Question

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

• A. 0
• B. 5
• C. 20
• D. -20

Answer 1

Answer: (D)

-20

Given, $f^{\prime }\left(4\right)=5$

Now, $\underset{x\to 2}{lim}\frac{f\left(4\right)-f\left(x^2\right)}{x-2}$ $\left[\frac{0}{0}form\right]$

Apllying L-hospitals rule, we get,

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

$=\frac{-f^{\prime }\left(4\right)\cdot 2\times 2}{1}$

$=-\left(5\right)\times 4=-20$