proving that a function doesn't satisfy lipschitz condition

Question

Prove that f(x)=$\sqrt{1-x^2}$ does not satisfy a Lipchitz condition on [-1,1]. My solution: Suppose on the contrary that f does satisfy a lipschitz condition on [-1,1], then $$|f(x)-f(y)|\leq M|x-y| $$ In particular, let y=1. $$|\sqrt{1-x^2}|\leq M|x-1|$$ $$ \frac{\sqrt{1-x^2}}{1-x}\leq M$$ However $$\lim_{x\to1}\frac{\sqrt{1-x^2}}{1-x}= \infty $$ It's just a sketch, but is this a valid proof?

Answer 1

You can make it more explicit with no need to talk about limits. If you consider $x=1-1/n$, then $$ M>\frac{\sqrt{1-(1-1/n)^2}}{1/n}=n\,\sqrt{2/n-1/n^2}=\sqrt n\,\sqrt{2-1/n}>\sqrt n. $$ As this can be done for every $n>0$, the desired number $M$ cannot exist.