# When L'Hopital's rule fails (again)

## I just realized that there may be a case where L'Hopital's rule fails, specifically $$\lim_{x \to \infty} \frac{e^x}{e^x}$$ which evaluates to an indeterminate form, specifically $\frac{\infty}{\infty}$. Sure, we can cancel the $e^x$s, but when we use L'Hopital's, we get $$\lim_{x \to \infty} \frac{(e^x)^\prime}{(e^x)^\prime}$$ Since the derivative of $e^x$ is $e^x$, we have $$\lim_{x \to \infty} \frac{e^x}{e^x}$$ which is our original limit. Therefore, L'Hopital's fails to work in this example. Question: Does L'Hopital's rule actually fail in this example, or am I understanding it wrong? I just realized that there may be a case where L'Hopital's rule fails, specifically

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

which evaluates to an indeterminate form, specifically $$\frac{\infty}{\infty}$$. Sure, we can cancel the $$e^x$$s, but when we use L'Hopital's, we get

$$\lim_{x \to \infty} \frac{(e^x)^\prime}{(e^x)^\prime}$$

Since the derivative of $$e^x$$ is $$e^x$$, we have

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

which is our original limit. Therefore, L'Hopital's fails to work in this example.

Question: Does L'Hopital's rule actually fail in this example, or am I understanding it wrong?