MCQ
$\mathop {\lim }\limits_{x \to - 2} \frac{{{{\sin }^{ - 1}}(x + 2)}}{{{x^2} + 2x}}$ is equal to
- A$0$
- B$\infty $
- ✓$-1/2$
- DNone of these
✓
Answer
Correct option: C.
$-1/2$
c
(c) $y = \mathop {\lim }\limits_{x \to - 2} \frac{{{{\sin }^{ - 1}}(x + 2)}}{{{x^2} + 2x}}$, $\left( {\frac{0}{0}{\rm{form}}} \right)$
(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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