Answer
\[\lim_{h\rightarrow 0}\frac{f(x+h)-2f(x)+f(x-h)}{h^2}=f''(x)\]
Work Step by Step
Given that :- $f''$ is continuous
Let \[l=\lim_{h\rightarrow 0}\frac{f(x+h)-2f(x)+f(x-h)}{h^2}\;\;\;...(1)\]
Since $f''$ is continuous so (1) is $\frac{0}{0}$ form
Using L' Hopital's rule
\[l=\lim_{h\rightarrow 0}\frac{\{f(x+h)-2f(x)+f(x-h)\}'}{(h^2)'}\]
\[l=\lim_{h\rightarrow 0}\frac{f'(x+h)-0-f'(x-h)}{2h}\;\;\;...(2)\]
Since $f''$ is continuous so (2) is $\frac{0}{0}$ form
Using L' Hopital's rule
\[l=\lim_{h\rightarrow 0}\frac{\{f'(x+h)-f'(x-h)\}'}{(2h)'}\]
\[l=\lim_{h\rightarrow 0}\frac{f''(x+h)+f''(x-h)}{2}\]
Since $f''$ is continuous
\[l=\frac{f''(x+0)+f''(x-0)}{2}\]
\[\Rightarrow l=f''(x)\]
Hence, \[\lim_{h\rightarrow 0}\frac{f(x+h)-2f(x)+f(x-h)}{h^2}=f''(x)\]
