Show that $\lim\limits_{n \to \infty} b \cdot (1-\sqrt{3})^n=0 \quad \forall b \in \mathbb{R}$

Question

I want to show the following limit:

$$ \lim_{n \to \infty} b \cdot (1-\sqrt{3})^n=0 \quad \forall b \in \mathbb{R} $$

It makes sense that $\lim\limits_{n \to \infty}(1-\sqrt{3})^n=0$ since $|1-\sqrt{3}|<1$ and so $b \cdot \lim\limits_{n \to \infty} (1-\sqrt{3})^n=b \cdot 0$.

But how do I then show:

$$ \lim_{n \to \infty} (1-\sqrt{3})^n=0 $$

Answer 1

If I understand you correctly you would like to see that

• $0< |q| < 1 \Rightarrow \lim_{n \to \infty} q^n = 0$

A possible way to see this is as follows:

• Note that $|q^n| = |q|^n$
• $0< |q| < 1 \Rightarrow 0 < |q|^{n+1} < |q|^n$ for all $n \in \mathbb{N}$
• $\Rightarrow $ it exists $ L =\lim_{n \to \infty} |q|^n $ and $L \geq 0$

So, it is enough to show that $L = 0$: $$|q|^{n+1} = |q| \cdot |q|^{n} \stackrel{n \to \infty}{\longrightarrow} L = |q|L \Rightarrow L(1-|q|) = 0 \stackrel{1-|q|>0}{\Rightarrow} L = 0$$