Prove that $\lim \limits_{x\to \infty}e^{\frac{\ln(x)}{x}}=1$ How to prove that $\lim \limits_{x\to \infty}e^{\frac{\ln(x)}{x}}=1$?
I know that $x$ grows much faster to infinity then $\ln(x)$, therefore the limit equivalent to $e^0 = 1$ 
but that's not a rigorous proof.
 A: $$\lim_{x \to \infty}\frac{\ln{x}}{x}=\lim_{x \to \infty}\frac{1}{x}=0$$
by L Hopital's Rule
A: Take the function inside $\lim \limits_{x\to \infty}e^{\frac{\ln(x)}{x}}$ and let $y$ be that function,
$$y = e^\frac{\ln(x)}{x}$$
Apply logarithm to both sides,
$$\ln(y) = \frac{\ln(x)}{x}$$
Applying the limits,
$$\lim \limits_{x\to \infty} \ln(y) = \lim \limits_{x\to \infty}\frac{\ln(x)}{x}$$
Since in $\dfrac{ln(x)}{x}$, as $x\to \infty$, both numerator and denominator goes to infinity (one of the requirement before we can apply L'Hospital Rule), we can take the L'Hospital Rule,
$$\lim \limits_{x\to \infty} \ln(y) = \lim \limits_{x\to \infty}\frac{\dfrac{1}{x}}{1}$$
$$\lim \limits_{x\to \infty} \ln(y) = \lim \limits_{x\to \infty}\frac{1}{x} = 0$$
Now we get rid of the log,
$$e^{\lim \limits_{x\to \infty} \ln(y)} = e^{\lim \limits_{x\to \infty}\frac{1}{x}} = e^0$$
$$\lim \limits_{x\to \infty} y = 1$$
Finishing the proof.
A: It is well known that the exponential function $e^x$ increases more faster then every $x^a$ , $a \in \mathbb R$. Since ln(x) is strict monotone increasingly this relation still holds for $ln(e^x)$ and $ln(x^a)$ for $a=1$ we know now that $x$ incrases more faster then $ln(x)$ we obtain:
$\lim_{x \to \infty} e^{\frac{ln(x)}{x}}=e^{\lim_{x \to \infty} \frac{ln(x)}{x} }=e^0=1 $
