SATHEE: Limit Continuity and Differentiability 1 Question 5

Limit Continuity and Differentiability 1 Question 5

6. $lim_{x \to 0} \frac{x \cot (4 x)}{\sin^{2} x \cot^{2} (2 x)}$ is equal to

(2019 Main, 11 Jan II)

(a) 0

(b) 1

(c) 4

(d) 2

Show Answer

Answer:

Correct Answer: 6. (b)

Solution:

Key Idea Use property of greatest integer function $[x] = x - x$ .

$lim_{x \to 0^{+}} x \frac{1}{x} + \frac{2}{x} + \dots + \frac{15}{x}$

We know, $[x] = x - x$

$∴ \frac{1}{x} = \frac{1}{x} - \frac{1}{x}$

Similarly, $\frac{n}{x} = \frac{n}{x} - \frac{n}{x}$

$∴$ Given limit $= lim_{x \to 0^{+}} x \frac{1}{x} - \frac{1}{x} + \frac{2}{x} - \frac{2}{x} + \dots \frac{15}{x} - \frac{15}{x}$

$= lim_{x \to 0^{+}} (1 + 2 + 3 + \dots + 15) - x \frac{1}{x} + \frac{2}{x} + \dots + \frac{15}{x}$

$= 120 - 0 = 120$

$∵ 0 \leq \frac{n}{x} < 1, therefore$

$0 \leq x \frac{n}{x} < x \Rightarrow lim_{x \to 0^{+}} x \frac{n}{x} = 0$

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