find $\lim_{x\rightarrow \infty } \frac{a^x}{x^{\alpha}}$

Question

Given that $a> 1$, $\alpha \in \mathbb{R} $,then find $$\lim_{x\rightarrow \infty } \frac{a^x}{x^{\alpha}}$$

My attempts :

$\frac{a^x}{x^{\alpha}} \le \frac{1}{x^\alpha}$

now i take $\lim_{x\rightarrow \infty } \frac{1}{x^{\alpha}} = \frac{1}{\infty} = 0$

Is its correct ??

thanks in advance

Why is $a^{x}/x^{\alpha} \le 1/x^{\alpha} \implies a^{x} \le 1$? $a^{x}$ is increasing for $a > 1$. — Matthew Cassell, Aug 13 '18 at 07:39
Related https://math.stackexchange.com/questions/2489665/prove-that-if-k-in-mathbb-n-and-a1-then-lim-limits-n-to-infty-fra/ — rtybase, Aug 13 '18 at 07:55

score 2 · Accepted Answer · answered Aug 13 '18 at 07:51

2

As pointed out by Mattos your first inequality is wrong. $\frac {a^{x}} {x^{\alpha}}=\frac {e^{x\log a}} {x^{\alpha}} \geq \frac {(x \log a)^{n}} {(n!)x^{\alpha}} \to \infty$ if we take $n > \alpha $.

answered Aug 13 '18 at 07:51

Kavi Rama Murthy

311,013

find $\lim_{x\rightarrow \infty } \frac{a^x}{x^{\alpha}}$

My attempts :

1 Answers1