# 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}&=\lim_{x\to 0}\dfrac{2\sinh^2 x(1+\cos x)}{\sin^2 x(1+\cosh x)}\\ &=\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.

• Have you noticed which function's square has its limit sought?
– J.G.
Oct 18, 2019 at 22:07
• Write $\dfrac{\sinh x}{\sin x}$ as $\dfrac{\sinh x}{x}\cdot\dfrac{x}{\sin x}$.
– Blue
Oct 18, 2019 at 22:17

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 $\lim\limits_{x \to 0} \sinh(x)/x =1$.

• If you know that then you can easily solve your problem Oct 18, 2019 at 22:36
• thanks the proof was interesting. i didn't think of using complex numbers but that's perfectly legal.
– user706791
Oct 18, 2019 at 22:44

Just to explain what my comment meant, we're trying to compute $$\lim_{x\to0}\frac12f^2(x)=\frac12\left(\lim_{x\to0}f(x)\right)^2$$ with$$f(x):=\frac{e^{x/2}-e^{-x/2}}{\sin\frac{x}{2}}=\frac{\frac{e^{x/2}-1}{x/2}-\frac{e^{-x/2}-1}{x/2}}{\frac{\sin\frac{x}{2}}{x/2}}.$$Since for any function $$g$$ we define $$g^\prime(0):=\lim_{x\to0}\frac{g(x)-g(0)}{x}=\lim_{x\to0}\frac{g(x/2)-g(0)}{x/2}$$ (this is not a use of the full-blown L'Hôpital's rule),$$\lim_{x\to0}f(x)=\frac{\lim_{x\to0}\frac{e^{x/2}-1}{x/2}-\lim_{x\to0}\frac{e^{-x/2}-1}{x/2}}{\lim_{x\to0}\frac{\sin\frac{x}{2}}{x/2}}=\frac{\left.\frac{de^x}{dx}\right|_0-\left.\frac{de^{-x}}{dx}\right|_0}{\left.\frac{d\sin x}{dx}\right|_0}=\frac{1-(-1)}{1}=2.$$Hence$$\lim_{x\to0}\frac12f^2(x)=\frac122^2=2.$$

Directly with the original limit and L'Hospital's Rule:

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

$$e^x+e^{-x}-2=\dfrac{(e^x-1)^2}{e^x}$$

$$\lim_{x\to0}\dfrac{e^x+e^{-x}-2}{1-\cos x}=\lim_{x\to0}\dfrac{(1+\cos x)}{e^x}\left(\lim_{x\to0}\dfrac{e^x-1}{\sin x}\right)^2$$

Finally $$\lim_{h\to0}\dfrac{e^h-1}h=1$$

and $$\lim_{u\to0}\dfrac{\sin u}u=1$$

• @fordjones, Please find the updated post Oct 19, 2019 at 0:42