Limit Continuity and Differentiability 1 Question 14
16. $\lim _{x \rightarrow 1} \frac{\sqrt{1-\cos 2(x-1)}}{x-1}$
(1998, 2M)
(a) exists and it equals $\sqrt{2}$
(b) exists and it equals $-\sqrt{2}$
(c) does not exist because $x-1 \rightarrow 0$
(d) does not exist because left hand limit is not equal to right hand limit
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Answer:
Correct Answer: 16. (c)
Solution:
- $\lim _{x \rightarrow 0} \tan \frac{\pi}{4}+x$
$$ \begin{aligned} & =\lim _{x \rightarrow 0} \frac{\tan \frac{\pi}{4}+\tan x}{1-\tan \frac{\pi}{4} \tan x}=\lim _{x \rightarrow 0} \frac{1+\tan x}{1-\tan x} \\ & =\lim _{x \rightarrow 0} \frac{\left[(1+\tan x)^{1 / \tan x}\right]^{\tan x / x}}{\left[(1-\tan x)^{-1 / \tan x}\right]^{-\tan x / x}}=\frac{e^{1}}{e^{-1}}=e^{2} \end{aligned} $$
