Answer
See below.
Work Step by Step
Theorem 4. Reworded in plain language, it states that:
if we find a FUNCTION $f(x)$ such that, after a certain index $n_{0},$ all terms of the sequence are in fact function values $f(n)$ of the term's index, then the existence of a limit
$\displaystyle \lim_{x\rightarrow\infty}f(x)=L$
implies that
$\displaystyle \lim_{n\rightarrow\infty}a_{n}=L.$
$\displaystyle \lim_{x\rightarrow\infty}f(x)$ is the limit of a function
$\displaystyle \lim_{n\rightarrow\infty}a_{n}$ is the limit of a sequence, if they exist.
L'Hopital's rule applies to limits of FUNCTIONS, and we can use it to find the limit of a SEQUENCE.
Example: Define the general term of $\{a_{n}\}$ as
$a_{n}=e^{-n}\ln(n)$
Then, the function $f(x)=e^{-x}\ln x$ fits the requirements of the theorem.
We can rewrite $f(x)$ as $ f(x)=\displaystyle \frac{\ln x}{e^{x}}, \quad$ and searching for the limit
$\displaystyle \lim_{x\rightarrow\infty}f(x)$ =$\displaystyle \lim_{x\rightarrow\infty}\frac{\ln x}{e^{x}}\quad$
we have $\displaystyle \frac{\infty}{\infty}$
so we can apply L'Hopital's rule (differentiate up and down):
$=\displaystyle \lim_{x\rightarrow\infty}\frac{[\ln x]'}{[e^{x}]'}=\lim_{x\rightarrow\infty}\frac{1/x}{e^{x}}=\lim_{x\rightarrow\infty}\frac{1}{xe^{x}}=$
The numerator is constant, and the denominator is the product of two large numbers the ratio approaches 0
$=0$
Now, by Theorem 4, since $\displaystyle \lim_{x\rightarrow\infty}f(x)=0$, it follows that
$\displaystyle \lim_{n\rightarrow\infty}a_{n}=0$
