mathop {lim }limits_{x to 0} frac{{{{sin }^{ | vedclass

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(D) Given limit: $\mathop {\lim }\limits_{x 	o 0} \frac{{{{\sin }^{ - 1}}x - {{	an }^{ - 1}}x}}{{{x^3}}}$,which is of the form $\left( \frac{0}{0} \

Vedclass · Accepted Answer

$1/2$

Vedclass · Answer

(D) Given limit: $\mathop {\lim }\limits_{x 	o 0} \frac{{{{\sin }^{ - 1}}x - {{	an }^{ - 1}}x}}{{{x^3}}}$,which is of the form $\left( \frac{0}{0} ight)$.Applying $L'	ext{Hospital's rule}$,we differentiate the numerator and denominator with respect to $x$:$\mathop {\lim }\limits_{x 	o 0} \frac{{\frac{1}{{\sqrt {1 - {x^2}} }} - \frac{1}{{1 + {x^2}}}}}{{3{x^2}}}$,which is still of the form $\left( \frac{0}{0} ight)$.Applying $L'	ext{Hospital's rule}$ again:$= \mathop {\lim }\limits_{x 	o 0} \frac{{\frac{d}{{dx}} \left( (1 - {x^2})^{-1/2} - (1 + {x^2})^{-1} ight)}}{{6x}}$$= \mathop {\lim }\limits_{x 	o 0} \frac{{-\frac{1}{2}(1 - {x^2})^{-3/2}(-2x) - (-1)(1 + {x^2})^{-2}(2x)}}{{6x}}$$= \mathop {\lim }\limits_{x 	o 0} \frac{{x(1 - {x^2})^{-3/2} + 2x(1 + {x^2})^{-2}}}{{6x}}$$= \mathop {\lim }\limits_{x 	o 0} \frac{1}{6} \left[ (1 - {x^2})^{-3/2} + 2(1 + {x^2})^{-2} ight]$$= \frac{1}{6} [1 + 2] = \frac{3}{6} = \frac{1}{2}$.

$\mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^{ - 1}}x - {{\tan }^{ - 1}}x}}{{{x^3}}}$ is equal to

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