Question

If $f\left(2\right)=2$ and $f^{\prime }\left(2\right)=1$, then $\underset{x\to 2}{lim}\frac{2x^2-4f\left(x\right)}{x-2}$ is equal to.

• A. $4$
• B. $-4$
• C. $2$
• D. $-2$

Answer 1

The correct option is A $4$

Now, $\underset{x\to 2}{lim}\frac{2x^2-4f\left(x\right)}{x-2}$ $\left[\frac{0}{0}form\right]$
By L'Hospital's rule, $=\underset{x\to 2}{lim}\frac{4x-4f^{\prime }\left(x\right)}{1}$
$=\frac{4\times 2-4f^{\prime }\left(2\right)}{1}$
$=8-4\times 1$
$=4$ $[\because f^{\prime }(2)=1]$