# Use the Sandwich theorem (Squeeze theorem) to find this limit:

$\lim_\limits{n\to \infty} \frac{1}{n!}$

I really don't know how to solve this. I usually use this theorem to solve limits of oscillating functions such as sine and cosine. But for this function I only start from the premise that it will always be greater than zero.

• I made some improvements to the MathJax in your question (which you can view by clicking on "edited x ago"). The main one is that you should use \infty to get the infinity symbol. May 7 '18 at 14:17
• thank you. I am very bad at writing here. May 7 '18 at 14:18
• Do you HAVE to use the sandwich lemma? Because you can say straightaway that $\frac{1}{n!}\to 0$ as $n\to\infty$ May 7 '18 at 14:22
• Yes I have to. I know that the limit of the function when n tends to infinity is 0 but I have to explain it using the theorem. That's why I asked for help. May 7 '18 at 14:27
• @MichalDvořák There are lots of limits that you can say straightaway are $0$. However when you're at the level of being asked this exercise you're probably expected to give justifications from basic results you know (e.g. showing $\frac1n \to 0$ is an application of the Archimedean property of $\mathbb{R}$). Admittedly it's been a long time since I did any of this stuff but it seems like a justification without squeeze theorem would be longer and would probably essentially do the same kind of thing anyway? May 7 '18 at 14:31

As you guessed, $\forall n\in \mathbb{N}$:
$$0< \frac{1}{n!}$$
$$\frac{1}{n!}\leq\frac{1}{n}$$
Since $\frac{1}{(n-1)!}\leq 1$ for $n\in \mathbb{N}^*$.
$$\frac{1}{n}\geq\frac{1}{n!}>0, \forall n \in \mathbb R^+$$ and $$\lim_{x\rightarrow\infty}\frac{1}{n}=\lim_{x\rightarrow\infty}0=0$$ so $$\lim_{x\rightarrow\infty} \frac{1}{n!}=0$$ by squeeze theorem.