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$$\displaystyle \operatorname{lim}_{x\to \infty}(\operatorname{log}x)^{\frac{1}{x}}$$

$t=\frac1x \Rightarrow \displaystyle \operatorname{lim}_{t\to 0^+}(\operatorname{log}\frac1t)^t\Rightarrow(-1)^t(\operatorname{log}t)^t$

In my opininion the limit doesnot exist.Is it Correct?

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    $\begingroup$ Hint: $$(\log x)^{1/x}=\exp (\frac{\log\log x}{x})$$ $\endgroup$ – BAI May 14 '18 at 7:50
  • $\begingroup$ it exists and is $1$ . btw , i used a graph to verify this. I'm not sure how to approach this in a algebraic way. $\endgroup$ – The Integrator May 14 '18 at 7:50
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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: $$\lim_{x\to\infty} (\ln x)^{1/x}=\lim_{t\to\infty} t^{1/e^t}=\lim_{t\to\infty} \exp\left(\frac{\ln t}{e^t}\right)=\exp\left(\lim_{t\to\infty} \frac{\ln t}{e^t}\right)=\exp(0)=1.$$

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  • $\begingroup$ Thank you understood. $\endgroup$ – DRPR May 14 '18 at 11:46
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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.

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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 $y=e^{\log(y)}\to 1$

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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 $\left(\frac{\log x}{x}\right)^{\frac{\log x}{x}}\to 1$

  • $\frac1{\log x}\to 0$

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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 $$ \lim_{x\to\infty}(\log x)^{1/x}=e^0=1 $$

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  • $\begingroup$ Your answer looks correct. But $(-3)^3=(-1)^3(3)^3.$Then why cant i do the same with$(-log\, t)^t$? $\endgroup$ – DRPR May 14 '18 at 9:24
  • $\begingroup$ @DRPR And what's $(-1)^{\sqrt{2}}$ or, more generally, $x^y$ where $x<0$ and $y$ is irrational? By “general exponentiation” I mean “with any real exponent”. $\endgroup$ – egreg May 14 '18 at 9:26

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