MCQ
$\mathop {\lim }\limits_{x \to 0} \frac{{x\cos x - \log (1 + x)}}{{{x^2}}} = . . .$
- ✓$1/2$
- B$0$
- C$1$
- Dએકપણ નહી.
Applying $ L-$ Hospital’s rule, we have
$\mathop {\lim }\limits_{x \to 0} \,\frac{{\cos x - x\sin x - \frac{1}{{x + 1}}}}{{2x}}$, $\left( {\frac{0}{0}{\rm{form}}} \right)$
$ = \mathop {\lim }\limits_{x \to 0} \,\frac{{ - \sin x - \sin x - x\cos x + \frac{1}{{{{(x + 1)}^2}}}}}{2} = \frac{1}{2}$.
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