A) \[\sqrt{2}\]
B) \[-\sqrt{2}\]
C) \[\frac{1}{\sqrt{2}}\]
D) Limit does not exist
Correct Answer: D
Solution :
| [d] \[\underset{x\to 0}{\mathop{\lim }}\,\frac{x}{\sqrt{1-\cos x}}\] |
| \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{x}{\sqrt{1-\left( 1-2{{\sin }^{2}}\frac{x}{2} \right)}}\] |
| \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{x}{\sqrt{2{{\sin }^{2}}\frac{x}{2}}}=\frac{1}{2}\underset{x\to 0}{\mathop{\lim }}\,\frac{x}{\left| \sin \frac{x}{2} \right|}\] |
| L.H.L\[=f(0-0)=\underset{h\to 0}{\mathop{\lim }}\,\frac{x}{\left| \sin \frac{x}{2} \right|}\] |
| \[=-\frac{1}{\sqrt{2}}\underset{x\to 0}{\mathop{\lim }}\,\frac{2\left( \frac{h}{2} \right)}{\sin \frac{h}{2}}\] |
| \[=\frac{1}{\sqrt{2}}\times 2\times 1\] \[\left( \because \underset{\theta \to 0}{\mathop{\lim }}\,\frac{\theta }{\sin \theta }=1 \right)\] |
| RHL \[=f(0+0)=\underset{h\to 0}{\mathop{\lim }}\,f(0+h)\] |
| \[=\frac{1}{\sqrt{2}}\underset{h\to 0}{\mathop{\lim }}\,\frac{2\left( \frac{h}{2} \right)}{\sin \frac{h}{2}}=\frac{1}{\sqrt{2}}\times 2\times 1\] |
| \[LHL\ne RHL=\sqrt{2}\] |
| Therefore limit does not exist. |
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