# Square root limit with $\epsilon-\delta$

$$\lim_{x\to 0^+} \sqrt{x} = 0$$

Proof:

The definition for a "right" limit is:

$$\forall \epsilon > 0, \exists \delta_{\epsilon} > 0$$ such that $$\forall x \in \mathcal{D}$$ (where $$\mathcal{D}$$ is the domain of the function) for which it is true that $$0 < x - x_0 < \delta$$ we have $$|f(x) - \ell | < \epsilon$$.

So we have

$$\sqrt{x}| < \epsilon$$

Now we know the square root outputs only positive numbers (in its principal branch, right?) so the absolute value is useless. Whence $$x < \epsilon^2$$

At this point I can take $$\delta = \epsilon^2$$ such that we have $$0 < x < \epsilon^2$$ to prove $$|\sqrt{x} - 0| < \epsilon$$

Is this correct?

Now let's check for $$x\to 0^-$$ which doesn't exist. Following the similar definition (what changes for left limits is that we now have $$0 < x_0 - x < \delta$$):

$$|\sqrt{x} - 0 | < \epsilon$$

which implies that for $$0 < x_0 - x < \delta$$ I can prove $$|\sqrt{x}| < \epsilon$$. But the would mean

$$0 < -x < \delta \longrightarrow 0 > x > \delta$$

This is impossible for if $$\delta = \epsilon^2$$ as before, I cannot ever verify $$0 > x > \delta$$. So the limit doesn't exist.

Is this correct?

Eventually, how would I prove there is no limit as $$x\to 0$$ without specifying the direction?

• do you know that the domain of square root doesn't contain negative numbers. so it wrong to write $\sqrt{x}$ if $x<0$ Dec 2, 2021 at 12:00
• @AdityaDwivedi Yes, I was demanding a sort of "proof" as if I did not know it :) Dec 2, 2021 at 12:00
• you can't prove a wrong statement, however I see what you are saying Dec 2, 2021 at 12:01

Since $$\sqrt x$$ is undefined when $$x<0$$, it makes no sense to talk about $$\lim_{x\to0^-}\sqrt x$$. And it also follows that $$\lim_{x\to0^+}\sqrt x=\lim_{x\to0}\sqrt x$$, since both assertions $$\lim_{x\to0^+}\sqrt x=0$$ and $$\lim_{x\to0}\sqrt x=0$$ mean$$(\forall\varepsilon>0)(\exists\delta>0):|x|<\delta\wedge x>0\implies\left|\sqrt x\right|<\varepsilon.$$
• Isn't it wrong to say the function has limit in $0$? I know that for a function to have a limit at $x_0$ it has to have both defined limits at $x_0^+$ and $x_0^-$. That's also the definition of a continuous function, so we say $\sqrt{x}$ is continuous at $x = 0$ just because it makes no sense to talk about $\sqrt{x}$ when $x < 0$? Dec 2, 2021 at 19:20
• No. We say that the square root function has limit at $0$ (and that that limit is $0$) because $0$ is an accumulation point of its domain (which is $[0,\infty)$) and because$$(\forall\varepsilon>0)(\exists\delta>0):|x|<\delta\wedge x\in[0,\infty)\implies|\sqrt x|<\varepsilon.$$ Dec 2, 2021 at 19:30