Question

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

• A. $2$
• B. $-2$
• C. $-4$
• D. $3$

Answer 1

The correct option is C $-4$

$\underset{x\to 2}{lim}\frac{xf\left(2\right)-2f\left(x\right)+2f\left(2\right)-2f\left(2\right)}{x-2}$
$=\underset{x\to 2}{lim}\frac{(x-2)f\left(2\right)-2[f\left(x\right)-f\left(2\right)]}{x-2}$
$=\underset{x\to 2}{lim}\frac{(x-2)f\left(2\right)}{(x-2)}-\underset{x\to 2}{lim}\frac{2[f\left(x\right)-f\left(2\right)]}{x-2}$
$=f\left(2\right)-2f^{\prime }\left(x\right)$
$=4-2\times 4=-4$
Hence, option 'C' is correct.