A) \[{{e}^{2}}\]
B) \[e\]
C) \[{{e}^{-1}}\]
D) 1
Correct Answer: D
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,{{\left( \frac{1+f(3+x)-f(3)}{1+f(2-x)-f(2)} \right)}^{\frac{1}{x}}}({{1}^{\infty }}form)\] \[\Rightarrow {{e}^{\underset{x\to 0}{\mathop{\lim }}\,\frac{f(3+x)-f(2-x)-f(3)+f(2)}{x(1+f(2-x)-f(2))}}}\] using L'Hopital \[\Rightarrow {{e}^{\underset{x\to 0}{\mathop{\lim }}\,\frac{f'(3+x)+f'(2-x)}{-xf'(2-x)+(1+f(2-x)-f(2))}}}\] \[\Rightarrow {{e}^{\frac{f'(3)+f'(2)}{1}}}=1\]
You need to login to perform this action.
You will be redirected in
3 sec
