# Proving $\lim_{n\to\infty}\frac{\sin nx}{n!}=0$ without the Sandwich Theorem?

So I want to find: $$\lim_{n\to \infty} \frac{\sin nx}{n!}$$

The solution I came up with includes the use of the Sandwich Theorem. So:

$$\forall x \in \mathbb R,\quad|\sin(nx)|\leq 1\quad\Rightarrow\quad -\left|\frac{1}{n!}\right| \leq \frac{\sin(nx)}{n!} \leq \left|\frac{1}{n!}\right|$$ However, $$\lim_{n\to \infty}-\left|\frac{1}{n!}\right|=\lim_{n\to \infty}\left|\frac{1}{n!}\right|=0$$ So using the Sandwich Theorem $$\lim_{n\to \infty}\frac{\sin(nx)}{n!}=0$$

The result is $$0$$, but is there another way to prove this, e.g. using L'Hospital's rule?

• You can't use L'Hopital's Rule since you can't differentiate $n!$ with respect to $n$ - it is not continuous. Jan 3, 2020 at 15:14
• Okay, thank you very much! Jan 3, 2020 at 15:15
• Anyway, L'Hospital's rule should generally be avoided. Jan 3, 2020 at 15:19
• Why should it be avoided? If you can explain of course, thanks for your time! Jan 3, 2020 at 15:20
• You can use the theorem on the product of infinitesimal to bounded Jan 3, 2020 at 15:23

You can just use the definition of the limit. Write $$\left|\frac{\sin\left(nx\right)}{n!}\right| \leq \frac{1}{n}.$$ Then, given $$\epsilon>0$$, for any $$n > N := \left[\frac{1}{\epsilon}\right]$$ we have $$\frac{1}{n} < \epsilon$$ and therefore $$\left|\frac{\sin\left(nx\right)}{n!} - 0\right| < \epsilon.$$

We claim that $$\lim_{n \to \infty}\frac{\sin(nx)}{n} = 0$$. If we prove this, then we get $$\lim_{n \to \infty}\frac{\sin(nx)}{n!} = \lim_{n \to \infty}\left(\frac{\sin(nx)}{n}\cdot\frac{1}{(n-1)!}\right) =\left(\lim_{n \to \infty}\frac{\sin(nx)}{n}\right)\cdot\left(\lim_{n \to \infty}\frac{1}{(n-1)!}\right)= 0 \cdot 0 = 0 .$$

Proof of claim. (to be continued --- I have to avoid sandwitch for this part too --- will require some thought)

• It's going to be difficult to avoid using bounds on $\sin(nx)$ (or maybe $\cos(nx)$) somewhere. Note that if $x$ is not real, the limit does not exist. Jan 3, 2020 at 15:43
• @RobertIsrael: I think you are right. Jan 3, 2020 at 15:48

$$\lim_{n\to \infty} \frac{\sin nx}{n!}=0$$$$⇔$$$$∀ε>0,∃N∈ℕ^+, ∀n∈ℕ^+(n>N⇒\left|\frac{\sin\left(nx\right)}{n!}\right|<ε)$$

$$n>N⇒\left|\frac{\sin\left(nx\right)}{n!}\right|<\frac{\left|nx\right|}{n!}=\frac{\left|x\right|}{\left(n-1\right)!}\le\frac{\left|x\right|}{2^{\left(n-2\right)}}<ε$$

$$n>N⇒2^{\left(n-2\right)}>\frac{\left|x\right|}{ε}$$

Since $$\log_{a}\left(x\right)$$ is increasing over its domain for $$a>1$$, so we have:

$$n>N⇒n>\log_{2}\left(\frac{\left|x\right|}{ε}\right)+2$$

Hence take $$N\ge\log_{2}\left(\frac{\left|x\right|}{ε}\right)+2$$ and the result follows.

Another way would be using Stirling approximation.