If \[f(x)=\frac{\sin ({{e}^{x-2}}-1)}{in(x-1)},\] then \[\underset{x\to 2}{\mathop{\lim }}\,f(x)\] is equal to

If \[f(x)=\frac{\sin ({{e}^{x-2}}-1)}{in(x-1)},\] then \[\underset{x\t

1.	If \[f(x)=\frac{\sin ({{e}^{x-2}}-1)}{in(x-1)},\] then \[\underset{x\to 2}{\mathop{\lim }}\,f(x)\] is equal to
A.	\[-2\]
B.	\[-1\]
C.	\[0\]
D.	1
Answer» E.

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If \[f(x)=\frac{\sin ({{e}^{x-2}}-1)}{in(x-1)},\] then \[\underset{x\to 2}{\mathop{\lim }}\,f(x)\] is equal to

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