Evaluation of limit $\lim_{x\to0}\frac{e^x + e^{-x} -2}{x^2}$

Question

I've been given to evaluate the limit, $$\lim_{x\to0}\frac{e^x + e^{-x} -2}{x^2}$$

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Here's my approach to the problem: $$\lim_{x\to0}\frac{e^x + e^{-x} -2}{x^2}$$ $$\implies\lim_{x\to0}\frac{(e^x -1)+(e^{-x} -1)}{x^2}$$ $$\implies\lim_{x\to0}\frac{(e^x -1)}{x^2}+\lim_{x\to0}\frac{(e^{-x} -1)}{x^2}$$ $$\implies\lim_{x\to0}\frac{(e^x -1)}{x}\cdot\frac{1}x+\lim_{x\to0}\frac{(e^{-x} -1)}{x}\cdot \frac1x$$ $$\implies\lim_{x\to0}\frac{1}x-\lim_{x\to0}\frac{(e^{-x} -1)}{-x}\cdot \frac1x$$ $$\implies\lim_{x\to0}\frac{1}x-\frac1x$$ $$\implies0$$

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But solving using another method, $$\lim_{x\to0}\frac{e^x + e^{-x} -2}{x^2}$$ $$\implies\lim_{x\to0}\frac{e^x + \dfrac1{e^x} -2}{x^2}$$ $$\implies\lim_{x\to0}\frac{e^{2x}-2e^x + 1}{e^x\cdot x^2}$$ $$\implies\lim_{x\to0}\frac{(e^{x}-1)^2}{x^2}\cdot\frac1{e^x}$$ $$\implies\lim_{x\to0}\frac{1}{e^x}$$ $$\implies 1$$

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Why are these two methods giving two different results? I know that the first method is incorrect, but don't actually know what's the mistake. Can anyone help me?

Answer 1

While it is true that that you can break up limits $\lim_\limits{x\to a} f(x) g(x) = \lim_\limits{x\to a} f(x) \lim_\limits{x\to a} g(x)$ and $\lim_\limits{x\to a} f(x) + g(x) = \lim_\limits{x\to a} f(x)+\lim_\limits{x\to a} g(x)$
You can only do it if $\lim_\limits{x\to a} f(x)$ and $\lim_\limits{x\to a} g(x)$ exist and are finite.

Where you have: $\lim_\limits{x\to 0} \frac {e^x - 1}{x}\frac {1}{x}$ you cannot evaluate the first factor as the limit does not exist for the second factor.

This line is not correct either:

$\lim_\limits{x\to 0} \frac {(e^x - 1) + (e^{-x}-1)}{x^2}= \lim_\limits{x\to 0} \frac {(e^x - 1)}{x^2}+ \lim_\limits{x\to 0}\frac {(e^{-x} - 1)}{x^2} $

So, what can you do?

Apply L'Hopital's rule: Since the denominator has an $x^2$ term to it, you are going to need to do it twice.

$\lim_\limits{x\to 0} \frac {e^x + e^{-x}-2}{x^2} = \lim_\limits{x\to 0} \frac {e^x + e^{-x}}{2} = 1$

Restate the exponentials as Taylor series: This is very much like applying L'Hopital's rule, at least it is from a theoretical point of view, but once you get the hang of it, I think, it is much easier to do on complicated limits.

$\lim_\limits{x\to 0} \frac {e^x + e^{-x}-2}{x^2} = \lim_\limits{x\to 0} \frac {(1 + x + \frac {x^2}{2}) + (1 - x + \frac {x^2}{2}) - 2 + O(x^3)}{x^2} = 1$

Get creative:

When I see $e^x + e^{-x}$ I am thinking $2\cosh x.$ And once in this frame of mind what can I do to get rid of the $-2$ hanging at the end....$e^{x} + e^{-x} - 2 = 2\sinh^2 \frac x2$

$\lim_\limits{x\to 0} \frac {2\sinh^2 \frac {x}{2}}{x^2} = 1$

Answer 2

$$\lim_{x\to0}\frac{1}{x}$$ is not defined. So you can't make arithmetic operations with it like sums.

Answer 3

HINT

I would start with noticing that

\begin{align*} \lim_{x\to 0}\frac{e^{x} + e^{-x} - 2}{x^{2}} & = \lim_{x\to 0}\frac{(e^{x/2} - e^{-x/2})^{2}}{x^{2}}\\\\ & = \lim_{x\to 0}\left(\frac{e^{x/2} - e^{-x/2}}{x}\right)^{2}\\\\ & = \lim_{x\to 0}\left[\frac{(e^{x/2} -1)}{x} - \frac{(e^{-x/2} - 1)}{x}\right]^{2} \end{align*}

Can you recognize the last expressions as derivatives?

EDIT

Based on the request on the comments, here is an improvement of the answer: \begin{align*} \lim_{x\to 0}\frac{e^{x/2} - 1}{x} = \lim_{x\to 0}\frac{e^{x/2} - e^{0/2}}{x - 0} = (e^{x/2})'\Bigr|_{\substack{x=0}} = \frac{1}{2} \end{align*}

Similarly, we do also have that \begin{align*} \lim_{x\to 0}\frac{e^{-x/2} - 1}{x} = \lim_{x\to 0}\frac{e^{-x/2} - e^{-0/2}}{x - 0} = (e^{-x/2})'\Bigr|_{\substack{x=0}} = -\frac{1}{2} \end{align*}

Can you take it from here?