Other approaches to evaluate $\lim_{h\to 0} \frac{4^{x+h}+4^{x-h}-4^{x+\frac12}}{h^2}$ 
$$\lim_{h\to 0} \frac{4^{x+h}+4^{x-h}-4^{x+\frac12}}{h^2}=?$$

I evaluated the limit by using the Hopital rule,$$\lim_{h\to 0} \frac{4^{x+h}+4^{x-h}-4^{x+\frac12}}{h^2}=4^x\lim_{h\to0}\frac{4^h+4^{-h}-2}{h^2}=4^x\lim_{h\to0}\frac{\ln(4)(4^h-4^{-h})}{2h}=4^x\lim_{h\to0}\frac{(\ln4)^2(4^h+4^{-h})}{2}=4^x\times4(\ln2)^2=4^{x+1}(\ln2)^2$$
I want to learn other ideas to solving this problem, so can you please evaluate the limit with other approaches?

A possible way is using the Taylor expansion
$$2\cosh t = e^t+e^{-t} = 2+t^2 + o(t^2)$$
Hence,
\begin{eqnarray*}\frac{4^{x+h}+4^{x-h}-4^{x+\frac12}}{h^2}
& = & 4^x\frac{e^{h\ln 4}+e^{-h\ln 4}-2}{h^2}\\
& = & 4^x\cdot \frac{2+h^2\ln^2 4 + o(h^2)-2}{h^2} \\
& = & 4^x\cdot\ln^2 4 + o(1) \\
& \stackrel{h\to 0}{\longrightarrow} & 4^x\cdot\ln^2 4
\end{eqnarray*}
