A) \[-2\]
B) \[-1\]
C) \[0\]
D) 1
Correct Answer: D
Solution :
| [d] \[f(x)=\frac{\sin \left( {{e}^{x-2}}-1 \right)}{In(x-1)}\] |
| \[\underset{x\to 2}{\mathop{\lim }}\,\frac{\sin \left( {{e}^{x-2}}-1 \right)}{In(x-1)}=L\] |
| It is \[\frac{0}{0}\](undefined) condition so using Hospital?s rule |
| \[\Rightarrow L=\underset{x\to 2}{\mathop{\lim }}\,\left[ \frac{\{sin\left( {{e}^{x-2}}-1 \right)\}}{\{In(x-1)\}} \right]\] |
| \[\Rightarrow L=\underset{x\to 2}{\mathop{\lim }}\,\frac{\cos ({{e}^{x-2}}-1).{{e}^{(x-2)}}}{1/(x-1)}\] |
| \[\Rightarrow L=\underset{x\to 2}{\mathop{\lim }}\,\cos \left( {{e}^{2-2}}-1 \right){{e}^{2-2}}.(2-1)\] |
| \[\Rightarrow L=\cos (0){{e}^{0}}.1\Rightarrow L=1\] |
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