How to deal with an indefinite L'Hôpital operation Today in our AP Calculus class, we learned what is called L'Hôpital's rule for finding the limit of indefinite limits $\infty/\infty$ or $0/0$. The operation works by continuing to take the derivative of the limit until the answer is not indeterminate. How would one "identify ahead of time" and if possible, solve for a limit that would be continuously deriving without achieving an indeterminate answer?
For example something along the lines of $\lim_{x \to \infty} \dfrac{e^{x}}{e^{x}}$
Wouldn't this example be indefinitely deriving to achieve the answer?
 A: The answer by RRL gives very good examples where L'Hospital's Rule fails (or is not useful) so I will not dwell upon them. I would rather like to focus on the technique of applying L'Hospital's Rule to evaluate certain limits.
Before we start to apply this rule, it is important to know the conditions under which it is applicable:


*

*If "$f(x) \to 0, g(x) \to 0$" or "$g(x) \to \infty$" or "$g(x) \to -\infty$" then one can think of applying the rule to calculate limit of ration $f(x)/g(x)$.

*However the rule will be useful only when in addition to the first condition above the limit of the ratio $f'(x)/g'(x)$ also exists.


One hopes that the evaluation of the limit of $f'(x)/g'(x)$ would be simpler than evaluating the limit of $f(x)/g(x)$. Note however that using L'Hospital's rule successively is almost never a good idea, because repeated differentiation can lead to complicated functions. One expects that after applying the rule, the ratio $f'(x)/g'(x)$ can be simplified (via algebraic manipulation) so as to make use of standard limits and thereby evaluate its limit.
A: You run into this problem if $(f^2)' = (g^2)'$ and
$$\frac{f'(x)}{g'(x)} = \frac{g(x)}{f(x)} .$$
For example, try applying L'Hospital's rule to
$$\lim_{x \to \infty} \frac{x}{\sqrt{x^2+1}}.$$
In this case, application of the rule is inconclusive, but by other means you can show this limit equals $1$.
Another example is 
$$\lim_{x \to \infty} \frac{e^x+e^{-x}}{e^x - e^{-x}}.$$
Here we have
$$\lim_{x \to \infty} \frac{e^x+e^{-x}}{e^x - e^{-x}}=\lim_{x \to \infty} \frac{1+e^{-2x}}{1 - e^{-2x}}=1.$$
Another type of problem occurs when the limit of the ratio of derivatives does not exist.  Consider, for example $f(x) = x^2 \sin(1/x)$ and $g(x) = x.$
Then
$$\frac{f'(x)}{g'(x)}= 2x \sin(1/x) - \cos(1/x),$$
and the limit as $x \to 0$ does not exist.
However it is easy to show that
$$\lim_{x \to 0} \frac{f(x)}{g(x)} = \lim_{x \to 0} x \sin(1/x) = 0.$$
A: If you factor out $e^x$, you get $\lim_{x \to ∞} 1 = 1$
