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

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(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

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$-2$

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

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