A) \[{{e}^{\alpha }}\]
B) \[-\alpha \]
C) \[{{e}^{1}}^{-\alpha }\]
D) \[{{e}^{1}}^{+\alpha }\]
Correct Answer: C
Solution :
[C] \[Limit={{e}^{\underset{n\to \infty }{\mathop{\lim }}\,\,\,n\left( {{\left( \frac{n}{n+1} \right)}^{\alpha }}+\,\sin \frac{1}{n}-1 \right)}}\] \[={{e}^{\underset{n\to \infty }{\mathop{\lim }}\,\left( \frac{\sin \,\,(1/n)}{1/n}=\frac{{{(1\,+\,\,(1/n))}^{\alpha }}-1}{1/n}.\frac{1}{{{(1\,+\,\,(1/n))}^{\alpha }}} \right)}}\] \[={{e}^{1-\alpha }}\]
You need to login to perform this action.
You will be redirected in
3 sec
