mathop {lim }limits_{x to - 2} frac{{{{sin }^ | vedclass

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Question

(C) Let $y = \mathop {\lim }\limits_{x 	o - 2} \frac{{{{\sin }^{ - 1}}(x + 2)}}{{{x^2} + 2x}}$.Since the limit is in the $\frac{0}{0}$ form,we apply

Vedclass · Accepted Answer

$-1/2$

Vedclass · Answer

(C) Let $y = \mathop {\lim }\limits_{x 	o - 2} \frac{{{{\sin }^{ - 1}}(x + 2)}}{{{x^2} + 2x}}$.Since the limit is in the $\frac{0}{0}$ form,we apply $L'	ext{Hospital's rule}$.$y = \mathop {\lim }\limits_{x 	o - 2} \frac{\frac{d}{dx}(\sin^{-1}(x+2))}{\frac{d}{dx}(x^2+2x)}$$y = \mathop {\lim }\limits_{x 	o - 2} \frac{\frac{1}{\sqrt{1-(x+2)^2}}}{2x+2}$Substituting $x = -2$:$y = \frac{\frac{1}{\sqrt{1-0}}}{2(-2)+2} = \frac{1}{-4+2} = -\frac{1}{2}$.

$\mathop {\lim }\limits_{x \to - 2} \frac{{{{\sin }^{ - 1}}(x + 2)}}{{{x^2} + 2x}}$ is equal to

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