Prove that $f:(-1,1) \to \mathbb{R}, \quad x \to \frac{1}{\sqrt{1-|x|}}$ is local lipschitz continuous

Question

Is there a short proof why the function

\begin{align} f:(-1,1) \to \mathbb{R}, \quad x \to \frac{1}{\sqrt{1-|x|}} \end{align}

is local lipschitz continuous?

I have written it as $f(x) = \frac{1}{\sqrt{1-x}}$ for $x \geq 0$ and $f(x) = \frac{1}{\sqrt{1+x}}$ for $x \leq 0$ and thought about showing that $f$ is continuously differentiable. Sadly, that's not the case.

Are there other ideas how one can prove the claim?

Answer 1

Hint: $\frac 1 {\sqrt {1-y}}-\frac 1 {\sqrt {1-x}}=\frac {\sqrt {1-x}-\sqrt {1-y}} {\sqrt {1-x}\sqrt {1-y}}=\frac {y-x} {\sqrt {1-x}\sqrt {1-y} (\sqrt {1-x}+\sqrt {1-y})} $. Now fix $x$ and show that the denominator stays away from $0$ for $y$ close enough to $x$. [$1-y >1-x -|x-y|$. Take $|x-y| <\frac {1-x} 2$].