Question

Let $f\left(x\right)$ be a differential function and $f^{\prime }\left(4\right)=5$. Then ${lim}_{x\to 2}\frac{f\left(4\right)-f\left(x^2\right)}{x-2}$ equals

• A. $0$
• B. $5$
• C. $20$
• D. $-20$

Answer 1

Answer: (D)

$-20$

$\underset{x\to 2}{lim}\frac{f\left(4\right)-f\left(x^2\right)}{x-2}=\underset{x\to 2}{lim}\frac{f\left(4\right)-f\left(4\right)}{2-2}\frac{0}{0}$ form.

L hospital rule

$=\underset{x\to 2}{lim}\frac{0-f^{\prime }\left(x^2\right)\left(2x\right)}{1-0}=\frac{-4f^{\prime }\left(4\right)}{1}=-(4\times 5)$

$=-20$