$\lim_{x \to 0} \frac{\log \left(\cosh\left(x^2-xc\right) \right)}{x^2}=\frac{c^2}{2}$ 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. 

Denote $a=x^2-cx$ for simplicity. Then
\begin{align}
\frac{\ln(\cosh a )}{x^2}&=\frac{\ln(\cosh^2a)}{2x^2}=\color{blue}{\frac{\ln(1+\sinh^2a)}{2x^2}}=\frac12\cdot\frac{\ln(1+\sinh^2a)}{\sinh^2a}\cdot\left(\frac{\sinh a}{x}\right)^2=\\
&=\frac12\cdot\frac{\ln(1+\sinh^2a)}{\sinh^2a}\cdot\left(\frac12\cdot\left[\frac{e^a-1}{a}+\frac{e^{-a}-1}{-a}\right]\cdot\frac{a}{x}\right)^2\to\frac{c^2}{2}.
\end{align}

P.S. If you want a lower bound then using the inequality $e^a-1\ge a$ we can estimate
$$
\sinh^2 a=\left(\frac12\left[\frac{e^a-1}{a}+\frac{e^{-a}-1}{-a}\right]\right)^2a^2\ge a^2
$$
and continue from the blue expression above as
$$
\color{blue}{\frac{\ln(1+\sinh^2a)}{2x^2}}\ge\frac{\ln(1+a^2)}{2x^2}=\frac12\cdot\frac{\ln(1+a^2)}{a^2}\cdot\frac{a^2}{x^2}\to \frac{c^2}{2}.
$$
