A) -2
B) -1
C) 0
D) 1
Correct Answer: D
Solution :
We have,\[\underset{x\to 2}{\mathop{\lim }}\,f(x)=\underset{x\to 2}{\mathop{\lim }}\,\frac{\sin ({{e}^{x-2}}-1)}{\log (x-1)}\] \[=\underset{x\to 2}{\mathop{\lim }}\,\left\{ \frac{\sin ({{e}^{x-2}}-1)}{{{e}^{x-2}}-1} \right..\frac{{{e}^{x-2}}-1}{x-2}.\] \[\left. \frac{x-2}{\log [1+(x-1)]} \right\}\] \[=1\times 1\times 1=1\]
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