Using the squeeze theorem

[Albi]

Hey everyone, I'm trying to use the squeeze theorem to solve the following limit$$\lim_{x\rightarrow \infty}\frac{x^a}{x!} , a \in \mathbb{R}$$ but I'm finding it difficult to squeeze it from the right side, since I cannot find something greater or equal than $\frac{x^a}{x!}$ can someone give me some strategies on how to approach these kinds of problems?

 

[blamocur]

Not sure how to squeeze it, but an alternative approach would be to show that for some $c < 1$ and large enough $x$: $f(x+1) \leq c f(x)$.

 

[Steven G]

Not that this will help, but $\displaystyle x^a\ge \dfrac {x^a}{x!}\ and\ x^a\ge \dfrac {x^a}{(x-k)!}\ for\ some\ k$

 

[Steven G]

Actually it should be clear that if a<1, then $\displaystyle \dfrac {x^a}{x!}<\dfrac {x}{x!} = \dfrac{1}{(x-1)!}$