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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

$1/2$

Vedclass · Answer

(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

Similar Questions

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

The value of the limit $\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - {e^{ - x}} - 2x}}{{x - \sin x}}$ is

$\mathop {\lim }\limits_{x \to 1} {\left( {\frac{4}{\pi }{{\tan }^{ - 1}}x} \right)^{\frac{1}{{({x^2} - 1)}}}}$ is equal to -

If $\lim _{x \rightarrow 0} \frac{a x e^{x}-b \log (1+x)}{x^{2}}=3$, then the values of $a$ and $b$ are, respectively:

$\lim _{x \rightarrow 0} \frac{\cos 2x - \cos 3x}{\cos 4x - \cos 5x} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module