Evaluate $\lim_{x\to \infty}(\log x)^{1/x}$ $$\displaystyle \operatorname{lim}_{x\to \infty}(\operatorname{log}x)^{\frac{1}{x}}$$
$t=\frac1x \Rightarrow \displaystyle \operatorname{lim}_{t\to 0^+}(\operatorname{log}\frac1t)^t\Rightarrow(-1)^t(\operatorname{log}t)^t$
In my opininion the limit doesnot exist.Is it Correct?
 A: Rewrite the functions as $e^\frac{\log  \log x }{x}$, since $\exp$ is a continuous functions, you can interchange $\exp$ and limit operator and get $1$ as the limit. 
A: Consider
$$y=\sqrt[x]{\log (x)}\implies \log(y)=\frac 1x \log(\log(x))$$ $x$ goes much faster than $\log(x)$ and still faster than $\log(\log(x))$.
So, $\log(y)\to 0$ and $y=e^{\log(y)}\to 1$
A: Note that by standard limit $t^t\to 1$ as $t\to 0^+$ we have that
$$(\log x)^{\frac{1}{x}}=x^{\frac1x} \left[\left(\frac{\log x}{x}\right)^{\frac{\log x}{x}}\right]^\frac1{\log x}\to 1$$
indeed


*

*$x^{\frac1x}$ with $y=\frac 1x \to 0^+ \implies\frac1{y^y}\to 1$

*$\left(\frac{\log x}{x}\right)^{\frac{\log x}{x}}$ with $t=\frac{\log x}{x}\to 0$ form above then $\left(\frac{\log x}{x}\right)^{\frac{\log x}{x}}\to 1$

*$\frac1{\log x}\to 0$
A: You can't do
$$
(-\log t)^t=(-1)^t(\log t)^t
$$
because general exponentiation is only defined for positive base and here $-1$ and $\log t$ are negative, as $t\to0^+$.
When confronted with $f(x)^{g(x)}$ it's most of the times convenient to compute the limit of $\log(f(x)^{g(x)}=g(x)\log f(x)$ and then use the properties of $\exp$. In this case
$$
\lim_{x\to\infty}\log((\log x)^{1/x})=
\lim_{x\to\infty}\frac{\log\log x}{x}
$$
There are several ways to finish this up. You can observe that $\log\log x<\log x$ and
$$
\lim_{x\to\infty}\frac{\log x}{x}=\lim_{t\to\infty}\frac{t}{e^t}=0
$$
Or you can directly apply l'Hôpital:
$$
\lim_{x\to\infty}\frac{\log\log x}{x}
=
\lim_{x\to\infty}\frac{1}{x\log x}=0
$$
Thus
$$
\lim_{x\to\infty}(\log x)^{1/x}=e^0=1
$$
A: Note: $(-\ln 0.5)^{0.5}=((-1)\cdot(-\ln 2))^{0.5}\ne (-1)^{0.5}\cdot (-\ln 2)^{0.5}$, because the last is undefined.
If you want to make a change, let: $\ln x=t \Rightarrow x=e^t$. Then:
$$\lim_{x\to\infty} (\ln x)^{1/x}=\lim_{t\to\infty} t^{1/e^t}=\lim_{t\to\infty} \exp\left(\frac{\ln t}{e^t}\right)=\exp\left(\lim_{t\to\infty} \frac{\ln t}{e^t}\right)=\exp(0)=1.$$
