# Evaluating limit using logarithms.

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.

• The limit of the logarithm is $0$. So what's the original limit? – David Mitra Feb 6 '14 at 21:53
• Could you use L'Hospital? – user122283 Feb 6 '14 at 21:58
• Could you try rewriting $(\ln x)^{1/x}$ as $\exp(\ln(\ln x)/x)$? – TooTone Feb 6 '14 at 22:06
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.
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}