Question

If $f$ is differentiable in $(0,6)$ and $f^{\prime }\left(4\right)=5$, then $\underset{x\to 2}{lim}\frac{f^{\prime }\left(4\right)-f\left(x^2\right)}{2-x}=$

• A. $5$
• B. $\frac{5}{4}$
• C. $20$
• D. $10$

Answer 1

Answer: (C)

$20$

$\underset{x\to 2}{lim}\frac{f^{{}^{\prime }}\left(4\right)-f\left(x^2\right)}{2-x}$

Since $f\left(x\right)$ is differentiable in (0,6)

$\therefore \underset{x\to 2}{lim}{f}^{{}^{\prime }}\left(y\right)$ exists.

where $f\left(y\right)=\frac{f^{{}^{\prime }}\left(4\right)-f\left(x^2\right)}{2-x}$

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

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

$=20\left(f^{{}^{\prime }}\right(4)=5)$