A) \[\frac{1}{4}\]
B) \[-\frac{1}{4}\]
C) \[\frac{1}{8}\]
D) \[-\frac{1}{8}\]
Correct Answer: D
Solution :
If \[f(x)\] is continuous at \[x=0,\] then \[f(0)\,=\,\underset{x\to 0}{\mathop{\lim }}\,f(x)\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{2-\sqrt{x+4}}{\sin 2x}\], \[\left( \frac{0}{0}\text{ }\,\text{form} \right)\] Using L?Hospital?s rule, \[f(0)=\underset{x\to 0}{\mathop{\lim }}\,\frac{\left( -\frac{1}{2\sqrt{x+4}} \right)}{2\cos 2x}=-\frac{1}{8}\].
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