calculating limit exponential function

Question

$$\lim_{x \rightarrow \infty} \frac{e^x+e^{-x}}{e^x-e^{-x}}$$

It was required to first try with l'hospital, show that it does not work, and then find another way. But it doesn't even fulfill the rules to apply l'Hospital? How can I even try it?

what I have done: $$\lim_{x \rightarrow \infty} \frac{e^x+e^{-x}}{e^x-e^{-x}}= \lim_{x \rightarrow \infty} \frac{e^{-x}(e^{2x}+1)}{e^{-x}(e^{2x}-1)}=\lim_{x \rightarrow \infty} \frac{(e^{2x}-1)+2}{(e^{2x}-1)} = 1+\lim_{x \rightarrow \infty} \frac{2}{(e^{2x}-1)} = 1 $$

since $\lim_{x \rightarrow \infty} \frac{2}{(e^{2x}-1)} $ = 0. Could someone check whether my approach is correct? and tell me how I could nicely argue that l'Hospital will not work here?

Answer 1

Your approach is just fine.

If you try a direct approach through L'Hopital's Rule, you get\begin{align}\lim_{x\to\infty}\frac{e^x+e^{-x}}{e^x-e^{-x}}&=\lim_{x\to\infty}\frac{e^x-e^{-x}}{e^x+e^{-x}}\\&=\lim_{x\to\infty}\frac{e^x+e^{-x}}{e^x-e^{-x}}\\&=\lim_{x\to\infty}\frac{e^x-e^{-x}}{e^x+e^{-x}}\\&=\vdots\end{align}and so on, until the end of time.

Answer 2

The dominant terms up and down when $x$ is very large are both $e^x$ which cancel to give the limit as 1.

Answer 3

$\lim \limits _ {x \to \infty} \frac{e^x+e^{-x}}{e^x-e^{-x}}$ take $e^x$ outside both in numerator and denominator, then we get the answer 1 immediately.