# How can I explain $\lim_{x \to \infty} \frac{e^x+e^{-x}}{e^x-e^{-x}}$ using L'Hôpital's Rule?

I have given a problem in limits to my students:

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

Most of the students used direct substitution and identified that it is an indeterminate form $\frac{\infty}{\infty}$ and tried to apply l'Hôpital's Rule. They observed that it will fail, since the numerator and denominator get swapped, which again leads to the same indeterminate form.

So I gave the solution as $$\lim_{x \to \infty}\frac{e^x+e^{-x}}{e^x-e^{-x}}=\lim_{x \to \infty}\frac{1+e^{-2x}}{1-e^{-2x}}=1.$$

But students' query was: is there any way to solve this limit using l'Hôpital's rule. Can any one give some comments on this?

• If you let $L$ denote the limit in question, then an application of L'Hopital's Rule shows that if $L$ exists, then $L = 1/L$. Since $L \geq 0$ is clear, it follows that $L = 1$. But this argument assumes the existence of the limit in the first place. Jan 18 '16 at 5:59
• the question is why on earth would you want to use L'Hopitals on this?
– abel
Jan 19 '16 at 0:43
• Mr abel you cant question like that to students, hope you got it Jan 19 '16 at 3:32
• Actually you can get students to think about the idea that L'Hopital's rule is not always the best plan, even if it is an indeterminate form. This is the perfect opportunity to do so! Perhaps say, "I suppose there might be a way, but since this approach works, maybe it's better not to do L'Hopital's rule and use this. Now here's some pointers to noticing how to do a limit..." Jan 19 '16 at 4:25
• I notice that the same problem occurs with $\displaystyle\lim\limits_{x \to \infty} \frac{e^x}{e^x},$ except in this case the algebraic manipulation needed to get yourself unstuck is simpler. Maybe you can also use this as an example of what DavidButlerUofA suggests in his comment just above my comment. Jan 20 '16 at 16:19

As noted, l'Hopital directly fails. So you have to transform somehow first. For example, $$\frac{e^x+e^{-x}}{e^x-e^{-x}} = \frac{e^{2x}+1}{e^{2x}-1}$$ then l'Hopital works. But of course the same method $$\frac{e^x+e^{-x}}{e^x-e^{-x}} = \frac{1+e^{-2x}}{1-e^{-2x}} \to \frac{1+0}{1-0}$$ solves it without l'Hopital at all.