_ x - 2 ^ - 1 (x + 2) x^2 + 2x is equal to - Vidyadip

MCQ

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

A
$0$
B
$\infty $
✓
$-1/2$
D
None of these

✓

Answer

Correct option: C.

$-1/2$

c
(c) $y = \mathop {\lim }\limits_{x \to - 2} \frac{{{{\sin }^{ - 1}}(x + 2)}}{{{x^2} + 2x}}$, $\left( {\frac{0}{0}{\rm{form}}} \right)$

Using $ L-$ Hospital’s rule

==> $y = \mathop {\lim }\limits_{x \to - 2} \frac{{\left( {\frac{1}{{\sqrt {1 - {{(x + 2)}^2}} }}} \right)}}{{2x + 2}}$

==> $y = \frac{1}{{ - 4 + 2}} = - \frac{1}{2}$.

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