MCQ
$\mathop {\lim }\limits_{x \to 0} x\log (\sin x) = $
- A$-1$
- ✓${\log _e}1$
- C$1$
- DNone of these
✓
Answer
Correct option: B.
${\log _e}1$
b
(b)$\mathop {\lim }\limits_{x \to 0} x\log \sin x = \mathop {\lim }\limits_{x \to 0} \,\log \,{(\sin x)^x} = \log \,[\mathop {\lim }\limits_{x \to 0} \,\,{(\sin x)^x}]$
(b)$\mathop {\lim }\limits_{x \to 0} x\log \sin x = \mathop {\lim }\limits_{x \to 0} \,\log \,{(\sin x)^x} = \log \,[\mathop {\lim }\limits_{x \to 0} \,\,{(\sin x)^x}]$
$ = \log \,\left[ {\mathop {\lim }\limits_{x \to 0} \,{{(1 + \sin x - 1)}^{\frac{{x(\sin x - 1)}}{{\sin x - 1}}}}} \right]$
$ = {\log _e}[{e^{\mathop {\lim }\limits_{x \to 0} \,x(\sin x - 1)}}] = {\log _e}1.$
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