Limit of $\left(\frac{\ln x}{x}\right)^\frac{1}{x}$

Question

Find $$\lim_{x \to \infty}\bigg(\frac{\ln x}{x}\bigg)^\frac{1}{x}$$ I tried substituting $x=1/t$ but it gave a dead end(according to me).

Answer 1

For $x > 1$,

$$0 < \frac{1}{x}\ln\left(\frac{\ln x}{x}\right) < \frac{1}{x}\cdot \frac{\ln x}{x}.$$

The right hand side tends to $0$ as $x \to \infty$, because $0 \le (\ln x)/x \le 1$ for all $x \ge 1$ and $1/x \to 0$ as $x \to \infty$. Hence, by the squeeze theorem,

$$\lim_{x\to \infty} \frac{1}{x}\ln \left(\frac{\ln x}{x}\right) = 0.$$

Exponentiating results in

$$\lim_{x\to \infty} \left(\frac{\ln x}{x}\right)^{1/x} = e^0 = 1.$$

Answer 2

L'Hospital rule is not the alpha and omega of limits computation! It works as far as linear approximations work. $$\bigg(\frac{\ln x}{x}\bigg)^\frac{1}{x}=\mathrm e^{\tfrac{\ln\ln x -\ln x}x}.$$

We know that $\,\ln x =_{\infty}o(x),\enspace \ln\ln x=_{\infty}o(\ln x)\,$ hence $\,\ln\ln x=_{\infty}o(x)$, so $$\lim_{x\to\infty}\frac{\ln\ln x -\ln x}x=0$$ which proves $\lim\limits_{x\to\infty}\bigg(\dfrac{\ln x}{x}\bigg)^\tfrac{1}{x}=1$.