Using the squeeze theorem

[Albi]

Hey everyone, I'm trying to use the squeeze theorem to solve the following limit[math]\lim_{x\rightarrow \infty}\frac{x^a}{x!} , a \in \mathbb{R}[/math] but I'm finding it difficult to squeeze it from the right side, since I cannot find something greater or equal than [imath]\frac{x^a}{x!}[/imath] 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 [imath]c < 1[/imath] and large enough [imath]x[/imath]: [imath]f(x+1) \leq c f(x)[/imath].

 

[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 \(\displaystyle \dfrac {x^a}{x!}<\dfrac {x}{x!} = \dfrac{1}{(x-1)!}\)