Show $\lim_{n\rightarrow \infty}(1-x^{a \log(n)}) =0$ when $a <\frac{-1}{\log(x)}$

Question

I'd like to show that $$\lim_{n\rightarrow \infty}(1-x^{a \log(n)}) =1$$ when $a >\frac{-1}{\log(x)}$ and $$\lim_{n\rightarrow \infty}(1-x^{a \log(n)}) =0$$ when $a <\frac{-1}{\log(x)}$

Where $x \in (0,1)$. I have managed to show the first limit using $a = \frac{-1}{\log(x)}+\epsilon$ where $\epsilon>0$. However this approach did not work for the second one (taking $a = \frac{-1}{\log(x)}-\epsilon$).

My idea was since we want the limit to be $0$, we would want $\lim_{n\rightarrow \infty} x ^{a\log(n)}=1$. This only happens when $\lim_{n\rightarrow \infty} a \log(n) = 0$. Substituting $a = \frac{-1}{\log(x)}-\epsilon$ does not help solving this limit though. I was wondering if someone maybe could give me a hint on how to approach this.

Thanks in advance

Answer 1

When $0 < x < 1$ then \begin{aligned} \lim_{n\to\infty}\big( 1 - x^{a\log n} \big) = \left\{ \begin{array} & 0 & \text{if } a = 0 \\ 1 & \text{if } a > 0 \\ -\infty & \text{if } a < 0 \end{array} \right. \end{aligned}