Prove that $g$ is continuous at $x=0$ Given, $g(x) = \frac{1}{1-x} + 1$. I want to prove that $g$ is continuous at $x=0$. I specifically want to do an $\epsilon-\delta$ proof. Related to this Is $g(x)\equiv f(x,1) = \frac{1}{1-x}+1$ increasing or decreasing? differentiable $x=1$?.  
My work: Let $\epsilon >0$ be given. The challenging part for me is to pick the $\delta$. 
Let $\delta = 2 \epsilon$ and suppose that $|x-0| = |x| < \delta$ and $|\frac{1}{1-x}| < \frac{1}{2}$.
So $$|g(x) - g(0)|$$ 
$$=|\frac{2-x}{1-x} - 2|$$
$$=|\frac{x}{1-x}|$$
$$=|\frac{1}{1-x}||x|$$
$$<\frac{1}{2} 2 \epsilon = \epsilon$$.
So $g$ is continuous at $x=0$. Is my proof correct? 
EDIT: Rough work on how I picked $\delta$. Suppose $|x-0|<\delta$ and since $\delta \leq 1$, we have $|x|<1$, so $-1<x<1$ Then this implies $0<1-x<2$. The next step I'm not too sure of, I have $0<\frac{1}{1-x}<\frac{1}{2} \implies |\frac{1}{1-x}|<\frac{1}{2}$.

Hint. See my comment to your post. 
After doing your algebra, as you have shown,
$$\left|\dfrac{1}{1-x}+1-2\right| = \left|x\right|\left|\dfrac{1}{1-x}\right|\text{.}$$
We have $|x| < \delta$. Let $\delta = 1/2$, then
$$|x| < \delta \implies -\delta < x < \delta \implies 1-\delta < 1-x < 1 + \delta \implies \dfrac{1}{1+\delta} < \dfrac{1}{1-x} < \dfrac{1}{1-\delta}= 2\text{,}$$
thus implying $\left|\dfrac{1}{1-x}\right| < 2$ (notice $\dfrac{1}{1+\delta} = \dfrac{1}{1.5} \in (-2, 2)$, so we can make this claim). 
Choose $\delta := \min\left(\dfrac{\epsilon}{2}, \dfrac{1}{2}\right)$.
