A) \[4\]
B) \[0\]
C) \[2\]
D) \[\infty \]
Correct Answer: A
Solution :
\[\underset{x\to 2}{\mathop{\lim }}\,\frac{2{{x}^{2}}-4f'(x)}{x-2}=\underset{x\to 2}{\mathop{\lim }}\,\frac{4x-4f''(x)}{1}\] \[=8-4f'(2)=8-4=4\]You need to login to perform this action.
You will be redirected in
3 sec