Question

Let f(x) = 4 and f'(x) = 4, then ${lim}_{x\to 2}\frac{xf\left(2\right)-2f\left(x\right)}{x-2}$ equals

• A. 2
• B. -2
• C. -4
• D. 3

Answer 1

The correct option is C -4

$y={lim}_{x\to 2}\frac{xf\left(2\right)-2f\left(x\right)}{x-2}$
$\Rightarrow y={lim}_{x\to 2}\frac{xf\left(2\right)+2f\left(2\right)+2f\left(2\right)-2f\left(x\right)}{x-2}$
$\Rightarrow y={lim}_{x\to 2}\frac{-2f\left(x\right)+2f\left(2\right)+xf\left(2\right)-2f\left(2\right)}{(x-2)}$
$\Rightarrow y={lim}_{x\to 2}-2\frac{\left[f\right(x)-f(2\left)\right]}{x-2}+{lim}_{x\to 2}\frac{f\left(2\right).(x-2)}{(x-2)}$
$\Rightarrow y=-2{lim}_{x\to 2}\frac{f\left(x\right)-f\left(2\right)}{x-2}+f\left(2\right)$
$\Rightarrow y=-2{lim}_{x\to 2}{f}^{\prime }\left(x\right)+f\left(2\right)=-8+4=-4$