Does the limit $\lim_{x\to 0} \sqrt{x^3 - x^2}$ exist or not?

Question

I am having some arguments with a friend about the following limit:

$$\lim_{x\to 0} \sqrt{x^3 - x^2}$$

FACTS: the domain of the function is $x\in \{0\}\cup [1,\ +\infty)$ and $0$ is an isolated point.

My friend says the limit doesn't exist, whilst to me it is $0$. Who is right?

In my opinion, if we take $\epsilon > 0$ and $\delta = 1/2$, we calculate $|f(x) - f(0)| < 0$ hence less than $\epsilon$, so for any positive $\epsilon$ we have positive $\delta$ such that whenever $x$ is in $\delta$-neighbourhood of $0$, the quantity $|f(x) - f(0)|$ is less than $\epsilon$ so $f$ is continuous at $0$.

Am I right or wrong?

Can someone pease make a limpid clarification of the existence (or not) of this limit? Thank you in advance!

Answer 1

Of course, any answer will depend on exactly which definition of a limit you are using. As was pointed out in the comments, one usually doesn't take limits at isolated points in the domain. From a definitions perspective, one could argue that that ANY limit is vacuously the limit of the function at $0$ although perhaps that's not very helpful in an argument with your friend. Perhaps it would be more enlightening to expand the domain of the function to include complex numbers as this will indeed give you the limit of zero. Consider the transformation $z=re^{i\theta}$. Then for small enough $r$ we have

$$|\sqrt{z^3-z^2}|=|z|\sqrt{z-1}=r\left|\sqrt{re^{i\theta}-1}\right|\leq 2r$$

Thus

$$0\leq\lim_{z\to 0}|\sqrt{z^3-z^2}|=\lim_{r\to 0^+}r\left|\sqrt{re^{i\theta}-1}\right|\leq \lim_{r\to 0^+}2r=0$$

Answer 2

I'm not sure about delta-epsilon proof, but I know a different way to compute the limit. $$\lim_{x\to 0}\sqrt{x^3 - x^2} = \lim_{x\to 0}\sqrt{x^2( x-1)}$$ Now if $x\to 0^+$ or $x\to 0^-$, in both of these cases, $(x-1)$ will be negative and hence, both, left-hand as well as right-hand limits are undefined. As a result, the overall limit doesn't exist.