Question

Let $f\left(2\right)=4$ and $f^{\prime }\left(2\right)=4$. Then $\underset{x\to 2}{lim}\frac{xf\left(2\right)-2f\left(x\right)}{x-2}$ is given by

• A. $2$
• B. $-2$
• C. $-4$
• D. $3$

Answer 1

Answer: (C)

$-4$

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

Using L-hospital's rule,

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

$=f\left(2\right)-2f^{\prime }\left(2\right)=4-2\times 4=-4$

Hence, option 'C' is correct.