Showing that $f: D_1(0) \subset \mathbb{R}^2 \to \mathbb{R}, f(x) = \frac{1}{1- \|x\| }$ is continuous

Question

I understand that I need to show that for all $x_0 \in D_1(0)$,

$$\lim_{x\to x_0} \frac{1}{1- \|x\| } = \frac{1}{1- \|x_0\| }$$

But I have trouble bounding the denominator,

$$\bigg| \frac{1}{1- \|x\| } - \frac{1}{1- \|x_0\| } \bigg| = \bigg| \frac{\|x\| - \|x_0\|}{(1- \|x\|)(1- \|x_0\|)} \bigg| \stackrel{\text{triangle inequality}}{\leq} \frac{\|x - x_0\|}{|1-(\|x_0\| + \|x\|) + \|x_0\|\|x\||}$$

Answer 1

Hint: drop your last inequality. That is not useful. $|\|x\|-\|x_0\|| \leq \|x-x_0\|$ and $1-\|x\| \geq 1-\|x_0\|-\|x-x_0\|>\frac {1-\|x_0\|} 2$ if $\|x-x_0\|$ is sufficiently small.