Doubt on a limit with parameter I have to study the value of the following limit:
$$\lim_{n \to +\infty} n^{\alpha}(\sqrt[n]{n+1} - 1)$$
for $\alpha \in \mathbb{R}$. So I made the substitution $\displaystyle x = \frac{1}{n}$ and got the limit:
$$\lim_{x \to 0^+} \frac{1}{x^{\alpha}}\left(\left(\frac{1}{x} +1\right)^{\displaystyle\frac{1}{x}}-1\right) = \lim_{x \to 0^+} \frac{1}{x^{\alpha}}\left(e^{\displaystyle\frac{1}{x}\log{\left(\frac{1}{x}+1\right)} }-1\right)$$
now I would like to expand in Taylor series, but I can't because the internal function doesn't vanish, how could I go on?

We recall that $\dfrac{\ln(n+1)}{n}\rightarrow 0$ $(n\rightarrow\infty)$.
Hence $\lim\limits_{n\rightarrow\infty}\dfrac{\dfrac{\ln(n+1)}{n}}{\sqrt[n]{n+1}-1}=$
$$\lim\limits_{n\rightarrow\infty}\dfrac{\dfrac{\ln(n+1)}{n}}{e^{\dfrac{\ln(n+1)}{n}}-1}=1$$
So it suffices to determine $$\lim\limits_{n\rightarrow\infty}\dfrac{\ln(n+1)}{n^{1-\alpha}}$$
We then recall that L'Hopital rule implies that for fixed $t$,
$$\lim\limits_{x\rightarrow+\infty}\dfrac{\ln(x+1)}{x^{t}}=\begin{cases} 0 & t>0 \\ +\infty& t\leq 0 \end{cases}$$
