Limit of $\sqrt{x-x^2}$ as $x$ approaches $1$ I'm certain that the result is 0, but my book does not quite agree with me, infact it says that the limit is undefined, is the book wrong? Looking at the domain of f it should be correct supposing the limit as 0.. Am I doing something wrong?
 A: Assume $$\lim_{x\to 1}\sqrt{x-x^2}=0.$$
Then, $$\lim_{x\to 1^+}\sqrt{x-x^2}=\lim_{x\to 1^-}\sqrt{x-x^2}=0.$$
But noting $$\sqrt{x-x^2}=\sqrt{x(1-x)}$$  we have that for $x> 1$, $x(1-x)$ is negative, and thus $\sqrt{x(1-x)}$ is not a real number. 
As a result, $\lim_{x\to 1^+}\sqrt{x-x^2}$ does not exist in $\mathbb{R}$, and so the same goes for $\lim_{x\to 1}\sqrt{x-x^2}.$ 
A: No, your book is not wrong.. x approaches to 1, that means x can be approach to 1 either from left side or right side. Let, x approaches from right side. i.e. $x>1$, and then $x<x^2$ and so $x-x^2<0$, so $\sqrt {x-x^2}$ doesn't exists, so limit won't exist. But if x approaches to 1 from left side, i.e. $x<1$, then limit will be $0$. 
A: $$\lim_{x\to1^-}\sqrt{x-x^2}=0$$
as expected.  But taking $\epsilon\gt 0$, we have $x=1+\epsilon\implies x-x^2=1+\epsilon-1-2\epsilon-\epsilon^2=-\epsilon-\epsilon^2,$ which is always negative for $\epsilon\gt 0$.  Taking the square root of such a number results in a non-real number quantity and is therefore outside the range of the function.
