-1

If there exists $\;\lim\limits_{x\to c}\sqrt{f(x)}\;,\;$ then there exists $\;\lim\limits_{x\to c} f(x)\;.$

If it is true, prove it. If it false, give a counterexample.

Could you also do the converse?

Any help would be appreciate.

Martin R
  • 113,040
Dalton Kozak
  • 15
  • 3
    These are equivalent because both $x\mapsto\sqrt{x}$ and $x\mapsto x^2$ are continuous for $x\ge0$. If $f(x)\lt0$ then neither implications really make sense. – Peter Foreman Oct 12 '20 at 19:32
  • 1
    Continuous functions commute with taking limits, so you get the same result if you take the limit first and then apply a continuous function, or if you apply a contentious function to the argument of the limit first and take the limit afterwards. – Gyro Gearloose Oct 12 '20 at 19:46
  • Are you familiar with the theorem that if $\sqrt$ is continuous then $\lim \sqrt{f(x)}= \sqrt{\lim f(x)}$. – fleablood Oct 12 '20 at 20:56

3 Answers3

2

There is a general theorem for the algebra of limits that

$$\lim_{x\to c}(F(x)G(x))=\left(\lim_{x\to c}F(x)\right)\left(\lim_{x\to c}G(x)\right)$$

if both limits on the right hand side exist. Letting $F(x)=G(x)=\sqrt{f(x)}$, we get

$$\lim_{x\to c}f(x)=\left(\lim_{x\to c}\sqrt{f(x)}\right)^2$$

In particular, the limit exists.

Barry Cipra
  • 79,832
0

According to the more general definition of limit that is

$$\lim_{x\rightarrow c} f(x) = L \iff \forall \varepsilon >0\, \exists \delta > 0: \forall x\in D\quad 0<\vert x-c\vert <\delta \implies \vert f(x)-L\vert <\varepsilon $$

we have that for $f(x)=\operatorname{sign} (x-c)$

$$\lim\limits_{x\to c}\sqrt{f(x)} =1 \quad \lim\limits_{x\to c}f(x)=N.E.$$

What is always true independently from the definition is that

$$\lim\limits_{x\to c}\sqrt{f(x)} =L \implies \lim\limits_{x\to c}\left|f(x)\right|=L^2$$

Refer also to the related

user
  • 154,566
  • Thank you, what do you mean by "sign"? Can you define what you mean? – Dalton Kozak Oct 12 '20 at 19:58
  • @DaltonKozak Refer to that sign function. – user Oct 12 '20 at 20:00
  • 1
    This answer makes no sense. – Kosh Oct 12 '20 at 20:39
  • @Kosh What is the point which is no sense? – user Oct 12 '20 at 20:42
  • 2
    @user, the function $\sqrt{\text{sign}(x-c)}$ is not defined for $x\lt c$. – Barry Cipra Oct 12 '20 at 20:44
  • Actually $\lim\limits_{x\to c}\sqrt{f(x)}$ is $\lim\limits_{x\to c^+}\sqrt{f(x)}$ and $\lim\limits_{x\to c^+}f(x)$ does exist ! – Angelo Oct 12 '20 at 20:47
  • 1
    @BarryCipra Yes indeed but according to the given definition of limit th elimit exists, it is the same for $\lim x \to 0 \log x$, we exclude form the domain the points where the function is not defined. – user Oct 12 '20 at 20:47
  • But we write $\lim\limits_{x\to 0^+}\log x$ not $\lim\limits_{x\to 0}\log x$ indeed the function is defined for $x>0$. – Angelo Oct 12 '20 at 20:50
  • @Angelo It depends upon the definition we are referring to. It is always a point of discussion here :) Take also a look to the given link. – user Oct 12 '20 at 20:53
  • @user, ah, I think I see now what you're getting at: the "$\forall x\in D$" is crucial. – Barry Cipra Oct 12 '20 at 20:57
  • @BarryCipra Yes indeed, it seems this is the "more general" definition of limit that is the definition which allows to define limits also for problematic cases. In the given link this point is discussed in great detail with some different point of view! – user Oct 12 '20 at 21:00
  • If $f(x)\ge0$ for all $x\in(a,b)$ and $a<c<b$, there are not counterexamples. – Angelo Oct 12 '20 at 21:02
  • @Angelo Yes of course, with these assumption it works well! My point is that when we state a general property or a theorem for limits we should always indicate the definition we are referring to. – user Oct 12 '20 at 21:05
  • But I think that even though the OP did not write it explicitly, the function $f(x)$ is non-negative for $x>c$ and for $x<c$, otherwise he would write $\lim\limits_{x\to c^+}\sqrt{f(x)}$. For that reason, I said your answer makes no sense. Of course your answer is formally correct according to the definition of limit. – Angelo Oct 12 '20 at 21:10
0

Let $\;a,\;b,\;c\in\mathbb{R}\;$ such that $\;a<c<b\;.$

Let $\;f(x):(a,b)\to\mathbb{R}\;$ be a function such that

$f(x)\ge0\;$ for all $\;x\in(a,b)\;.$

If there exists $\;\lim\limits_{x\to c}\sqrt{f(x)}=l\in\mathbb{R}\;,\;$ then

$\lim\limits_{x\to c} f(x)=\lim\limits_{x\to c}\left[\sqrt{f(x)}\cdot\sqrt{f(x)}\;\right]=\\\qquad\quad=\lim\limits_{x\to c}\sqrt{f(x)}\cdot\lim\limits_{x\to c}\sqrt{f(x)}=l^2\;.$

Conversely, if there exists $\;\lim\limits_{x\to c}f(x)=l\in\mathbb{R}\;,\;$ then $\;l\ge0.$

If $\;l=0\;,\;$ then $\;\lim\limits_{x\to c}\sqrt{f(x)}=0\;,\;$ indeed

$0\le f(x)<\epsilon^2\implies 0\le\sqrt{f(x)}<\epsilon\;.$

If $\;l>0\;,\;$ then $\;\lim\limits_{x\to c}\sqrt{f(x)}=\sqrt{l}\;,\;$ indeed

$0\le l-\epsilon^2<f(x)<l+\epsilon^2\implies\\\implies\sqrt{l}-\epsilon\le\sqrt{l-\epsilon^2}<\sqrt{f(x)}<\sqrt{l+\epsilon^2}\le\sqrt{l}+\epsilon\;.$

Angelo
  • 12,328