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
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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 $=\lim _{x \rightarrow 0^{+}} x \frac{1}{x}-\frac{1}{x}+\frac{2}{x}-\frac{2}{x}+\ldots \frac{15}{x}-\frac{15}{x}$
$=\lim _{x \rightarrow 0^{+}}(1+2+3+\ldots+15)-x \frac{1}{x}+\frac{2}{x}+\ldots+\frac{15}{x}$
$=120-0=120$
$$ \because 0 \leq \frac{n}{x}<1 \text {, therefore } $$
$$ 0 \leq x \frac{n}{x}<x \Rightarrow \lim _{x \rightarrow 0^{+}} x \frac{n}{x}=0 $$
