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

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?

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

1 Answer 1

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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$$(\forall\varepsilon>0)(\exists\delta>0):|x|<\delta\wedge x>0\implies\left|\sqrt x\right|<\varepsilon.$$

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  • $\begingroup$ 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 $x < 0$? $\endgroup$ Dec 2, 2021 at 19:20
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    $\begingroup$ 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$$(\forall\varepsilon>0)(\exists\delta>0):|x|<\delta\wedge x\in[0,\infty)\implies|\sqrt x|<\varepsilon.$$ $\endgroup$ Dec 2, 2021 at 19:30
  • $\begingroup$ Right, accumulation point! Thank you! $\endgroup$ Dec 2, 2021 at 20:42

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