A) 0
B) \[\frac{1}{2}\]
C) \[-\frac{1}{2}\]
D) None of these
Correct Answer: C
Solution :
Apply L-Hospital?s rule, we get \[\underset{x\to 0}{\mathop{\lim }}\,\,\frac{\cos x-\frac{1}{1-x}}{2x}=\underset{x\to 0}{\mathop{\lim }}\,\,\frac{-\sin x-\frac{1}{{{(1-x)}^{2}}}}{2}=-\frac{1}{2}\]. Aliter : \[\underset{x\to 0}{\mathop{\lim }}\,\,\frac{\sin x+\log \,(1-x)}{{{x}^{2}}}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{\left( x-\frac{{{x}^{3}}}{3\,\,!}+\frac{{{x}^{5}}}{5\,\,!}-... \right)}{{{x}^{2}}}+\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{\left( -x-\frac{{{x}^{2}}}{2}-\frac{{{x}^{3}}}{3}-\frac{{{x}^{4}}}{4}-... \right)}{{{x}^{2}}}\] \[\left( \because \sin x=x-\frac{{{x}^{3}}}{3\,!}+\frac{{{x}^{5}}}{5\,!}-.. \right.\] and \[\left. \log \,(1-x)=-x-\frac{{{x}^{2}}}{2}-\frac{{{x}^{3}}}{3}-.. \right)\] \[=\underset{x\to 0}{\mathop{\lim }}\,\,\frac{\frac{-{{x}^{2}}}{2}-{{x}^{3}}\left( \frac{1}{3\,\,!}+\frac{1}{3} \right)-\frac{{{x}^{4}}}{4}...}{{{x}^{2}}}=-\frac{1}{2}.\]
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