Hints for limit $\lim_{x\to\infty}\left(\frac{x(x^\frac{1}{x}-1)}{\ln x}\right)$

Question

$$\lim_{x\to\infty}\left(\frac{x(x^\frac{1}{x}-1)}{\ln x}\right)$$

The only thing that came in my mind was to rewrite $x$ as $\frac{1}{\frac{1}{x}}$ and then use $\lim_{x\to 0}\left(\frac{a^x - 1}{x}\right)=\ln a$ but that would give me $\ln x$ which tends to $\infty$ so I would have $\frac{\infty}{\infty}$.

Answer 1

$$\lim_{x\to\infty} \frac{x(x^{1/x}-1)}{\ln x}=\lim_{x\to \infty}\frac{e^{\frac{\ln x}x}-1}{\frac{\ln x}x}=\lim_{y\to 0^+}\frac{e^y-1}{y}=1$$

because $\frac{\ln x}{x}\to0^+$ as $x\to\infty$.

Answer 2

hint

If you put $$t=\frac{\ln(x)}{x}$$

your limit becomes

$$\lim_{t\to 0}\frac{e^t-1}{t}$$