Is $\lim_{n \to \infty} (1 - n^{-a}x)^n = 0$ when $0 < a < 1$?

Question

I'm trying to find the value of the following when $0<a<1$, $\forall x > 0$

$$ \lim_{n \to \infty} (1 - n^{-a}x)^n $$

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What I tried:

$$ \begin{aligned} \lim_{n \to \infty} (1 - n^{-a}x)^n &= \lim_{n \to \infty} (1 - n^{-a}x)^{n^{a}n^{-a+1}} \\ &= \Big[ \lim_{n \to \infty} (1 - n^{-a}x)^{n^{a}} \Big]^{n^{-a+1}} \\ &= \lim_{n \to \infty}\Big( e^{-x} \Big)^{n^{-a+1}} \\ &= \infty \\ &\Big(\because \lim_{n \to \infty} n^{-a+1} \to \infty \text{ when } 0 < a < 1 \Big) \end{aligned} $$

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But I confuse, since whenever I try approximating it with Wolfram Alpha with $a = \frac{1}{2}$, I get 0. Is there anyone to help me out?

(In Wolfram Alpha, I typed the following)

lim (1 - n^(-(1/2)) * x)^n n->infinity

Answer 1

For $x> 0:$ Consider the logarithm of the expression:

(1).For $y\ne 0$ we have $\lim_{y\to 0}\frac {\log (1+y)}{y}=1.$

(2). For $n$ large enough that $|n^{-a}x|<1$ we have $$\log ((1-xn^{-a})^n)=n\cdot (-xn^{-a})\cdot \frac {\log (1-xn^{-a})} {-xn^{-a}}.$$ By (1), this $\to -\infty$ as $n\to \infty,$ when $0<a<1$ and $x> 0.$