A) 0
B) \[\infty \]
C) ?1/2
D) None of these
Correct Answer: C
Solution :
\[y=\underset{x\to -2}{\mathop{\lim }}\,\frac{{{\sin }^{-1}}(x+2)}{{{x}^{2}}+2x}\], \[\left( \frac{0}{0}\text{form} \right)\] Using L-Hospital?s rule Þ \[y=\underset{x\to -2}{\mathop{\lim }}\,\frac{\left( \frac{1}{\sqrt{1-{{(x+2)}^{2}}}} \right)}{2x+2}\]Þ \[y=\frac{1}{-4+2}=-\frac{1}{2}\].
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