Evaluating the limit $\lim_{x\to+\infty}\frac{x\left(x^{1/x}-1\right)}{\ln(x)}$ 
Evaluate the limit
  $$\lim_{x\to+\infty}\frac{x\left(x^{1/x}-1\right)}{\ln(x)}$$

This is what I tried, with Taylor Series:
$$\lim_{x\to+\infty}\frac{x\left(x^{1/x}-1\right)}{\ln(x)}=\lim_{x\to+\infty}\frac{x\left(1+\frac{\ln(x)}{x}+O\left(\left(\frac{1}{x}\right)^2\right)-1\right)}{\ln(x)}=\lim_{x\to+\infty}\frac{x\left(\frac{\ln(x)}{x}+O\left(\left(\frac{1}{x}\right)^2\right)\right)}{\ln(x)}=1.$$
Is this right? Is there another way?
 A: An alternative computation that doesn't require Taylor series at all:
$$\lim_{x \to \infty} \frac{x^{1/x} - 1}{\ln x^{1/x}} = \lim_{t \to 1} \frac{t - 1}{\ln t} = \left(\frac{d}{dt}\big|_{t = 1} \ln t\right)^{-1} = 1.$$
The only thing we've used is that $x^{1/x} \to 1$ as $x \to \infty$.
A: First you could show that $\dfrac{\ln x}{x}$ tends to $0$. This is easily done by employing L'hospital's rule.
Nos with $t= \dfrac{\ln x}{x}$ the limit becomes $\displaystyle \lim_{t \rightarrow 0}  \dfrac{e^t-1}{t}= 1$ 
A: The final result look right, but I am not sure on your procedure.
Here is an approach that doesn't involve L'Hôspital's Rule and Taylor Series:
$$
\begin{aligned}
&{\displaystyle \lim_{x\to+\infty}\frac{x\left(x^{\frac{1}{x}}-1\right)}{\ln(x)}}\to\small{\begin{bmatrix}
&t=1/x&\\
&t\to0&
\end{bmatrix}}\to{\displaystyle \lim_{t\to0}\left(\frac{\left(\frac1{t}\right)^t-1}{t\ln\left(\frac1{t}\right)}\right)}={\displaystyle \lim_{t\to0}\left(\frac{e^{-t\ln{(t)}}-1}{-t\ln\left(t\right)}\right)}\to\\ \\
&\to\small{\begin{bmatrix}
&s=-t\ln\left(t\right)&\\
&s\to0&
\end{bmatrix}}\to{\displaystyle \lim_{t\to0}\left(\frac{e^{s}-1}{s}\right)}=1.
\end{aligned}$$
Where 
$\lim_\limits{x\to0}\frac{e^{x}-1}{x},\lim_\limits{x\to0}x\ln(x)$ are special - known - limits that you can easily check in any calculus book.
