A) equals \[\sqrt{2}\]
B) equals\[-\sqrt{2}\]
C) equals \[\frac{1}{\sqrt{2}}\]
D) does not exist
Correct Answer: D
Solution :
| \[\underset{x\to 2}{\mathop{\lim }}\,\frac{\sqrt{1-\cos \left\{ 2\left( x-2 \right) \right\}}}{x-2}\]\[=\underset{x\to 2}{\mathop{\lim }}\,\frac{\sqrt{2}\left| \sin (x-2) \right|}{x-2}\] |
| \[\underset{\left( \operatorname{at}\,x=2 \right)}{\mathop{\operatorname{L}.H.L}}\,=-\underset{x\to 2}{\mathop{\lim }}\,\frac{\sqrt{2}\sin \left( x-2 \right)}{\left( x-2 \right)}=-1\] |
| \[\underset{\left( \operatorname{at}\,x=2 \right)}{\mathop{\operatorname{R}.H.L}}\,=\underset{x\to 2}{\mathop{\lim }}\,\frac{\sqrt{2}\sin \left( x-2 \right)}{\left( x-2 \right)}=1.\] |
| Thus \[\underset{\left( \operatorname{at}x=2 \right)}{\mathop{\operatorname{L}.H.L}}\,\ne \underset{\left( \operatorname{at}\,x=2 \right)}{\mathop{R.H.L}}\,\] Hence, \[\underset{x\to 2}{\mathop{\lim }}\,\frac{\sqrt{1-\cos \left\{ 2\left( x-2 \right) \right\}}}{x-2}\]does not exist. |
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