,

$f (x) = {\begin{cases} x^{2} - 1, & 0 < x < 2 \\ 2 x + 3, & 2 \leq x < 3 \end{cases}$

$\begin{aligned} ∴ lim_{x \to 2^{-}} f (x) & = lim_{x \to 2^{-}} (x^{2} - 1) \\ = lim_{h \to 0} [(2 - h)^{2} - 1] = lim_{h \to 0} (4 + h^{2} - 4 h - 1) \\ = lim_{h \to 0} (h^{2} - 4 h + 3) = 3 \end{aligned}$

and $lim_{x \to 2^{+}} f (x) = lim_{x \to 2^{+}} (2 x + 3)$

$= lim_{h \to 0} [2 (2 + h) + 3] = lim_{h \to 0} (4 + 2 h + 3) = 7$

So, the quadratic equation whose roots are 3 and 7 is $x^{2} - (3 + 7) x + 3 \times 7 = 0$ i.e., $x^{2} - 10 x + 21 = 0$ .

66. $lim_{x \to 0} \frac{\tan 2 x - x}{3 x - \sin x}$ is equal to

(a) 2

(b) $\frac{1}{2}$

(d) $\frac{1}{4}$

Show Answer

Solution

(b) Given,

$\begin{aligned} lim_{x \to 0} \frac{\tan 2 x - x}{3 x - \sin x} = lim_{x \to 0} \frac{x [\frac{\tan 2 x}{x} - 1]}{x [3 - \frac{\sin x}{x}]} \\ = \frac{lim_{x \to 0} 2 \times \frac{\tan 2 x}{2 x} - 1}{3 - lim_{x \to 0} \frac{\sin x}{x}} = \frac{2 - 1}{3 - 1} = \frac{1}{2} \end{aligned}$

67. If $f (x) = x - [x], \in R$ , then $f^{'} \frac{1}{2}$ is equal to

(a) $\frac{3}{2}$

(b) 1

(d) -1

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Solution

(b) Given, $f (x) = x - [x]$

Now, first we have to check the differentiability of $f (x)$ at $x = \frac{1}{2}$ .

$∴ L f^{'} \frac{1}{2} = L H D = lim_{h \to 0} \frac{f (\frac{1}{2} - h) - f (\frac{1}{2})}{- h}$

$= lim_{n \to 0} \frac{(\frac{1}{2} - h) - (\frac{1}{2} - h) - \frac{1}{2} + \frac{1}{2}}{h} = lim_{h \to 0} \frac{\frac{1}{21} - h - 0 - \frac{1}{2} + 0 = 1}{h}$

and $R f^{'} \frac{1}{2} = R H D lim_{h \to 0} \frac{f (\frac{1}{2} + h) - f (\frac{1}{2})}{h}$

$= lim_{h \to 0} \frac{\frac{1}{2} + h - \frac{1}{2} + h - \frac{1}{2} + \frac{1}{2}}{h} = lim_{h \to 0} \frac{\frac{1}{2} + h - 0 - \frac{1}{2} + 0}{h} = 1$

$∵ L H L = R H D$

$∴ f^{'} \frac{1}{2} = 1$

68. If $y = \sqrt{x} + \frac{1}{\sqrt{x}}$ , then $\frac{d y}{d x}$ at $x = 1$ is equal to

(a) 1

(b) $\frac{1}{2}$

(d) 0

Show Answer

Solution

(d) Given,

$\begin{aligned} Now, \\ ∴ \frac{d y}{d x} = \frac{1}{2} - \frac{1}{2} = 0 \end{aligned}$

$y = \sqrt{x} + \frac{1}{\sqrt{x}}$

69. If $f (x) = \frac{x - 4}{2 \sqrt{x}}$ , then $f^{'} (1)$ is equal to

(a) $\frac{5}{4}$

(b) $\frac{4}{5}$

(d) 0

Show Answer

Solution

(a) Given,

$f (x)^{5} = \frac{x - 4}{2 \sqrt{x}}$

$Now, \begin{aligned} f^{'} (x) & = \frac{2 \sqrt{x} - (x - 4) \cdot 2 \cdot \frac{1}{2 \sqrt{x}}}{4 x} \\ = \frac{2 x - (x - 4)}{4 x^{3 / 2}} = \frac{2 x - x + 4}{4 x^{3 / 2}} \\ = \frac{x + 4}{4 x^{3 / 2}} \\ ∴ f^{'} (1) & = \frac{1 + 4}{4 \times (1)^{3 / 2}} = \frac{5}{4} \end{aligned}$

70. If $y = \frac{1 + \frac{1}{x^{2}}}{1 - \frac{1}{x^{2}}}$ , then $\frac{d y}{d x}$ is equal to

(a) $\frac{- 4 x}{(x^{2} - 1)^{2}}$

(b) $\frac{- 4 x}{x^{2} - 1}$

(d) $\frac{4 x}{x^{2} - 1}$

Show Answer

Solution

(a) Given,

$\begin{aligned} y & = \frac{1 + \frac{1}{x^{2}}}{1 - \frac{1}{x^{2}}} \Rightarrow y = \frac{x^{2} + 1}{x^{2} - 1} \\ \frac{d y}{d x} & = \frac{(x^{2} - 1) 2 x - (x^{2} + 1) (2 x}{(x^{2} - 1)^{2}} \\ \frac{d y}{d x} & = \frac{2 x (x^{2} - 1 - x^{2} - 1)}{(x^{2} - 1)^{2}} \\ = \frac{2 x (- 2)}{(x^{2} - 1)^{2}} = \frac{- 4 x}{(x^{2} - 1)^{2}} \end{aligned}$

$∴ \frac{d y}{d x} = \frac{(x^{2} - 1) 2 x - (x^{2} + 1) (2 x)}{(x^{2} - 1)^{2}} [by quotient rule]$

71. If $y = \frac{\sin x + \cos x}{\sin x - \cos x}$ , then $\frac{d y}{d x}$ at $x = 0$ is equal to

(a) -2

(b) 0

(d) Does not exist

Show Answer

Solution

(a) Given, $y = \frac{\sin x + \cos x}{\sin x - \cos x}$

$\begin{aligned} ∴ \frac{d y}{d x} & = \frac{(\sin x - \cos x) (\cos x - \sin x) - (\sin x + \cos x) (\cos x + \sin x)}{(\sin x - \cos x)^{2}} [by quotient rule] \\ = \frac{- (\sin x - \cos x)^{2} - (\sin x + \cos x)^{2}}{(\sin x - \cos x)^{2}} \\ = \frac{- [(\sin x - \cos x)^{2} + (\sin x + \cos x)^{2}]}{(\sin x - \cos x)^{2}} \\ = \frac{- [\sin^{2} x + \cos^{2} x - 2 \sin x \cos x + \sin^{2} x + \cos^{2} x + 2 \sin x \cos x]}{(\sin x - \cos x)^{2}} \\ = \frac{- 2}{(\sin x - \cos x)^{2}} \\ ∴ \frac{d y}{d x} & = - 2 \end{aligned}$

72. If $y = \frac{\sin (x + 9)}{\cos x}$ , then $\frac{d y}{d x}$ at $x = 0$ is equal to

(a) $\cos 9$

(b) $\sin 9$

(d) 1

Show Answer

Solution

(a) Given, $y = \frac{\sin (x + 9)}{\cos x}$

$\begin{aligned} ∴ \frac{d y}{d x} & = \frac{\cos x \cos (x + 9) - \sin (x + 9) (- \sin x)}{(\cos x)^{2}} [by quotient rule] \\ = \frac{\cos x \cos (x + 9) + \sin x \sin (x + 9)}{\cos^{2} x} \\ ∴ {\frac{d y}{d x}}_{x = 0} & = \frac{\cos 9}{1} \\ = \cos 9 \end{aligned}$

73. If $f (x) = 1 + x + \frac{x^{2}}{2} + \dots + \frac{x^{100}}{100}$ , then $f^{'} (1)$ is equal to

(a) $\frac{1}{100}$

(b) 100

(d) Does not exist

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Solution

(b) Given,

$f (x) = 1 + x + \frac{x^{2}}{2} + \dots + \frac{x^{100}}{100}$

$\begin{aligned} ∴ f^{'} (x) = 0 + 1 + 2 \times \frac{x}{2} + \dots + 100 \frac{x^{99}}{100} \\ f^{'} (x) = 1 + x + x^{2} + \dots + x^{99} \\ f^{'} (1) = 1 + 1 + 1 + \dots + 1 (100 times) \\ = 100 \end{aligned}$

74. If $f (x) = \frac{x^{n} - a^{n}}{x - a}$ for some constant $a$ , then $f^{'} (a)$ is equal to

(a) 1

(b) 0

(d) Does not exist

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Solution(d) Given,

$f (x) = \frac{x^{n} - a^{n}}{x - a}$

$\begin{matrix} ∴ & f^{'} (x) = \frac{(x - a) n x^{n - 1} - (x^{n} - a^{n}) (1)}{(x - a)^{2}} [by quotient rule] \\ \Rightarrow & f^{'} (x) = \frac{n x^{n - 1} (x - a) - x^{n} + a^{n}}{(x - a)^{2}} \\ Now, & f^{'} (a) = \frac{n a^{n - 1} (0) - a^{n} + a^{n}}{(x - a)^{2}} \\ \Rightarrow & f^{'} (a) = \frac{0}{0} \end{matrix}$

So, $f^{'} (a)$ does not exist,

Since, $f (x)$ is not defined at $x = a$ .

Hence, $f^{'} (x)$ at $x = a$ does not exist.

75. If $f (x) = x^{100} + x^{99} + \dots + x + 1$ , then $f^{'} (1)$ is equal to

(a) 5050

(b) 5049

(d) 50051

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Solution

(a) Given,

$f (x) = x^{100} + x^{99} + \dots + x + 1$

$\begin{matrix} ∴ f^{'} (x) = 100 x^{99} + 99 x^{98} + \dots + 1 + 0 \\ = 100 x^{99} + 99 x^{98} + \dots + 1 \\ Now, \\ f^{'} (1) = 100 + 99 + \dots + 1 \\ = \frac{100}{2} [2 \times 100 + (100 - 1) (- 1)] \\ = 50 [200 - 99] \\ = 50 \times 101 \\ = 5050 \end{matrix}$

76. If $f (x) = 1 - x + x^{2} - x^{3} + \dots - x^{99} + x^{100}$ , then $f^{'} (1)$ is equal to

(a) 150

(b) -50

(d) 50

Show Answer

Solution

(d) Given, $f (x) = 1 - x + x^{2} - x^{3} + \dots - x^{99} + x^{100}$

$\begin{array}{r} f^{'} (x) = 0 - 1 + 2 x - 3 x^{2} + \dots - 99 x^{98} + 100 x^{99} \\ = - 1 + 2 x - 3 x^{2} + \dots - 99 x^{98} + 100 x^{99} \\ ∴ f^{'} (1) = - 1 + 2 - 3 + \dots - 99 + 100 \\ = (- 1 - 3 - 5 - \dots - 99) + (2 + 4 + \dots + 100) ∵ S_{n} = \frac{n}{2} 2 a + (n - 1) d \\ = - \frac{50}{2} [2 \times 1 + (50 - 1) 2] + \frac{50}{2} [2 \times 2 + (50 - 1) 2] \\ = - 25 [2 + 49 \times 2] + 25 [4 + 49 \times 2] \\ = - 25 (2 + 98) + 25 (4 + 98) \\ = - 25 \times 100 + 25 \times 102 \\ = - 2500 + 2550 \\ = 50 \end{array}$

Fillers

77. If $f (x) = \frac{\tan x}{x - π}$ , then $lim_{x \to π} f (x) =$ ……

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Solution

Given, $f (x) = \frac{\tan x}{x - π} = lim_{x \to π} \frac{\tan x}{x - π} = lim_{π \to x \to 0} \frac{- \tan (π - x)}{- (π - x)} [∵ lim_{x \to 0} \frac{\tan x}{x} = 1]$

$= 1$

78. $lim_{x \to 0} \sin m x \cot \frac{x}{\sqrt{3}} = 2$ , then $m =$ ……

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Solution

Given, $lim_{x \to 0} \sin m x \cot \frac{x}{\sqrt{3}} = 2$

$\begin{aligned} \Rightarrow lim_{x \to 0} \frac{\sin m x}{m x} \cdot m x \cdot \frac{1}{\tan \frac{x}{\sqrt{3}}} = 2 \\ \Rightarrow lim_{x \to 0} \frac{\sin m x}{m x} \cdot m x \cdot \frac{\frac{x}{\sqrt{3}}}{\tan \frac{x}{\sqrt{3}}} \cdot \frac{1}{\frac{x}{\sqrt{3}}} = 2 \\ \Rightarrow lim_{x \to 0} \frac{\sin m x}{m x} \cdot lim_{x \to 0} \frac{\frac{x}{\sqrt{3}}}{\tan \frac{x}{\sqrt{3}}} \cdot lim_{x \to 0} \frac{m x}{\frac{x}{\sqrt{3}}} = 2 \\ \begin{array}{c} \Rightarrow & \sqrt{3} x = 2 \end{array} \\ ∴ m = \frac{2 \sqrt{3}}{3} \end{aligned}$

79. If $y = 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \dots$ , then $\frac{d y}{d x} =$ ……

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Solution

Given,

$\begin{aligned} y & = 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \frac{x^{4}}{4!} + \dots \\ \frac{d y}{d x} & = 0 + 1 + \frac{2 x}{2} + \frac{3 x^{2}}{6} + \frac{4 x^{3}}{4!} \\ = 1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{6} + \dots \\ = 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \dots \\ = y \end{aligned}$

80. $lim_{x \to 3^{+}} \frac{x}{[x]} =$ ……

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Solution

Given,

$\begin{aligned} lim_{x \to 3^{+}} \frac{x}{[x]} & = lim_{h \to 0} \frac{(3 + h)}{[3 + h]} \\ = lim_{h \to 0} \frac{(3 + h)}{3} = 1 \end{aligned}$

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Maths (Class-11) | NCERT Exemplar

Limits and Derivatives

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Long Answer Type Questions

Objective Type Questions

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Engineering Preparation

Medical Preparation

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Test Platform

Sample Mock Test

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