_ x 0 ^ - 1 x - ^ - 1 x x^3 is equal to - Vidyadip

MCQ

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

A
$0$
B
$1$
C
$-1$
✓
$1/2$

✓

Answer

Correct option: D.

$1/2$

d
(d) $\mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^{ - 1}}x - {{\tan }^{ - 1}}x}}{{{x^3}}}$, $\left( {\frac{0}{0}} \,\,form \,\, \right)$

Applying $ L-$ Hospital’s rule,

$\mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{{\sqrt {1 - {x^2}} }} - \frac{1}{{1 + {x^2}}}}}{{3{x^2}}}$, $\left( {\frac{0}{0}} \,\,form \,\, \right)$

$ = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{ - 1}}{2} \times \frac{{ - 2x}}{{{{(1 - {x^2})}^{3/2}}}} + \frac{{2x}}{{{{(1 + {x^2})}^2}}}}}{{6x}}$

$ = \mathop {\lim }\limits_{x \to 0} \frac{1}{6}\left[ {\frac{1}{{{{(1 - {x^2})}^{3/2}}}} + \frac{2}{{{{(1 + {x^2})}^2}}}} \right]\, = \,\frac{1}{2}$.

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