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Does $\lim_{x\to 0^-} \sqrt{-x}$ exist?

I tried reading again and again the theory, and asking to people, even i ask to some AIs, and I don't get it.

If the limit approaches to $0$ from the left, it supposed to be $-0,000000001$ or something like that; but if that is right, when I do the substitution it shouldn't be: $\sqrt{-(-0,000000001)} = \sqrt{0,000000001}$?, or what am I doing wrong?

Blue
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Jlposada
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    You don't care about the limit from the right here, since you want the limit as $x \to 0^-$, i.e. the limit from the left. When you approach zero from the left, note that you will have, for instance, $x=-0.1$, so the function value is $\sqrt{-(-0.1)} = \sqrt{0.1}$ – PrincessEev Apr 18 '24 at 00:10
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    Yes, and it's the same as $\lim_{t\to 0^+}\sqrt{t}$ by substitution $t = -x$. – Ennar Apr 18 '24 at 00:18
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    Your equation $\sqrt{-(-0,000000001)} = \sqrt{0,000000001}$ is perfectly correct. Why do you think you did something wrong? – David K Apr 18 '24 at 00:19
  • Okay, I got it, so it exists?, I mean if the sqrt is positive it could exist, right? – Jlposada Apr 18 '24 at 00:22
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    By definition $a^{-}$ means you are taking it from the LEFT and you are not taking it from the right. So $\lim_{x\to 0} \sqrt{-x}$ doe not exist because the limit from the right does not exist even though from the left it does. And $\lim_{x\to 0^+}\sqrt{-x}$ does not exist because the limit from the right doesn't exist and we don't care about the left. And $\lim_{x\to 0^-}\sqrt{-x}$ does exist because the left limit does and we don't give a flying frigate in a jelly donut what the right limit does or does not do. – fleablood Apr 18 '24 at 00:24
  • Okay, fnlly I got it at all. Thx so much. – Jlposada Apr 18 '24 at 00:30

3 Answers3

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Because $x\rightarrow0^-$ we have that $x<0$ or $-x>0$. So this is another way of stating the limit $\lim_{x\rightarrow0^+}\sqrt{x}$ and you know the value of this limit.

Kamal Saleh
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By definition the $a^-$ means from the left. We simply do not care what happens to the right. And $a^+$ means from the right and we don't care what happens on the left. And $a$ means we care about both and hope they agree.

So $\lim_{x\to 0^-}\sqrt{-x}$ means $x <0$ and we just plain do not care about what happens when $x$ is positive.

So $\lim_{x\to 0^-}\sqrt{-x}$ does exist and is equal to $0$. For all very small $x$ where $x < 0$ we have $\sqrt{-x}\to 0$ as $x\to 0^-$. We do not care about very small positive $x$. We only have to consider $x < 0$.

$\lim_{x\to 0}\sqrt{-x}$ does not exist because even though $\sqrt{-x}\to 0$ for very small negative $x$, $\sqrt{-x}$ is not defined for very small positive $x$ and $x\to 0$ must consider both $x < 0$ and $x > 0$>

That's just what $a^{-}$ means. We made up that notation precisely to describe left side limits when the right side limit does not matter.

fleablood
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Continuing on the comment of PrincessEev, which more or less solves the problem, you could actually carry on and use the definion of a limit of a function to answer your question.

Suppose $f:\mathbb {R} \rightarrow \mathbb {R}$ is a function defined on the real line, and there are two real numbers $a$ and A.

One would say that the limit of $f$, as $x$ approaches $a$ from below, is $A$, if for every real $\varepsilon > 0$, there exists a real $\delta > 0$ such that for all real $x$, $0 < a − x < \delta$ implies $|f(x) − A| < \varepsilon$.

Now, in your problem the probable limit is equal to zero, and the point on the $x$-axis at which it happens is $a=0$. Thus, the limit exists if: $$ \forall \varepsilon > 0, \,\exists \delta(\varepsilon)> 0, \,\forall x: 0 < −x < \delta \Rightarrow |\sqrt{-x}| < \varepsilon. $$

Since $ −x < \delta$, it follows that $|\sqrt{-x}| < \sqrt{\delta}$, hence, taking $\delta = \varepsilon^2$ we prove the limit from below indeed exists and it's zero. The limit from above isn't defined because the square root of the negative variable isn't defined on a positive half-axis.

Egor Larionov
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