Limit Continuity and Differentiability 1 Question 5

6. $\lim _{x \rightarrow 0} \frac{x \cot (4 x)}{\sin ^{2} x \cot ^{2}(2 x)}$ 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 $[x]=x-{x}$.

$$ \lim _{x \rightarrow 0^{+}} x \frac{1}{x}+\frac{2}{x}+\ldots+\frac{15}{x} $$

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

$\therefore \quad \frac{1}{x}=\frac{1}{x}-\frac{1}{x}$

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

$\therefore$ Given limit $=\lim _{x \rightarrow 0^{+}} x \frac{1}{x}-\frac{1}{x}+\frac{2}{x}-\frac{2}{x}+\ldots \frac{15}{x}-\frac{15}{x}$

$=\lim _{x \rightarrow 0^{+}}(1+2+3+\ldots+15)-x \frac{1}{x}+\frac{2}{x}+\ldots+\frac{15}{x}$

$=120-0=120$

$$ \because 0 \leq \frac{n}{x}<1 \text {, therefore } $$

$$ 0 \leq x \frac{n}{x}<x \Rightarrow \lim _{x \rightarrow 0^{+}} x \frac{n}{x}=0 $$



NCERT Chapter Video Solution

Dual Pane