Evaluating $\lim_{x\to 0}\frac{e^x+e^{-x}-2}{1-\cos x}$ without using L'Hospital's Rule Compute without using L'Hospital's Rule $$\lim_{x\to 0}\dfrac{e^x+e^{-x}-2}{1-\cos x}.$$


I thought of simplifying the limit as shown below.
\begin{align}
\lim_{x\to 0} \dfrac{e^x+e^{-x}-2}{1-\cos x}&amp;=\lim_{x\to 0}\dfrac{2\sinh^2 x(1+\cos x)}{\sin^2 x(1+\cosh x)}\\
&amp;=\space2\lim_{x\to 0}\dfrac{\sinh^2 x}{\sin^2 x}\cdot \lim_{x\to 0}\dfrac{1+\cos x}{1+\cosh x}
\end{align} But I'm a little stuck as to how to show that
$$\lim_{x\to 0}\dfrac{\sinh^2 x}{\sin^2 x}=1$$
so that the original limit is $2$. I can't just use a simple substitution.

Any help would be nice.



Edit: No use of L'Hospital's Rule is allowed. If I were allowed to use that, I would simply do the following:

$$\lim\limits_{x\to 0}\dfrac{e^x+e^{-x}-2}{1-\cos x} = \lim\limits_{x\to 0}\dfrac{e^x-e^{-x}}{\sin x}=\lim\limits_{x\to 0}\dfrac{e^x+e^{-x}}{\cos x}=2$$ 
That is literally trivial.
Divide numerator and denominator by $x^2$

And use the limit $\lim_{x \to 0} \frac{sinh (x)}{x}= 1$ and $\lim_{x \to 0} \frac{sin (x)}{x}=1$

Here you can find the proof of limit of sinh 
Prove that $\lim\limits_{x \to 0} \sinh(x)/x =1$.
Directly with the original limit and L'Hospital's Rule:

$$\lim_{x\to0}\frac{e^x+e^{-x}-2}{1-\cos x}\stackrel{L'H}=\lim_{x\to0}\frac{e^x-e^{-x}}{\sin x}\stackrel{L'H}=\lim_{x\to0}\frac{e^x+e^{-x}}{\cos x}=2$$