mathop {lim }limits_{x to 0} frac{{sqrt {1 + | vedclass

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(B) Let $L = \mathop {\lim }\limits_{x 	o 0} \frac{{\sqrt {1 + x} - \sqrt {1 - x} }}{{{{\sin }^{ - 1}}x}}$.Since the form is $\frac{0}{0}$,we apply $

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) Let $L = \mathop {\lim }\limits_{x 	o 0} \frac{{\sqrt {1 + x} - \sqrt {1 - x} }}{{{{\sin }^{ - 1}}x}}$.Since the form is $\frac{0}{0}$,we apply $L$-Hospital's rule:$L = \mathop {\lim }\limits_{x 	o 0} \frac{{\frac{d}{{dx}}(\sqrt {1 + x} - \sqrt {1 - x} )}}{{\frac{d}{{dx}}({{\sin }^{ - 1}}x)}}$$L = \mathop {\lim }\limits_{x 	o 0} \frac{{\frac{1}{{2\sqrt {1 + x} }} - (\frac{{ - 1}}{{2\sqrt {1 - x} }})}}{{\frac{1}{{\sqrt {1 - {x^2}} }}}}$$L = \mathop {\lim }\limits_{x 	o 0} \frac{{\frac{1}{{2\sqrt {1 + x} }} + \frac{1}{{2\sqrt {1 - x} }}}}{{\frac{1}{{\sqrt {1 - {x^2}} }}}}$Substituting $x = 0$:$L = \frac{{\frac{1}{2} + \frac{1}{2}}}{1} = \frac{1}{1} = 1$.

$\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + x} - \sqrt {1 - x} }}{{{{\sin }^{ - 1}}x}} = $

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