A) \[0\]
B) \[1\]
C) \[-1\]
D) does not exist
Correct Answer: D
Solution :
We have,\[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{x}}{\sqrt{4-\sqrt{x}}-\sqrt{h}}\] \[RHL=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{\sqrt{x}}{\sqrt{4-\sqrt{x}}-\sqrt{x}}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{\sqrt{0+h}}{\sqrt{4-\sqrt{h}}-\sqrt{h}}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{\sqrt{h}}{\sqrt{4-\sqrt{h}}-\sqrt{h}}\cdot \frac{\sqrt{4-\sqrt{h}}+\sqrt{h}}{\sqrt{4-\sqrt{h}}+\sqrt{h}}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{\sqrt{h}\left( \sqrt{4-\sqrt{h}}+\sqrt{h} \right)}{4-\sqrt{h}-h}=\frac{0}{4}=0\] \[LHL=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\frac{\sqrt{x}}{\sqrt{4-\sqrt{h}}-\sqrt{-h}}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{\sqrt{-h}}{\sqrt{4-\sqrt{h}}-\sqrt{-h}}\] This limit does not exist, since function it not defined for negative value of x. Hence,\[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{x}}{\sqrt{4-\sqrt{x}}-\sqrt{x}}\]does not exist.
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