A) \[\frac{3}{2}\]
B) 3
C) -3
D) -1
Correct Answer: C
Solution :
\[\underset{x\to 2}{\mathop{\lim }}\,\,\frac{{{e}^{3x-6}}-1}{\sin \,(2-x)}\,\,=\,\,\underset{x\to 2}{\mathop{\lim }}\,\,\frac{{{e}^{-3(2\,-\,x)}}-1}{\sin \,(2-x)}\] Put \[2-x=t\Rightarrow x=2-t\] \[\Rightarrow \,\,\,\,\underset{t\,\to \,0}{\mathop{\lim }}\,\frac{{{e}^{-\,3t}}-1}{\sin \,\,t}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( \frac{0}{0}\,\,\,Form \right)\] \[=\,\,\underset{t\to 0}{\mathop{\lim }}\,\frac{(-3).e{{\,}^{-\,3t}}}{\cos \,\,t}=-\,3\] [By L-Hospital?s rule]
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