I'm trying to find the following value.
For $x>0$ and $\alpha>0$,
$$ \lim_{n \to \infty} n(1-n^{-\alpha}x)^{n-1} - (n-1)(1-n^{-\alpha}x)^{n} $$
It's clear that when $\alpha=1$, the result is $exp(-x) (1+x)$. But how about when $0 < \alpha < 1$ or $\alpha > 1$?
What I tried:
$$ \begin{aligned} \lim_{n \to \infty} n(1-n^{-\alpha}x)^{n-1} - (n-1)(1-n^{-\alpha}x)^{n} & = \lim_{n \to \infty}(1-n^{-\alpha}x)^{n} \Big[ n (1-n^{-\alpha}x)^{-1} - (n-1) \Big] \\ \end{aligned} $$
I know $\lim_{n \to \infty}(1-n^{-\alpha}x)^{n} = 0 $ when $0 < \alpha < 1$, and $\lim_{n \to \infty}(1-n^{-\alpha}x)^{n} = 1 $ when $\alpha > 1$.
But what happens if they are combined with the remainder term?
For $x>0$ and $\alpha>0$, we can first perform some transformations on the limit:
\begin{align*} &\lim_{n\to\infty}n(1-n^{-\alpha}x)^{n-1}-(n-1)\cdot (1-n^{-\alpha}x)^n\\ &=\lim_{n\to\infty}n\cdot \left(1-\dfrac{x}{n^\alpha}\right)^{n-1} - (n-1)\cdot \left(1-\dfrac{x}{n^\alpha}\right)^n\\ &=\lim_{n\to\infty}\left(1-\dfrac{x}{n^\alpha}\right)^{n-1}\cdot \left(1+\dfrac{x}{n^{\alpha-1}}-\dfrac{x}{n^\alpha}\right) . \end{align*} We can consider the limits of the two factors separately. For the second factor, it is easy to see that
$$\lim_{n\to\infty} 1+\dfrac{x}{n^{\alpha-1}}-\dfrac{x}{n^\alpha}=\begin{cases}1 &\text{if }\alpha>1,\\ 1+x &\text{if }\alpha=1,\\ +\infty &\text{if }1>\alpha>0\end{cases}$$
For the first factor we have:
\begin{align*} \lim_{n\to\infty}\left(1-\dfrac{x}{n^\alpha}\right)^{n-1}&=\exp\left[\ln \left(\lim_{n\to\infty}\left(1-\dfrac{x}{n^\alpha}\right)^{n-1}\right)\right]&\\ &=\exp\left[\lim_{n\to\infty} \ln \left(1-\dfrac{x}{n^\alpha}\right)^{n-1}\right]&\\ &=\exp\,\lim_{n\to\infty} (n-1)\ln\left(1-\dfrac{x}{n^\alpha}\right)&\\ &=\exp\lim_{n\to\infty}\dfrac{\ln\left(1-\frac{x}{n^\alpha}\right)}{\frac{1}{n-1}}\\ &=\exp \left[\lim_{n\to\infty} \dfrac{\dfrac{1}{1-\dfrac{x}{n^\alpha}}\cdot \dfrac{-\alpha\cdot x}{n^{\alpha+1}}}{\dfrac{-1}{(n-1)^2}}\right] &\text{(By L'Hôpital's rule)}\\ &=\exp\left[\lim_{n\to\infty}\dfrac{\alpha\cdot x}{1-\dfrac{x}{n^\alpha}}\cdot \dfrac{(n-1)^2}{n^{\alpha+1}}\right]\\ &=\begin{cases}\exp(0)=1 &\text{if }\alpha>1,\\ \exp(\alpha\cdot x) &\text{if }\alpha=1,\\ +\infty &\text{if }1>\alpha>0. \end{cases} \end{align*}
Hence, putting things together, we conclude that $$\lim_{n\to\infty}n(1-n^{-\alpha}x)^{n-1}-(n-1)\cdot (1-n^{-\alpha}x)^n=\begin{cases}1\cdot 1=1, &\text{if }\alpha>1,\\ (1+x)\cdot e^{\alpha\cdot x} &\text{if }\alpha=1\\ +\infty &\text{if }1>\alpha>0. \end{cases}$$