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

$\underset{h\to 0}{lim}{\displaystyle \frac{f({x}_{0}+2h)2f({x}_{0}+h)+f({x}_{0})}{{h}^{2}}}$ 
$\mathrm{\hspace{0.33em}}=\underset{h\to 0}{lim}{\displaystyle \frac{{f}^{\prime}({x}_{0}+2h)\cdot 22{f}^{\prime}({x}_{0}+h)}{2h}}{\displaystyle \frac{{f}^{\prime}({x}_{0})}{h}}+{\displaystyle \frac{{f}^{\prime}({x}_{0})}{h}}$ 



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



$\mathrm{\hspace{0.33em}}=\mathrm{\hspace{0.33em}2}{f}^{\prime \prime}({x}_{0}){f}^{\prime \prime}({x}_{0})$ 



$\mathrm{\hspace{0.33em}}={f}^{\prime \prime}({x}_{0}).$ 
