A) \[{{e}^{-2}}\]
B) \[e\]
C) \[{{e}^{-1}}\]
D) \[1\]
Correct Answer: C
Solution :
If \[f(x)\] is continuous at \[x=2\] , then \[\underset{x\to 2}{\mathop{\lim }}\,{{(x-1)}^{\frac{1}{2x}}}=k\] Above is \[{{1}^{\infty }}\]form, \[\therefore k={{e}^{1}}\] Where \[l=\underset{x\to 2}{\mathop{\lim }}\,(x-1-1)\times \frac{1}{2-x}=\underset{x\to 2}{\mathop{\lim }}\,\frac{x-2}{2-x}=-1\] \[\Rightarrow k={{e}^{-1}}\]
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