SATHEE: Limit Continuity and Differentiability 1 Question 2

Limit Continuity and Differentiability 1 Question 2

3. 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 $k$ is

(a) $\frac{4}{3}$

(b) $\frac{3}{8}$

(2019 Main, 10 April I)

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Answer:

Correct Answer: 3. (d)

Solution:

Key Idea $lim_{x \to a} f (x)$ exist iff

$lim_{x \to a^{+}} f (x) = lim_{x \to a^{-}} f (x)$

At $x = 0$ ,

$\begin{aligned} RHL & = lim_{x \to 0^{+}} \frac{\tan (π \sin^{2} x) + (| x | - \sin (x [x]))^{2}}{x^{2}} & = lim_{x \to 0^{+}} \frac{\tan (π \sin^{2} x) + (x - \sin (x \cdot 0))^{2}}{x^{2}} \end{aligned}$

$∵ | x | = x$ for $x > 0$ and $[x] = 0$ for $0 < x < 1$

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

$\begin{aligned} = π lim_{x \to 0^{+}} \frac{\tan (π \sin^{2} x)}{π \sin^{2} x} \cdot lim_{x \to 0^{+}} \frac{\sin^{2} x}{x^{2}} + 1 \\ = π + 1 ∵ lim_{x \to 0} \frac{\tan x}{x} = 1 \\ and lim_{x \to 0} \frac{\sin x}{x} = 1 \end{aligned}$

and LHL

$\begin{aligned} = lim_{x \to 0^{-}} \frac{\tan (π \sin^{2} x) + (| x | - \sin (x [x])^{2}}{x^{2}} \\ = lim_{x \to 0^{-}} \frac{\tan (π \sin^{2} x) + (- x - \sin (x (- 1))^{2}}{x^{2}} \\ ∵ | x | = - x for x < 0 \\ and [x] = - 1 for - 1 < x < 0 \\ = lim_{x \to 0^{-}} \frac{\tan (π \sin^{2} x) + (x + \sin (- x))^{2}}{x^{2}} \\ = lim_{x \to 0^{-}} \frac{\tan (π \sin^{2} x) + (x - \sin x)^{2}}{x^{2}} \\ [∵ \sin (- θ) = - \sin θ] \\ = lim_{x \to 0^{-}} \frac{\tan (π \sin^{2} x) + x^{2} + \sin^{2} x - 2 x \sin x}{x^{2}} \\ = lim_{x \to 0^{-}} \frac{\tan (π \sin^{2} x)}{x^{2}} + 1 + \frac{\sin^{2} x}{x^{2}} - \frac{2 x \sin x}{x^{2}} \\ = lim_{x \to 0^{-}} \frac{\tan (π \sin^{2} x)}{π \sin^{2} x} \cdot \frac{π \sin^{2} x}{x^{2}} + 1 + \frac{\sin^{2} x}{x^{2}} - 2 \frac{\sin x}{x} \\ = lim_{x \to 0^{-}} \frac{\tan (π \sin^{2} x)}{π \sin^{2} x} \cdot lim_{x \to 0^{-}} \frac{π \sin^{2} x}{x^{2}} + \\ 1 + lim_{x \to 0^{-}} \frac{\sin^{2} x}{x^{2}} - 2 lim_{x \to 0^{-}} \frac{\sin x}{x} \\ = π + 1 + 1 - 2 = π \end{aligned}$