Evaluate the limit: mathop {lim }limits_{x to | vedclass

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

Question

(B) The given limit is of the form $\frac{0}{0}$ as $x 	o 0$. Applying $L$'Hopital's rule,we differentiate the numerator and the denominator with res

Vedclass · Accepted Answer

$-2$

Vedclass · Answer

(B) The given limit is of the form $\frac{0}{0}$ as $x 	o 0$. Applying $L$'Hopital's rule,we differentiate the numerator and the denominator with respect to $x$. Using the Leibniz integral rule for the numerator: $\frac{d}{dx} \int_0^x (	an^{-1} t)^2 dt = (	an^{-1} x)^2$. For the denominator: $\frac{d}{dx} (\sin x - x) = \cos x - 1$. So,the limit becomes $\mathop {\lim }\limits_{x 	o 0} \frac{(	an^{-1} x)^2}{\cos x - 1}$. Using the standard limits $\mathop {\lim }\limits_{x 	o 0} \frac{	an^{-1} x}{x} = 1$ and $\mathop {\lim }\limits_{x 	o 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$,we rewrite the expression: $\mathop {\lim }\limits_{x 	o 0} \frac{(	an^{-1} x)^2}{x^2} \cdot \frac{x^2}{\cos x - 1} = (1)^2 \cdot (-2) = -2$.

Evaluate the limit: $\mathop {\lim }\limits_{x \to 0} \frac{{\int\limits_0^x (\tan^{-1} t)^2 dt}}{{\sin x - x}}$

Similar Questions

If $\lim _{t}$ ${\rightarrow 0}\left(\int_0^1(3 x+5)^t d x\right)^{\frac{1}{t}}=\frac{\alpha}{5 e}\left(\frac{8}{5}\right)^{\frac{2}{3}}$,then $\alpha$ is equal to . . . . . .

$\mathop {\lim }\limits_{x \to 0} \frac{{{e^{\tan x}} - {e^x}}}{{\tan x - x}} = $

$\lim _{x \rightarrow 0} \frac{x 2^{x}-x}{1-\cos x}$ is equal to

$\mathop {\lim }\limits_{\alpha \to \pi /4} \frac{{\sin \alpha - \cos \alpha }}{{\alpha - \frac{\pi }{4}}} = $

$\mathop {\lim }\limits_{x \to \pi /4} \frac{{\sqrt 2 \cos x - 1}}{{\cot x - 1}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module