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Our teachers seem to ignore that, but if I take a limit of a function with square root, for example: $$\lim_{x \to 4}{(\sqrt x-2)}$$ is it appropriate to say that there are two answers? $0$ and $-4$?

Because in the solutions they gave us, nobody mentions the fact that $\sqrt {x^2}=|x|$, and they only say that the answer is the positive result. In the example - $0$.

Is it correct? if so, why?

Thank you.

Blue
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Netanel
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  • $\sqrt{x^2} = |x|$ is in the definition of square root. – OmG Jan 13 '19 at 10:41
  • no a limit of real numbers can only have one value. Note that $\sqrt{x}>0$ so there's no way you get a limit that is less than $-2$. – Yanko Jan 13 '19 at 10:44
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    No, by convention/definition, the square root function always returns a non-negative value, so $\sqrt{x} = \vert x\vert \geq 0$. Here, $\sqrt{4} = +2$ only. – KM101 Jan 13 '19 at 10:48
  • @KM101 I don't mean to be picky here, but I think you mean $\sqrt{x^{\color{red}{2}}}=|x|\geq0$ for which you did not write the $2$, but we all know what you mean ;) – Mr Pie Jan 13 '19 at 11:17
  • See this question and the linked questions for a discussion of the properties of the square root function. – N. F. Taussig Jan 13 '19 at 11:19
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    It's true that $x^2=4$ has two solutions: $x=\pm 2$. (More generally, $x^2=m$ has two solutions $x=\pm \sqrt{m}$. (Well, it's only one solution if $m=0$, but I digress...)) That is not the same thing as saying $\sqrt{4}$ has two values, $\pm 2$. When we write "$\sqrt{4}$", we mean the "principal square root of $4$"; that is, the non-negative number whose square is $4$; that number is $2$ and only $2$. (Things get a little messier among the complex numbers, but that's a discussion for another time.) – Blue Jan 13 '19 at 11:21
  • @user477343 Oh yes haha, I better pay more attention! :-) – KM101 Jan 13 '19 at 11:24
  • @KM101 nah, I think I have to start looking at the bigger picture xD – Mr Pie Jan 13 '19 at 12:32

2 Answers2

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This is a common misconception about the function $\sqrt{\cdot}\,:[0,\infty)\to\Bbb R$ as the "inverse" of $f(x)=x^2$.

In general, a function can have an inverse only when it is injective but $f(x)=x^2$ is not injective (e.g. $f(2)=f(-2) = 4$ ), so strictly speaking its inverse doesn't exist.

However, we can restrict its domain to $[0,\infty)$ and the function $f\lvert_{[0,\infty)}$ would be injective there, hence we can define $$\sqrt{x}:=f\lvert_{[0,\infty)}^{-1}.$$ The function $\sqrt{x}$ thus takes only one positive value for each input $x\ge 0$ (e.g. $\sqrt{4}=2$) even though there are two numbers such that $x^2=4$.

Remark: There are some cases where $\sqrt{\cdot}$ is defined to be multivalued, e.g. in complex analysis. However, $\sqrt{\cdot}$ wouldn't be a "function" in that case. Note also that we could very well define $$\sqrt{x}:=f\lvert_{(-\infty,0]}^{-1}$$ and we'd have $\sqrt{4} = -2$ according to this definition. This is not popular for obvious reasons.

BigbearZzz
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According to definition of principal square root this is impossible. The principal square root of $x\ge0$ is denoted by $\sqrt x$ and is always non-negative.

Mostafa Ayaz
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