What to do when l'Hopital's doesn't work I'm a first time Calc I student with a professor who loves using $e^x$ and logarithims in questions. So, loosely I know L'Hopital's rule states that when you have a limit that is indeterminate, you can differentiate the function to then solve the problem. But what do you do when no matter how much you differentiate, you just keep getting an indeterminate answer? For example, a problem like
$\lim _{x\to \infty }\frac{\left(e^x+e^{-x}\right)}{\left(e^x-e^{-x}\right)}$
When you apply L'Hopital's rule you just endlessly keep getting an indeterminate answer. With just my basic understanding of calculus, how would I go about solving a problem like that?
Thanks
 A: HINT: 
As for your problem, divide both numerator and denominator by $e^x$. You'll get your limit as $1$.
In mathematics, logic, representation and arrangement play an extremely vital role. So always check that you have arranged your expression properly. Else repeated applications of several powerful and helpful theorems might fail, not only in calculus but also in other mathematical topics as well.
A: why would you want to use l'hopitals on this? for $x$ large positive $e^{-x} = \frac 1{e^x}$ which is small. therefore $$\frac{e^x + e^{-x}}{e^x -e^{-x}} = \frac{e^x}{e^x}+\cdots \to 1 \text{ as } x \to \infty.$$
A: Factor out $e^x$: 
$\frac{\left(e^x+e^{-x}\right)}{\left(e^x-e^{-x}\right)}=\frac{1+e^{-2x}}{1-e^{-2x}}\rightarrow 1$ as $x\rightarrow \infty $.
In other situations, Taylor expansions and algebraic operations usually are sufficient. There are also many standard inequalities which can help with these problems.
A: $\frac{e^x+e^{-x}}{e^x-e^{-x}}
=\frac{e^x-e^{-x}+2e^{-x}}{e^x-e^{-x}}
=1+\frac{2e^{-x}}{e^x-e^{-x}}
=1+\frac{2}{e^{2x}-1}
\to 1
$
as
$x \to \infty
$.
