How do I factor equations involving $e^x$?

I was reviewing some of my notes from Calculus 1 so that I can prepare for Calculus 2 this fall, and I ran into one problem where I don't understand how the factoring works.

$$\lim_{x\to\infty} \frac{e^{1/x}+e^{-1/x}}{e^{1/x}-e^{-1/x}}$$

Then substituting... $$u = \frac {1}{x}$$

I know what the answer to this question is, but this step in the solution manual confused me:

$$\lim_{x\to\infty} \frac{e^u(1+e^{-2u})}{e^u(1-e^{-2u})}$$

My brain is still getting prepared to get back into doing math regularly after the summer vacation, but I'm hitting a serious wall trying to understand how this expression can be factored this way.

Any help in understanding this is greatly appreciated!

• First there isn't any $n$ in the expression after the lim symbol. But it also looks like you miscopied something from the solution manual. I think the top factor should be $(1+e^{-2u})$. That way when "multiplied out" the top becomes $e^u+e^{-u}$ as it should to match. – coffeemath Aug 10 '13 at 3:15
• @coffeemath good catch! Both typos. I will fix now – hax0r_n_code Aug 10 '13 at 3:17

HINT: If $u=1/x$ then $e^u(1+e^{-2u})=e^u+e^{-u}=e^\frac1x+e^{-\frac1x}$ which now matches the top of your expression. Similarly $e^u(1-e^{2u})$ matches the bottom of the expression. In the $u$ version the two $e^u$ factors cancel, and the expression becomes $$\frac{1+e^{-2/x}}{1-e^{-2/x}}.$$ This looks like, as $x \to \infty$, then $-2/x \to 0$ so that the exponential terms go to 1 and it diverges.
• Actually, it does not since $\lim_{x \to \infty} (1+e^{-\frac{2}{x}}) = 2 \neq 0$ – Doctor Dan Aug 10 '13 at 3:28