l'Hopitals rule - is my working correct? Is anyone able to help me with this question on l'Hopital's rule?

Use l'Hopital's rule to find the limit of the sequence $\{a_n\}_{n=1}^\infty$ with $n$-th term $\displaystyle a_n = \frac{\ln(n)}{e^n}$

Is my work correct? If not, how could I improve it?
$$\lim\limits_{n\to \infty} \frac{\ln(n)}{e^n} =
\lim\limits_{n\to \infty} \frac{1/n}{e^n} =
\lim\limits_{n\to \infty} \frac{1}{ne^n} =
\dfrac{1}{\infty} = 0$$
 A: It is correct, but a remark is necessary.

Since the functions $x\mapsto \log x$ and $x\mapsto e^x$ are defined for $x>0$ and, with l'Hôpital's theorem,
  $$
\lim_{x\to\infty}\frac{\log x}{e^x}\overset{(\mathrm{H})}{=}
\lim_{x\to\infty}\frac{1/x}{e^x}=\lim_{x\to\infty}\frac{1}{xe^x}=0
$$
  we can conclude that, whenever we have a sequence $a_n$ such that $\lim_{n\to\infty}a_n=\infty$,
  $$
\lim_{n\to\infty}\frac{\log a_n}{e^{a_n}}=0.
$$
  The particular case follows with $a_n=n$

There are cases where a sequence $b_n$ can be written as $b_n=f(n)$, where $f$ is defined on (a subset of) the real numbers, where
$$
\lim_{n\to\infty}b_n
$$
exists, but
$$
\lim_{x\to\infty}f(x)
$$
doesn't. A simple example is with $f(x)=\sin(2\pi x)$; the sequence $b_n=f(n)$ converges, because it's constant, but $f$ doesn't have a limit at infinity.
A: If $\lim\limits_{x\to \infty} f (x)=l$ and $x_n =f (n)$ then $\lim\limits_{x\to \infty}{x_n}=l$
The sequences are not differentiable.
Correct se l'Hopital's rule:
$$\lim\limits_{x\to \infty} \frac{\ln(x)}{e^x} =
\lim\limits_{x\to \infty} \frac{1/x}{e^x} =
\lim\limits_{x\to \infty} \frac{1}{xe^x} =
\dfrac{1}{\infty} = 0$$ 
For $x=n$ result $$\lim\limits_{n\to \infty} \frac{\ln(n)}{e^n} =0$$
