Limit Continuity and Differentiability 1 Question 3
4. $\lim _{x \rightarrow 0} \frac{\sin ^{2} x}{\sqrt{2}-\sqrt{1+\cos x}}$ equals
(2019 Main, 8 April I)
(a) $4 \sqrt{2}$
(b) $\sqrt{2}$
(c) $2 \sqrt{2}$
(d) 4
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Answer:
Correct Answer: 4. (a)
Solution:
- Given,
$$ \lim _{x \rightarrow 1^{+}} \frac{(1-|x|+\sin |1-x|) \sin \frac{\pi}{2}[1-x]}{|1-x|[1-x]} $$
Put $x=1+h$, then
$$ x \rightarrow 1^{+} \Rightarrow h \rightarrow 0^{+} $$
$\therefore \lim _{x \rightarrow 1^{+}} \frac{(1-|x|+\sin |1-x|) \sin \frac{\pi}{2}[1-x]}{|1-x|[1-x]}$
$=\lim _{h \rightarrow 0^{+}} \frac{(1-|h+1|+\sin |-h|) \sin \frac{\pi}{2}[-h]}{|-h|[-h]}$
$=\lim _{h \rightarrow 0^{+}} \frac{(1-(h+1)+\sin h) \sin \frac{\pi}{2}[-h]}{h[-h]}$
$(\because|-h|=h$ and $|h+1|=h+1$ as $h>0)$
$$ \begin{aligned} & =\lim _{h \rightarrow 0^{+}} \frac{(-h+\sin h) \sin \frac{\pi}{2}(-1)}{h(-1)} \\ & \left(\because[x]=-1 \text { for }-1<x<0 \text { and } h \rightarrow 0^{+} \Rightarrow-h \rightarrow 0\right) \\ & =\lim _{h \rightarrow 0^{+}} \frac{(-h+\sinh )}{-h} \sin \frac{-\pi}{2} \\ & =\lim _{h \rightarrow 0^{+}} \frac{\sin h}{h}-\lim _{h \rightarrow 0^{+}} \frac{h}{h} \\ & =\lim _{h \rightarrow 0^{+}} \frac{\sin h}{h}-\lim _{h \rightarrow 0^{+}} \frac{h}{h}=1-1=0 \because \lim _{h \rightarrow 0^{+}} \frac{\sin h}{h}=1 \end{aligned} $$
