Question

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

• A. $2$
• B. $-2$
• C. $-4$
• D. $4$

Answer 1

Answer: (C)

$-4$

$f\left(x\right)=4,f\left(x\right)=4$

$\underset{x\to 2}{lim}\frac{xf\left(2\right)-2f\left(x\right)}{x-2}=\frac{0}{0}$form

$=\underset{x\to 2}{lim}\frac{1.f\left(2\right)-2^{\prime }f\left(x\right)}{1}$

$=f\left(2\right)-2f^{\prime }\left(2\right)$

$=4-2\left(4\right)=-4$