Evaluate the following limit.
$$ \lim_{x\to \infty} (\ln\ x)^{\frac{1}{x}} $$
What i have tried:
$$ \ln\left[\lim_{x\to \infty} (\ln\ x)^{\frac{1}{x}}\right] $$
$$ \lim_{x\to \infty} \ln(\ln x)^{\frac{1}{x}} $$
$$ \lim_{x\to \infty} \frac{\ln(\ln x)}{x} $$
So as $ x$ approaches infinity, the limit goes to 0. But the answer in the book is 1.
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$\begingroup$You took the natural log $\ln$ of the limit to evaluate it easier, but you forgot to undo the natural log. It is just like how if you were to add $1$ to the limit to make it easier to calculate, you would have to subtract off $1$ in the end.
In this case, to "undo" a natural log, you take $e$ to the power of something. So after you took the natural log you calculated the limit to be $0$; then $$ e^0 = 1 $$ is your answer.
$\endgroup$ $\begingroup$I am still learning about limits, but I think the following approach ought to work.
For any $p,q$, $\;p^q = \exp(\ln(p)q)$. Hence $$\ln(x)^{1/x} = \exp\left(\frac{\ln(\ln x)}{x}\right)$$ And taking limits $$\begin{align}\lim_{x\to\infty} \ln(x)^{1/x} &= \lim_{x\to\infty} \left( \exp\left(\frac{\ln(\ln x)}{x}\right) \right)\\ &= \exp\left(\lim_{x\to\infty}\left(\frac{\ln(\ln x)}{x}\right) \right) \end{align}$$
and hopefully you can see the inner limit evaluates to 0.
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