$\lim_{x \to 0} \frac{\log \left(\cosh\left(x^2-xc\right) \right)}{x^2}=\frac{c^2}{2}$ 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. 
 A: 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!
A: 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
$$
\color{blue}{\frac{\ln(1+\sinh^2a)}{2x^2}}\ge\frac{\ln(1+a^2)}{2x^2}=\frac12\cdot\frac{\ln(1+a^2)}{a^2}\cdot\frac{a^2}{x^2}\to \frac{c^2}{2}.
$$
A: Note that below we have used well-known limits $$\lim _{ x\rightarrow 0 }{ { \left( 1+x \right)  }^{ \frac { 1 }{ x }  } } =e\\ \lim _{ x\rightarrow 0 }{ \frac { { e }^{ x }-1 }{ x }  } =1$$

$$\lim _{ x\to 0 } \frac { \log  \left( \cosh  \left( x^{ 2 }-xc \right)  \right)  }{ x^{ 2 } } =\lim _{ x\to 0 } \frac { \log  \left( \frac { { e }^{ 2x^{ 2 }-2xc }+1 }{ 2{ e }^{ x^{ 2 }-xc } }  \right)  }{ x^{ 2 } } =\\ =\lim _{ x\to 0 }{ \log  \left( \frac { { e }^{ 2x^{ 2 }-2xc }+1 }{ 2{ e }^{ x^{ 2 }-xc } }  \right)  } ^{ \frac { 1 }{ { x }^{ 2 } }  }=\lim _{ x\to 0 }{ \log  \left( 1+\frac { { 1+e }^{ 2x^{ 2 }-2xc }-2{ e }^{ x^{ 2 }-xc } }{ 2{ e }^{ x^{ 2 }-xc } }  \right)  } ^{ \frac { 1 }{ { x }^{ 2 } }  }=\\ =\log { { e }^{ \lim _{ x\rightarrow 0 }{ \frac { { 1+e }^{ 2x^{ 2 }-2xc }-2{ e }^{ x^{ 2 }-xc } }{ 2{ e }^{ x^{ 2 }-xc } } \cdot \frac { 1 }{ { x }^{ 2 } }  }  } } =\lim _{ x\rightarrow 0 }{ \frac { { 1+e }^{ 2x^{ 2 }-2xc }-2{ e }^{ x^{ 2 }-xc } }{ 2{ e }^{ x^{ 2 }-xc } } \cdot \frac { 1 }{ { x }^{ 2 } }  } =\lim _{ x\rightarrow 0 }{ \frac { { \left( { e }^{ x^{ 2 }-xc }-1 \right)  }^{ 2 } }{ { \left( x^{ 2 }-xc \right)  }^{ 2 } } \cdot \frac { { \left( x^{ 2 }-xc \right)  }^{ 2 } }{ { 2{ e }^{ x^{ 2 }-xc }x }^{ 2 } }  } =\\ =\lim _{ x\rightarrow 0 }{ \frac { { x }^{ 4 }-2c{ x }^{ 3 }+{ x }^{ 2 }{ c }^{ 2 } }{ { 2{ e }^{ x^{ 2 }-xc }x }^{ 2 } }  } =\lim _{ x\rightarrow 0 }{ \left( \frac { { x }^{ 2 } }{ 2{ e }^{ x^{ 2 }-xc } } -\frac { cx }{ { e }^{ x^{ 2 }-xc } } +\frac { { c }^{ 2 } }{ 2{ e }^{ x^{ 2 }-xc } }  \right) = }\color{blue} {\frac { { c }^{ 2 } }{ 2 }} \\  $$

A: 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} 
A: Notice first that
$${\log(\cosh(x^2-xc))\over x^2}={\log(\cosh(x^2-xc))\over(x^2-xc)^2}(x-c)^2$$
Since $(x-c)^2\to c^2$ as $x\to0$, it suffices to show that
$$\lim_{x\to0}{\log(\cosh(x^2-xc))\over(x^2-xc)^2}=\lim_{u\to0}{\log(\cosh u)\over u^2}={1\over2}$$
But
$${2\log(\cosh u)\over u^2}={\log(\cosh^2u)\over u^2}={\log(1+\sinh^2u)\over u^2}={\log(1+\sinh^2u)\over\sinh^2u}\left(\sinh u\over u\right)^2$$
so it suffices to show that
$$\lim_{u\to0}{\log(1+\sinh^2u)\over\sinh^2u}=\lim_{v\to0}{\log(1+v)\over v}=1\qquad\text{and}\qquad\lim_{u\to0}{\sinh u\over u}=1$$
But these final two limits are just the derivatives of the functions $\log(1+x)$ and $\sinh x$ evaluated at $0$.  We are not using L'Hopital's rule in obtaining those limits.
