Question

Let $f(2)=4$ and $f^{\prime }(2)=4$. Then $\underset{x\to 2}{\lim }\frac{xf(2)-2f(x)}{x-2}$ is given by

• A. 2
• B. -2
• C. -4
• D. 3

Answer 1

Answer: (C)

Apply L H Rule
We have, $\underset{x\to 2}{\lim }\frac{xf\left(2\right)-2f\left(x\right)}{x-2}\left(\frac{0}{0}\right)$
$=\underset{x\to 2}{\lim }f\left(2\right)-2f^{\prime }\left(x\right)=f\left(2\right)-2f^{\prime }\left(2\right)$
$=4-2\times 4=-4$