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The problem:

$\lim_{x\to 0}\sqrt{x^3-x}$

How I was thinking of solving it:

$\lim_{x\to 0}\sqrt{x^3-x} = \sqrt{0^3-0}=0$

My question:

Why does my math book say that the solution does not exist? How come it isn't zero?

Sien
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    The radicand (the part inside the radical) needs to be nonnegative. You get an inequality x^3-x>0 -> x^2-1>0 -> x^2>1. Therefore, x is not in that locality. Someone correct me if I'm wrong, that's just where my mind goes. –  Sep 20 '24 at 03:33
  • Could it possibly be because $\lim_{x\to 0-}\sqrt{x^3-x}$ does not exist? – Sien Sep 20 '24 at 03:35
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    \lim_{x\to0-}\sqrt{x} does not exist. in this case, we replace x with x^3-x, so x^3-x>0. This is similar to how x>0 for \sqrt{x}. So we get the domain problem. By the way, how do you compile the mathjax? –  Sep 20 '24 at 03:38
  • @Jasper has the right idea. To be explicit, try plugging $x = 1/10, 1/100$. Then, to deal with your $\lim_{x\to 0^-}$ question, try plugging $x=-1/10$... In general one needs $$x^3- x = x(x-1)(x+1)$$ to be positive for one to be able to take its square root. – peter a g Sep 20 '24 at 03:41
  • Assuming that you are dealing with the reals, the limit $\lim_{x\uparrow 0}\sqrt{x^3-x}$ exists, but the root is not defined for $x<0$. – copper.hat Sep 20 '24 at 03:42
  • @copper.hat still have an issue with your "not defined for ...$x<0$"... – peter a g Sep 20 '24 at 03:47
  • @peterag Yep, unfortunately my 5 mins passed, so I will have to live in MSE hell for eternity where open intervals are written as $]0,1[$. – copper.hat Sep 20 '24 at 03:49
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    When $0<x<1,$ $x^3-x<0.$ – Thomas Andrews Sep 20 '24 at 03:58
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    The book likely has defined limits to require the function to be defined in an open interval surrounding $0$, except perhaps at $0$. – Arturo Magidin Sep 20 '24 at 04:45
  • What exactly is the definition of $\lim_{x\to0} f(x)$ in the book? – David K Sep 20 '24 at 04:56

3 Answers3

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I hate this kind of problems because it is very imprecise. And in fact the book is plain wrong, or at least extremely misleading.

So first of all, why the book says what it says? That's because if $0<x<1$ then $x^3-x<0$, meaning $\sqrt{x^3-x}$ is not well defined for those $x$. So what the book wanted to say is "you cannot approach zero from above".

But this is misleading because the question is about domain really, not about the limit. The limit does exist inside the function domain (which includes $[-1,0]$ interval) and is zero there. Which indeed you can evaluate by substitution, since we are dealing with a composition of continuous functions.

So the real problem here is: limits are not entities on their own, they need domain, some surrounding space, and functions are not formulas only. Once we use proper definitions, e.g. function actually is domain + range + formula, then the answer becomes really simple.

freakish
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    IIRC there are some beginning textbooks that define limits only at the interior points of the domain of a function. If that is how the book defined a limit, the book's answer in this question is correct in that context, isn't it? But it is still true that in another (arguably more standard) context, the correct answer is that the limit is defined. (See my comment under the question asking what the definition is, to which has not yet been a response.) – David K Sep 23 '24 at 22:04
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We should ask: what is the domain of $f(x)=\sqrt{x^3-x}$?

and since $$x^3-x=x(x+1)(x-1)\ge 0 \implies -1\le x\le 0$$

as noticed in the comments

$$\lim_{x\to 0^-}\sqrt{x^3-x}=0$$

while $\lim_{x\to 0^+}\sqrt{x^3-x}$ is meaningless.

Note also that, according to the more general definition of limit, when we take the limit we are implicitely excluding the case $x\to 0^+$ and in this case the result

$$\lim_{x\to 0}\sqrt{x^3-x}=0$$

is perfectly fine.

Therefore the answer also depends upon to the definition you are referring to.

See also the related:

user
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The book is wrong and your reasoning is not sufficient.

They most probably meant that the limit does not exist because for small positive values of $x$ the square root is not defined, and the limit to $0^+$ would not exist. But this is wrong because in the definition of a limit only the points that belong to the domain must be considered, and in the given case, the limits to $0$ and to $0^-$ coincide and equal $0$.

When you say that the limit is $\sqrt{0^3-0}$, you don't address the existence of the limit, which demands that values of $x$ close to $0$ correspond to values of the function also close to $0$. You must explore the neighborhood, not a single point.

Justification of the limit:

$$|x-0|<\delta=\epsilon^2\implies |\sqrt{x^3-x}-0|<\epsilon$$