# The limit $\lim\limits_{n\to\infty}\frac{x^n}{n!}$

Evaluate $$\lim_{n\to\infty}\frac{x^n}{n!}$$ where $$x\in\mathbb R$$

This is a common limit and has been asked and answered many times here on this site. However, I present another approach with Stolz-Cesaro($$x\notin[-1,1]$$. When $$x\in[-1,1]$$, the limit is trivially zero):

$$L=\lim_{n\to\infty}\frac{x^n}{n!}=\lim_{n\to\infty}\frac{x^{n+1}-x^n}{(n+1)!-n!}$$

$$L=\lim_{n\to\infty}\frac{x^n}{n!}\frac{(x-1)}{n}=L\frac{(x-1)}{n}$$

The above equation holds for all $$x$$ in domain only when $$L=0$$. Hence the limit is just $$0$$.

I have never seen Stolz-Cesaro being used this way to solve a limit, hence the lack of surety. It would be great to get this verified!

• That assumes $L$ exists. Mar 19 at 15:35
• For the calculation of that limit see here. Mar 19 at 15:36
• @ThomasAndrews Yes, Jose pointed it out in their answer. Is there a fairly simple way to show that the limit exists without actually computing it? Mar 19 at 15:37
• $(n+1)! - n! \ne n!$ Mar 19 at 19:11
• @DatBoi When you take the limit, you cannot keep $n$ in $(x-1)/n$ and only let it $\to \infty$ in the first factor.
– Gary
Mar 20 at 1:47

You have not proved that the limit exists, what Stolz-Cesaro criterion says is the following:

Let $$\{a_n\}$$ and $$\{b_n\}$$ be sequences, with $$\{b_n\}$$ monotone increasing and divergent. Then, if $$\lim_{n\rightarrow \infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=\lambda,\ \lambda \in \mathbb{R} \Rightarrow \lim_{n\rightarrow \infty}\frac{a_n}{b_n}=\lambda$$ In order to use it you would have to calculate the limit after the second $$=$$ and then the criterion would guarantee that the limit you are actually trying to calculate is equal.

The most immediate way to see why it is zero is to notice that you are trying to calculate the limit of the general term of a convergent series (in this case the expansion of $$e^x$$), so it has to be zero.

• It would really help if you could elaborate! Mar 19 at 15:28
• @DatBoi Maybe, but the most inmediate way to see why it is zero is to notice that you are trying to calculate the limit of the general term of a convergent series (in this case the expansion of $e^x$), so it has to be zero.
– José
Mar 19 at 15:36
• @JoséAntonio, but how do you show that the series converges? The argument for that has to include a direct proof that the limit of the terms is zero. Mar 19 at 16:00
• @JoséAntonio, all the OP wants is that the general term converges to zero. If you look at the proof of convergence of the series using the ratio test, you will see that the ratio test itself proves directly that the terms converge to zero. There's no need to go through the entire proof that the series converges. The proof that $\frac{x^n}{n!}$ converges to zero is much easier than the proof that $\sum_{n=0}^\infty \frac{x^n}{n!}$ converges. Mar 19 at 17:53
• @Deane I see. Thanks!
– José
Mar 19 at 18:07

Alternative (much less elegant) approach than the one taken in the answer of José Antonio.

Given:
$$~x~$$ is some fixed (i.e. constant) real number.

It is sufficient to prove that
$$~\forall ~\epsilon > 0, ~\exists ~M \in \Bbb{Z^+},~$$ such that for all $$~\displaystyle n \geq 2M, ~\left| ~\frac{x^n}{n!} ~\right| < \epsilon.$$

For any fixed $$~M \in \Bbb{Z^+},~$$
let $$~f(k)~$$ denote $$(1 + 2M - k) \times k ~: 1 \leq k \leq M.$$

Then, $$~f'(k) = (1 + 2M) - 2k > 0 ~: ~1 \leq k \leq M.~$$
Therefore, $$~f(k)~$$ is strictly increasing on the interval $$~1 \leq k \leq M.$$
Therefore, $$(1 + 2M - k)k = f(k) \geq f(1) = 2M.$$

Then

$$\left| ~\frac{x^{2M}}{(2M)!} ~\right| = \prod_{k=1}^M \frac{x^2}{f(k)} \leq \left[ ~\frac{x^2}{2M} ~\right]^M = \left[ ~\frac{x^2}{M} ~\right]^M \times \frac{1}{2^M}. \tag1$$

Given a fixed $$~x \in \Bbb{R},~$$ and given a fixed $$~\epsilon > 0,~$$ choose $$~M \in \Bbb{Z^+},~$$ such that:

• $$~x^2 < M.$$
• $$~\dfrac{1}{2^M} < \epsilon.$$

For example, using $$~\lceil r\rceil~$$ to denote the ceiling of $$~r,~$$ you can set

$$M = 1 + \max\left\{ ~\left\lceil ~x^2 ~\right\rceil, ~~\left\lceil ~\frac{\ln(1/\epsilon)}{\ln(2)} ~\right\rceil ~\right\}. \tag2$$

With $$~M~$$ chosen by the constraint in (2) above, the RHS of (1) above is strictly $$~\displaystyle < 1 \times \frac{1}{2^M} < \epsilon.$$

Further, since $$~M \in \Bbb{Z^+},~$$ and $$~x^2 < M,~$$ you have that $$~|x| < M < 2M.$$

The problem can now be completed by induction.

Let $$~N~$$ be any positive integer such that :

• $$~|x| < N.$$

• $$\displaystyle \left| ~\frac{x^{N}}{N!} ~\right| < \epsilon.$$

Then:

• $$\displaystyle |x| < (N+1).$$

• $$\displaystyle \left| ~\frac{x^{N+1}}{(N+1)!} ~\right| = \left| ~\frac{x}{N+1} ~\right| \times \left| ~\frac{x^{N}}{N!} ~\right| < \epsilon.$$

So, if the two inequalities above hold for $$~N,~$$ they also hold for $$(N+1).$$

Therefore, with $$~M~$$ set by the constraints in (2) above, for all $$~n \in \Bbb{Z^+}~$$ such that $$~n \geq 2M,~$$ the two inequalities will hold for $$~n.$$