$\lim_{x \to 0} \frac{\log \left(\cosh\left(x^2-xc\right) \right)}{x^2}=\frac{c^2}{2}$ without L'Hospital's rule How to show that without using L'Hospital's rule
\begin{align}
\lim_{x \to 0}  \frac{\log \left(\cosh\left(x^2-xc\right) \right)}{x^2}=\frac{c^2}{2}
\end{align}
I was able to show the upper bound by using the bound $\cosh(x) \le e^{x^2/2}$
\begin{align}
\lim_{x \to 0} \frac{\log \left(\cosh\left(x^2-xc\right) \right)}{x^2} \le \lim_{x \to 0} \frac{\left(x^2-xc\right)^2}{2x^2}=\frac{c^2}{2}
\end{align}
My question: How finish this argument. 

We can proceed as follows
\begin{align} 
L&=\lim_{x\to 0}\frac{\log\cosh(x^{2}-cx)}{x^{2}}\notag\\
&=\lim_{x\to 0}\frac{\log\cosh(x^{2}-cx)}{\cosh(x^{2}-cx)-1}\cdot\frac{\cosh(x^{2}-cx)-1}{x^{2}}\notag\\
&=\lim_{x\to 0}\frac{2\sinh^{2}((x^{2}-cx)/2)}{x^{2}}\notag\\
&=\lim_{x\to 0}2\cdot\frac{\sinh^{2}((x^{2}-cx)/2)}{(x^{2}-cx)^{2}/4}\cdot\frac{x^{2}(x-c)^{2}}{4x^{2}}\notag\\
&=2\cdot 1\cdot\frac{c^{2}}{4}\notag\\
&=\frac{c^{2}}{2}\notag
\end{align} 
