How do you compute the limit $\lim\limits_{x \to \infty}\frac{e^{\sqrt{\ln x}}}{x}$ for integers $x$?

Question

Can the numerator be simiplifed? If not, how do you compute this limit? can you even compute this limit? my overall goal is to show that the numerator, $f(x)=e^{\sqrt{\ln x}}$, is $O(x)$ for integer $x$, so the limit achieves this (if it exists).

Answer 1

$$\lim\limits_{x\to\infty}\frac{e^\sqrt{\ln x}}{x} = \lim\limits_{x\to\infty}\frac{e^\sqrt{\ln x}}{e^{\ln x}} = \lim\limits_{x\to\infty}e^{\sqrt{\ln x}-\ln x} = e^{\lim\limits_{x\to\infty}\left(\sqrt{\ln x}-\ln x\right)} = e^{-\infty} = 0$$

Answer 2

$e^{\sqrt{\ln(x)}}$ is actually very small asymptotically, and negligible compared to any $x^a$ with $a>0$.

Let call $f(x)=\dfrac{e^{\sqrt{\ln x}}}{x^a}$, then you can use the substitution below to get rid of annoying square root and logarithm (note: D.Levine's answer is fine too, I just wanted to show the result following a different approach).

Indeed, it is more intuitive to examine $f(e^{x^2})=\dfrac{e^x}{e^{ax^2}}=\exp(x-ax^2)\to 0$

And since $e^{x^2}$ is strictly increasing to infinity, the limits behaves the same than with plain $x$.