second derivative as simple limit

If the real function $f$ is twice differentiable in a neighbourhood of  $x=x_0$,  then

$f^{\prime \prime }(x_0)=\underset{h\to 0}{lim}{\displaystyle \frac{f(x_0+2h)-2f(x_0+h)+f(x_0)}{h^2}}.$(1)

Proof.  The right hand side of the asserted equation is of the indeterminate form $\frac{0}{0}$.  Usinghttp://planetmath.org/node/2657l’Hôpital’s rule, we obtain

$\underset{h\to 0}{lim}{\displaystyle \frac{f(x_0+2h)-2f(x_0+h)+f(x_0)}{h^2}}$$\mathrm{\hspace{0.33em}}=\underset{h\to 0}{lim}{\displaystyle \frac{f^{\prime }(x_0+2h)\cdot 2-2f^{\prime }(x_0+h)}{2h}}-{\displaystyle \frac{f^{\prime }(x_0)}{h}}+{\displaystyle \frac{f^{\prime }(x_0)}{h}}$

$\mathrm{\hspace{0.33em}}=\mathrm{\hspace{0.33em}2}\underset{2h\to 0}{lim}{\displaystyle \frac{f^{\prime }(x_0+2h)-f^{\prime }(x_0)}{2h}}-\underset{h\to 0}{lim}{\displaystyle \frac{f^{\prime }(x_0+h)-f^{\prime }(x_0)}{h}}$

$\mathrm{\hspace{0.33em}}=\mathrm{\hspace{0.33em}2}{f}^{\prime \prime }(x_0)-f^{\prime \prime }(x_0)$

$\mathrm{\hspace{0.33em}}=f^{\prime \prime }(x_0).$