Question

If $f^{\prime }\left(c\right)$ exists and is non-zero then ${lim}_{h\to 0}\frac{f(c+h)+f(c-h)-2f\left(c\right)}{h}$ is equal to

• A. $f^{\prime }\left(c\right)$
• B. $0$
• C. $2f^{\prime }\left(c\right)$
• D. none of these

Answer 1

The correct option is B $0$

Given f'(c) exists i..e.
$\underset{h\to 0}{lim}\frac{f(c+h)-f\left(c\right)}{h}=\underset{h\to 0}{lim}\frac{f(c-h)-f\left(c\right)}{-h}$
Now, $\underset{h\to 0}{lim}\frac{f(c+h)+f(c-h)-2f\left(c\right)}{h}$
$=\underset{h\to 0}{lim}\frac{f(c+h)-f\left(c\right)}{h}+\frac{f(c-h)-f\left(c\right)}{h}$
$=f^{\prime }\left(c\right)-\underset{h\to 0}{lim}\frac{f(c-h)-f\left(c\right)}{-h}$
$=f^{\prime }\left(c\right)-f^{\prime }\left(c\right)=0$