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

Note that below we have used well-known limits $$\lim _{ x\rightarrow 0 }{ { \left( 1+x \right)  }^{ \frac { 1 }{ x }  } } =e\\ \lim _{ x\rightarrow 0 }{ \frac { { e }^{ x }-1 }{ x }  } =1$$

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

