Question

If $f$ is defined and continuous on $[3,5]$ and $f$ is differentiable at $x=4$ and $f^{\prime }\left(4\right)=6$ then the value of $\underset{x\to 0}{lim}\frac{f(4+x)-f(4-x)}{4x}$ is equal to

• A. $0$
• B. $2$
• C. $3$
• D. $4$
• E. $6$

Answer 1

Answer: (C)

$3$

We have,
$\underset{x\to 0}{lim}\frac{f(4+x)-f(4-x)}{4x}$
$\underset{x\to 0}{lim}\frac{f^{\prime }(4+x)+f^{\prime }(4-x)}{4}$ (by L'-hospital rule)
$=\frac{1}{4}\left\{f^{\prime }\right(4)+f^{\prime }(4\left)\right\}=\frac{2f^{\prime }\left(4\right)}{4}$
$=\frac{2\times 6}{4}(\because f^{\prime }(4)=6$, given)
$=\frac{12}{4}=3$