Question

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

• A. $2$
• B. $-2$
• C. $1$
• D. $3$

Answer 1

The correct option is A $2$

The value of $\underset{x\to 2}{lim}\frac{xf\left(2\right)-2f\left(x\right)}{x-2}$
$=\underset{x\to 2}{lim}\frac{xf\left(2\right)-2f\left(2\right)+2f\left(2\right)-2f\left(x\right)}{x-2}$
$=\underset{x\to 2}{lim}\frac{(x-2)f\left(2\right)}{x-2}-2\underset{x\to 2}{lim}\frac{f\left(x\right)-f\left(2\right)}{x-2}$
$=f\left(2\right)-2f^{\prime }\left(2\right)=4-2\times 1=2$