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

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.

• This seems a bit odd: Since $\log n \to \infty$ as $n \to \infty$, if $x^a > 1$ then your limit is $-\infty$. If $x^a = 1$ the limit is $0$ (as are all intermediate terms), and if $x^a < 1$ the limit is $1$. Is there something missing from the question? Commented Nov 8, 2020 at 14:47
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}