8

It is well known that

$$\lim_{x \to a} f(x) = L \iff \lim_{x \to a+}f(x) = L = \lim_{x \to a-}f(x)$$

Consider the function $\sqrt{.}: \mathbb{R}^+ \to \mathbb{R}$

Now, consider $\lim_{x \to 0} \sqrt{x}$

We can prove this limit is equal to $0$. Indeed, let $\epsilon > 0$. Choose $\delta = \epsilon^2$. Then, for $x \in \mathbb{R}^+$ satysfying $0 < |x| < \delta$ or equivalently $0 < x < \delta$, we have $\sqrt{x} < \sqrt{\delta} = \epsilon$, which establishes the result.

However, my confusion lies in the following: the limit from the left does not seem to exist, making the above theorem untrue. Where lies my mistake?

  • 3
    That function is continuous on its domain, which (as you know) does not include negative numbers. Your "well known" theorem about left and right limits is true only when the point is interior to the domain - an unmentioned part of the hypothesis (loosely speaking, but that's the idea). – Ethan Bolker Feb 05 '18 at 14:09
  • The equivalence is (only relevant and) only valid if the left and right-hand limits are defined, i.e. if $f$ is defined at least on $(a-\varepsilon,a+\varepsilon)$ for some $\varepsilon >0$. – StackTD Feb 05 '18 at 14:10
  • This is frustrating. I'm reading Spivak's calculus and he doesn't make those hypotheses. –  Feb 05 '18 at 14:11
  • When you are considering $\lim_{x\to a,\ x\in A}f(x)$ and $a\in A$ is an isolated point of $A$, then the limit is just $f(a)$. When you do $\lim_{x\to0^-}\sqrt{x}$ you are doing $\lim_{x\to0,\ x\in\text{Dom}(\sqrt{\cdot})\cap(-\infty,0]}\sqrt{x}$, but that intersection is only ${0}$ therefore, the limit from the right is $\sqrt{0}$. So, all is good. – orole Feb 05 '18 at 14:12
  • 1
    When I was undergraduate I used to prefer the notation $\lim_{x\to a,\ x\in A}$ to make it clear the set in which the $x$ is moving. It can get rid of those ambiguities in $x\to a^+$ and $x\to a^-$. The theorem should be written as $\lim_{x\to a,\ a\in\text{Dom}(f)}f(x)=L$ if and only if $\lim_{x\to a,\ x\in A\cap\text{Dom}(f)}f(x)=L$ and $\lim_{x\to a,\ x\in A^{c}\text{Dom}(f)}f(x)=L$, where $A$ is any set such that $a$ is in the closure of $A\cap \text{Dom}(f)$ and of $A^c\cap \text{Dom}(f)$. – orole Feb 05 '18 at 14:18
  • So, my example provides a counterexample to the theorem if we drop the assumption that $a$ has to be an interior point? –  Feb 05 '18 at 14:26
  • @Math_QED No, it doesn't. In your case $\lim_{x\to0^-}\sqrt{x}$, which is really $\lim_{x\to0,\ \text{Dom}(\sqrt{\cdot})\cap(-\infty,0]}\sqrt{x}=\sqrt{0}=0$. It is just a matter of carefully checking what the limit from the left is. – orole Feb 05 '18 at 14:46
  • But, the definition of limit says that (in this case) $x \neq 0$? So why does your interval contain that point. By the way, the condition $0 < -x < \delta$ is satisfied for all $x \in dom( \sqrt{.})$ (as there are none). So the implication $0 < -x < \delta \implies |f(x) - L| < \epsilon$ is true for any number $L$. Something's not right –  Feb 05 '18 at 14:52
  • @Math_QED There is no such $\delta$. Whatever $\delta$ you pick around $0$, there will always be an $x$ in there for which $0 < -x < \delta$ is true but $|f(x) - L| < \epsilon$ is false, since $f(x)$ is not defined for those $x$, making that implication false. – JuliusL33t Feb 05 '18 at 15:54
  • Ah I see, but I still can't see how $\lim_{x \to 0-} \sqrt{x} = 0$ –  Feb 05 '18 at 15:57

3 Answers3

8

Actually, what you wrote is not well known, and is in fact not even true.

What is well known is the fact that

Let $f$ be a function $\color{red}{\text{defined on some open neighborhood of $a$}}$. Then, $$\lim_{x \to a} f(x) = L \iff \lim_{x \to a+}f(x) = L = \lim_{x \to a-}f(x)$$

Always read the fine print.

5xum
  • 126,227
  • 6
  • 135
  • 211
  • 2
    Spivak's book does not mention this. –  Feb 05 '18 at 14:11
  • 1
    @Math_QED It doesn't mention it because it is not part of the theorem. It is not necessary. – orole Feb 05 '18 at 14:20
  • 1
    @orole That seems contradictory to me. 5xum says "read the fine print" but you say "it's not part of the theorem". To me, that implies that the "fine print" isn't in the theorem. – Brian J Feb 05 '18 at 16:44
1

In fact the limit exists in $x_0$ if and only if for all $x$ in the domain of the function we have$$\forall \epsilon>0\qquad,\qquad \exists \delta\qquad |x-x_0|<\delta\to |f(x)-f(x_0)|<\epsilon$$and the domain of $f(x)=\sqrt x$ in $x_0=0$ is only $\{x\ge 0\}$ where only the right limit needs to exist. This also can be generalized further to higher dimensions where all limits from all valid directions in a sufficiently small neighborhood of some point over a metric space of metric $d(.,.)$ must exist and be equal i.e.$$\forall \epsilon>0\qquad,\qquad \exists \delta\qquad d(x,x_0)<\delta\to d(f(x),f(x_0))<\epsilon$$

Mostafa Ayaz
  • 33,056
1

You should be careful while applying the $\varepsilon-\delta $ definition it says. Let $a$ be an accumulation (or limit point ) of $Dom(f)$ then $f$ is continuous at $a$ if

$$\forall~~\varepsilon>0, \exists~\delta>0:\color{blue}{\text{for every $x\in Dom(f),$ } 0<|x-a|<\delta }\implies |f(x)-f(a)|<\varepsilon $$

in your case $x<0$ is not in the domain of $\sqrt{\cdot}$

And then the notion of limit from the left at $x=0$ does not apply here,

Guy Fsone
  • 25,237
  • 1
    Where exactly do I wrongly apply the definition? –  Feb 05 '18 at 14:16
  • 1
    @Math_QED Read the last sentence – Guy Fsone Feb 05 '18 at 14:18
  • I'm not talking about continuity... Do you mean my proof that $\lim_{x \to 0} \sqrt{x} = 0$ is wrong? That's the only place where I use the definition. –  Feb 05 '18 at 14:20
  • 1
    @Math_QED you should add that $x\in Dom(\sqrt{\cdot})$ then the issue is gone. since i that case we cannot talk about the limit from the left. Yes your prooof is correct but don't forget the Dom(f) – Guy Fsone Feb 05 '18 at 14:22
  • I wrote $\mathbb{R}^+$, which is the domain I listed, didn't I? –  Feb 05 '18 at 14:23