The value of mathop {lim }limits_{x to 0} fra | vedclass

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Question

(A) Given limit: $\mathop {\lim }\limits_{x 	o 0} \frac{{x\cos x - \log (1 + x)}}{{{x^2}}}$,which is of the form $\frac{0}{0}$.Applying $L'Hospital's

Vedclass · Accepted Answer

$1/2$

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(A) Given limit: $\mathop {\lim }\limits_{x 	o 0} \frac{{x\cos x - \log (1 + x)}}{{{x^2}}}$,which is of the form $\frac{0}{0}$.Applying $L'Hospital's$ rule,we differentiate the numerator and denominator with respect to $x$:$\mathop {\lim }\limits_{x 	o 0} \frac{{\frac{d}{dx}(x\cos x - \log (1 + x))}}{{\frac{d}{dx}(x^2)}} = \mathop {\lim }\limits_{x 	o 0} \frac{{\cos x - x\sin x - \frac{1}{{1 + x}}}}{{2x}}$.This is still of the form $\frac{0}{0}$. Applying $L'Hospital's$ rule again:$\mathop {\lim }\limits_{x 	o 0} \frac{{\frac{d}{dx}(\cos x - x\sin x - \frac{1}{{1 + x}})}}{{\frac{d}{dx}(2x)}} = \mathop {\lim }\limits_{x 	o 0} \frac{{-\sin x - (\sin x + x\cos x) + \frac{1}{{{{(1 + x)}^2}}}}}{2}$.Evaluating the limit as $x 	o 0$:$= \frac{{-\sin(0) - (\sin(0) + 0 \cdot \cos(0)) + \frac{1}{{{{(1 + 0)}^2}}}}}{2} = \frac{{0 - 0 + 1}}{2} = \frac{1}{2}$.

The value of $\mathop {\lim }\limits_{x \to 0} \frac{{x\cos x - \log (1 + x)}}{{{x^2}}}$ is

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