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?

 A: 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
$$ \left|\frac{\sin\left(nx\right)}{n!} - 0\right| < \epsilon. $$
A: 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)
A: $$ \lim_{n\to \infty} \frac{\sin nx}{n!}=0$$$$⇔$$$$∀ε>0,∃N∈ℕ^+, ∀n∈ℕ^+(n>N⇒\left|\frac{\sin\left(nx\right)}{n!}\right|<ε)$$
$$n>N⇒\left|\frac{\sin\left(nx\right)}{n!}\right|<\frac{\left|nx\right|}{n!}=\frac{\left|x\right|}{\left(n-1\right)!}\le\frac{\left|x\right|}{2^{\left(n-2\right)}}<ε$$
$$n>N⇒2^{\left(n-2\right)}>\frac{\left|x\right|}{ε}$$
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.
