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

PRIMER:
In This Answer , I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the exponential and logarithm functions satisfies the inequalities 
$$1+x\le e^x\le \frac{1}{1-x} \tag 1$$
for $x<1$, and
$$\frac{x-1}{x}\le \log(x)\le x-1 \tag 2$$
for $x>0$,respectively.
It can also be shown that 
$$|\sinh(x)|\ge |x| \tag 3$$
for all $x$. 

METHODOLOGY $1$: DEVELOPING A USEFUL LOWER BOUND 

The OP has expressed a preference to finding a useful lower bound for the function of interest.  To that end, we proceed.

If we wish to find a lower bound then we can apply $(2)$ and $(3)$ and proceed to write
$$\begin{align}
\log(\cosh(x^2-cx)) &=\log\left(1-(1-\cosh(x^2-cx))\right)\\\\
&=\log\left(1+2\sinh^2\left(\frac{x^2-cx}{2}\right)\right)\\\\
&\ge \frac{2\sinh^2\left(\frac{x^2-cx}{2}\right)}{1+2\sinh^2\left(\frac{x^2-cx}{2}\right)}\\\\
&\ge \frac{2\left(\frac{x^2-cx}{2}\right)^2}{1+2\sinh^2\left(\frac{x^2-cx}{2}\right)}\tag 4
\end{align}$$
Then, dividing $(4)$ by $x^2$ we find that
$$\frac{\log(\cosh(x^2-cx))}{x^2}\ge \frac12 \frac{(x-c)^2}{1+2\sinh^2\left(\frac{x^2-cx}{2}\right)} \tag 5$$
The limit of the right-hand side of $(5)$ approaches $\frac{c^2}2$ as $x\to 0$

METHODOLOGY $2$: APPLYING TAYLOR'S THEOREM

The OP has stated that application of Taylor's is an acceptable way forward.  To that end, we proceed.

Note from Taylor's Theorem, 
$$\cosh(s)=1+\frac12s^2+O(s^4)\tag6$$
and 
$$\log(1+t)=t-\frac12t^2+O(t^4) \tag7$$
Using $(6)$ and $(7)$ yields
$$\begin{align}
\log(\cosh(x^2-cx))&=\log\left(1+\frac12(x^2-cx)^2+O((x^2-cx)^4)\right)\\\\
&=\frac12(x^2-cx)^2+O((x^2-cx)^4)\tag 8
\end{align}$$
Dividing $(8)$ by $x^2$ as taking the limit as $x\to 0$, we obtain the coveted limit
$$\lim_{x\to 0}\frac{\log(\cosh(x^2-cx))}{x^2}=\lim_{x\to 0}\frac{\frac12(x^2-cx)^2+O((x^2-cx)^4)}{x^2}=\frac{c^2}{2}$$
as was to be shown!
