Question

Let $f\left(2\right)=4,f^{\prime }\left(2\right)=4$, then $\underset{x\to 2}{lim}\frac{xf\left(2\right)-2f\left(x\right)}{x-2}$ is

• A. $2$
• B. $-2$
• C. $-4$
• D. $3$

Answer 1

Answer: (C)

$-4$

On applying the limit, since the function is of $\frac{0}{0}$ form,
We apply L-Hospital's Rule to get,
$\underset{x\to 2}{Lim}f\left(2\right)-2f^{\prime }\left(x\right)$
$=f\left(2\right)-2f^{\prime }\left(2\right)$
$=4-2\left(4\right)$
$=4-8$
$=-4$