MCQ
$\mathop {\lim }\limits_{x \to - 2} \frac{{{{\sin }^{ - 1}}(x + 2)}}{{{x^2} + 2x}} = . . .$
- A$0$
- B$\infty $
- ✓$-1/2$
- Dએકપણ નહી.
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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