Why does L'Hôpital's rule work for sequences? Say, for the classic example, $\frac{\log(n)}{n}$, this sequence converges to zero, from applying L'Hôpital's rule.  Why does it work in the discrete setting, when the rule is about differentiable functions?
Is it because at infinity, it doesn't matter that we relabel the discrete variable, $n$, with a continuous variable, $x$, and instead look at the limit of  $\frac{\log(x)}{x}$?
But then what about the quotients of sequences that go to the indeterminate form $\frac{0}{0}$?  Why is it OK to use L'Hôpital's rule, as $n$ goes to zero?
I haven't found anything on Wikipedia or Wolfram about the discrete setting.
Thanks.
 A: Your explanation is not really precise, it does matter whether you use the discrete or continuous variable. However, there is a theorem in mathematical analysis that states that the following is equivalent:


*

*$\lim_{x \rightarrow c} f(x) = A$

*For every sequence $\{x_n\}$ such that $\forall n \in \mathbb{N} : x_n \in D(f), x_n \neq c$ and that $\lim_{n \rightarrow \infty} x_n = c$ it is true that
$\lim_{n \rightarrow \infty} f(x_n) = A$.


In simpler words, once you know the limit of a function in continuous variable, like $\lim_{x \rightarrow \infty} \frac{\log x}{x} = 0$, you also know the limit of any sequence you get by "picking out" points of this function's domain, in your case specifically you take the sequence $\{x_n\} = \{n\}$. Notice that the conditions of the second statement are met, since $\frac{\log x}{x}$ is defined for every $n$, $\lim_{n \rightarrow \infty} n = \infty$ and $n \neq \infty$ for every $n$ as well. 
This also solves the problem for the limits of indeterminate form "$\frac{0}{0}$", or any other. 
Hope that helps :),
Epsiloney
A: There IS a L'Hospital's rule for sequences called Stolz-Cesàro theorem. If you have an indeterminate form, then:
$$\lim\limits_{n\to\infty} \frac{s_n}{t_n}=\lim\limits_{n\to\infty} \frac{s_n-s_{n-1}}{t_n-t_{n-1}}$$
So for your example:
$$\lim\limits_{n\to\infty}\frac{\ln(n)}{n}=\lim\limits_{n\to\infty}\frac{\ln\left(\frac{n}{n-1}\right)}{n-n+1}=\lim\limits_{n\to\infty}\ln\left(\frac{n}{n-1}\right)=0$$
But that isn't your question. Your question is, why do people "differentiate"? Basically because the real case covers the discrete case. 
Recall the definition of limits for real and discrete cases.
Definition. A sequence, $s_n\colon \Bbb{N}\to \Bbb{R},$ converges to $L$ as $n\to\infty$, written $\lim\limits_{n\to\infty} s_n=L$ iff for all $\epsilon>0$ there is some $N$ such that for all $n\in \Bbb{N}$ with $n>N$, $|s_n-L|<\epsilon$.
Definition. A function, $f(x) \colon \Bbb{R}\to \Bbb{R}$ converges to $L$ as $x\to\infty$, written $\lim\limits_{x\to\infty} f(x)=L$ iff for all $\epsilon>0$ there is some $X$ such that for all $x\in \Bbb{R}$ with $x>X$, $|f(x)-L|<\epsilon$.
So if $f(x)$ is a real valued function that agrees with a sequence, $s_n$ on integer values, then $\lim\limits_{x\to\infty} f(x)=L$ implies $\lim\limits_{n\to\infty} s_n=L$.
A: Fact: If a function $f:\Bbb R\to\Bbb R$ is continuous, $x_n\to x$ then
$f(x_n)\to f(x)$.
A: You can see here Stolz–Cesàro theorem for the "l'Hôpital's rule" for sequences.
A: This uses the fact that if $\displaystyle\lim_{x\to\infty}f(x)=L, \text{  then  } \lim_{n\to\infty}f(n)=L$ since
$\displaystyle\lim_{x\to\infty}f(x)=L\iff$ for every $\epsilon>0,$ there is an M such that if $x\ge M,$ then $\big|f(x)-L\big|<\epsilon$ and
$\displaystyle\lim_{n\to\infty}f(n)=L\iff$ for every $\epsilon>0,$ there is an N such that if $n\ge N,$ then $\big|f(n)-L\big|<\epsilon$ $\;\;$(with $n\in\mathbb{N}$)
