# L'Hopital's Rule, Factorials, and Derivatives

I have the following limit $\displaystyle\lim_{n\to\infty}\frac{e^n}{n!}$. Now if I try to solve this using this using L'hopital's rule, I won't be able to since I can't take the derivative of $n!$.

My question is why can't I take the derivative? Generally speaking, why can't I take the direct derivative of a factorial? I've seen other questions, but I just want a more simple answer (for someone at the Calculus 1 or 2 level).

Thank you in advance

• Perhaps this might help: math.stackexchange.com/questions/300526/… – Sherlock Holmes Oct 23 '14 at 2:08
• @SherlockHolmes Thank you, I saw that question as I was searching. I read about the Gamma function, but I don't fully understand. I think it's because I don't fully understand how factorials and derivatives work (or don't work) – Alti Oct 23 '14 at 2:13
• The comment above is about as good as you can hope for. Derivatives are a tool from calculus, for a derivative to exist the function must (at least) be continuous (over $\mathbb{R}$); because the factorial function is define over $\mathbb{N}$ we need a sensible way to define the notion of a derivative via a good extension to $\mathbb{R}$, the gamma function is just that. – Squirtle Oct 23 '14 at 2:15
• @Squirtle So the factorial function as we know it (defined over ℕ) is not continuous over ℝ? – Alti Oct 23 '14 at 2:29
• Of course it is not continuous over $\mathbb{R}$, because its not defined for non-integer values. Often physicists (and some mathematicians) will talk about defining things like $(\frac{1}{2})!$ but in fact this is abuse of notation and what they are really talking about is the gamma function as it acts on rationals (in this case $0.5$). – Squirtle Oct 23 '14 at 2:32

A derivative of the factorial function exists if you can define factorials of non-integers is a smooth way, and that can be done by using the fact that $n!=\int_0^\infty x^n e^{-x}\,dx$. But actually writing down a good expression for the derivative is another matter.
However, the limit is easy to show to be $0$. Think of what happens when $n$, on its way up to $\infty$, goes from $1000$ to $1001$, and observe that the pattern continues: The numerator gets multiplied by $e$, making it less than $3$ times as big, but the denominator gets multiplied by $1001$, so the whole thing gets multiplied by something smaller than $3/1001$. And at the next step, from $1001$ to $1002$, and all later steps, it gets multiplied by something even smaller. And this keeps happening over and over every time $n$ increases by $1$.
So the fraction must approach $0$.