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

Question

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

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.

Answer 1

As you guessed, $\forall n\in \mathbb{N}$:

$$ 0< \frac{1}{n!} $$

But you could also use:

$$ \frac{1}{n!}\leq\frac{1}{n} $$

Since $\frac{1}{(n-1)!}\leq 1$ for $n\in \mathbb{N}^*$.

Answer 2

$$\frac{1}{n}\geq\frac{1}{n!}>0, \forall n \in \mathbb R^+$$ 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.