A) ?2
B) ?1
C) 0
D) 1
Correct Answer: D
Solution :
\[\underset{x\to 2}{\mathop{\lim }}\,\,\,f(x)=\underset{x\to 2}{\mathop{\lim }}\,\,\frac{\sin \,({{e}^{x-2}}-1)}{\log \,(x-1)}\] \[=\underset{t\to 0}{\mathop{\lim }}\,\,\,\frac{\sin \,({{e}^{t}}-1)}{\log \,(1+t)}\], {Putting \[x=2+t\}\] \[=\underset{t\to 0}{\mathop{\lim }}\,\,\,\frac{\sin \,({{e}^{t}}-1)}{{{e}^{t}}-1}.\frac{{{e}^{t}}-1}{t}.\frac{t}{\log \,(1+t)}\] \[=\underset{t\to 0}{\mathop{\lim }}\,\,\,\frac{\sin \,({{e}^{t}}-1)}{{{e}^{t}}-1}.\left( \frac{1}{1\,\,!}+\frac{t}{2\,\,!}+... \right)\times \left[ \frac{1}{\left( 1-\frac{1}{2}t+\frac{1}{3}{{t}^{2}}-... \right)} \right]\] \[=1\,\,.\,\,1\,\,.\,\,1=1,\,\,\,\,(\because \,\,\text{As}\,\,t\to 0,\,\,{{e}^{t}}-1\to 0).\]
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