We know that one of the characterizations of the exponential function is:
$$e^x=\lim_{n\rightarrow\infty}\left(1+\frac{x}{n}\right)^{n}$$
Trivially, it follows that $\lim_{n\rightarrow\infty}\left(1-\frac{x}{n}\right)^{n}=e^{-x}$
I am wondering about
$$\lim_{n\rightarrow\infty}\left(1-\frac{x}{n^{1+a}}\right)^{n}$$
where $a$ is a real number. Is the following evaluation of the expression correct?
$$\begin{array}{rcl}\lim_{n\rightarrow\infty}\left(1-\frac{x}{n^{1+a}}\right)^{n}&=&\lim_{n\rightarrow\infty}\left(1-\frac{xn^{-a}}{n}\right)^{n}\\ &=&\lim_{n\rightarrow\infty}e^{-xn^{-a}}\\ &=&\left\{\begin{array}{rl}1,&a>0\\e^{-x},&a=0\\0,&a<0\end{array}\right. \end{array}$$
I am uncomfortable taking the second equality, not sure what the justification is...