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 $0<\frac{1}{1-x}<\frac{1}{2} \implies |\frac{1}{1-x}|<\frac{1}{2}$.

$$\begin{align}\left|\frac{1}{1-x}\right|<\frac{1}{2}\\
\Longleftrightarrow -\frac{1}{2}<\frac{1}{1-x}<\frac{1}{2}\\
\Longleftrightarrow \frac{x-1}{2}<1< \frac{1-x}{2}\\
\Longleftrightarrow x-1<2<1-x\\
\Longleftrightarrow x<3<2-x\end{align}$$
Which is clearly not true for an $x$ arbitrarily near zero. I think there could be a problem with your choice of $\delta$.
