A) 1/4
B) \[-1/4\]
C) 1/8
D) \[-1/8\]
Correct Answer: D
Solution :
Since,\[f(x)\]is continuous at\[x=0\] \[\therefore \] \[f({{0}^{-}})=f(0)=f({{0}^{+}})\] \[f({{0}^{-}})=\underset{h\to 0}{\mathop{\lim }}\,\frac{2-\sqrt{(0-h)+4}}{\sin 2(0-h)}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{2-\sqrt{(4-h)}}{-\sin 2h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{0-\frac{1}{2}{{(4-h)}^{-1/2}}(-1)}{-2\cos 2h}\] (using L-Hospital rule) \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{-1}{2.2.\sqrt{4-h}\cos 2h}\] \[=\frac{-1}{2.2.2}=-\frac{1}{8}\]
You need to login to perform this action.
You will be redirected in
3 sec
