1. Nov 5, 2015

### Steve Turchin

1. The problem statement, all variables and given/known data
If $lim_{x\to 0^+} f(x) = A$ and $lim_{x\to 0^-} f(x) = B$, find:
(a) $lim_{x\to 0^+} f(x^3-x)$
(b) $lim_{x\to 0^-} f(x^3-x)$

2. Relevant equations
None. Only the problem statement.

3. The attempt at a solution
(a) $lim_{x\to 0^+} f(x^3-x) = lim_{x\to 0^+} f(x) \cdot lim_{x\to 0^+} f(x^2-1) = -A$

This "attempt" definitely brings a false answer.I just couldn't think of any way to solve this. Guidance would be greatly appreciated! Thanks.

2. Nov 5, 2015

### andrewkirk

In your attempt you have assumed that $f(ab)=f(a)f(b)$, which will be false for most functions.

Instead, write $y=x^3-x$ and then work out $\lim_{x\to 0^+}y$ and $\lim_{x\to 0^-}y$. Express the answers as $0^+$ or $0^-$, to indicate approaching the limit from above or below, if that fits.

Then use that to work out $\lim_{y\to 0^+}f(y)$ and $\lim_{y\to 0^-}f(y)$.

3. Nov 5, 2015

### haruspex

I cannot see how that will help.
What are the possible answers for a?
As $x \to 0$, what does x3-x tend to?
If $x \to 0$, what extra do you need to know in order to decide $\lim_{x\to 0} f(x)$?

4. Nov 6, 2015

### andrewkirk

Just use the limit chain rule.
It requires some stronger conditions than the other limit rules, but the second of the two alternative conditions outlined in the link is satisfied by the function in the OP.

5. Nov 6, 2015

### haruspex

Your advice was to find the limiting values of x3-x as x tends to zero from each side. Since those two answers will be the same, it cannot help in solving the problem. Maybe you meant something else.

6. Nov 6, 2015

### andrewkirk

Sure it can. You just do the two cases separately, once with $f$ and $g$ having domains of $(0,\infty)$ and once with their having domains of $(-\infty,0)$. Then the limits are different. That's how I interpret statements like $\lim_{x\to 0^+}f(x)$. What I wrote was a short-hand for that. Perhaps the shorthand was a bit too short and clarity suffered.

7. Nov 6, 2015

### haruspex

You wrote:
Ok, let's do that.
$\lim_{x\to 0^+}y=0$, $\lim_{x\to 0^-}y=0$. Since f(0) is undefined, this appears to be a dead end.

8. Nov 6, 2015

### Steve Turchin

I think a possible answer would be the limit of $f \circ g$ as $g(x)=x^3-x$ .
$\lim_{x\to 0} x^3-x = 0$
I believe that I need to know whether or not the function is continuous at $x=0$.

Thanks for the replies. Here's what I tried next, using the chain limit rule which andrewkirk posted:

$g(x)=x^3-x \ \ \ \ f \circ g =f(x^3-x)$
(a) $\lim_{x\to 0^-}f(x)=B$ and $\lim_{x\to 0^+} g(x)=0^-$ (Since $x^3-x<0$ for $0<x<1 ) \Rightarrow \lim_{x\to 0^+}f(x^3-x)=B$

Is this reasonable?
same for (b):
$\lim_{x\to 0^+}f(x)=A$ and $\lim_{x\to 0^-} g(x)=0^+$ (Since $x^3-x>0$ for $0>x>-1 ) \Rightarrow \lim_{x\to 0^-}f(x^3-x)=A$

This is exactly what I thought at first. Which made me wonder if $g \circ f$ would work.

Last edited: Nov 6, 2015
9. Nov 6, 2015

### andrewkirk

That's the general idea, yes. To be fully rigorous, we need to operate with functions that restrict domains to one side of the y axis.

So for instance we write $f^\dagger\equiv f\bigr|_{(-\infty,0)}$ as the function $f$ with domain restricted to the negative reals, and $g^\ddagger\equiv g\bigr|_{(0,\infty)}$. Then we can formally say that $\lim_{x\to 0}g^\ddagger(x)=0$ and $\lim_{y\to 0}f^\dagger(y)=B$ without having to use the somewhat informal $0^+$ and $0^-$ notation. We can then use the limit chain rule to conclude that $\lim_{x\to 0} f^\ddagger(g^\dagger(x))=B$. For the chain rule to be applicable, we require there to be a neighbourhood of $x=0$ in which $g^\dagger(x)$ is always nonzero, but that is satisfied by the given function $x^3-x$ on the restricted domain $(0,\infty)$, so it all goes through.

For the other case, we swap the domains of $f^\ddagger$ and $g^\dagger$ and use the same approach to get the result $A$ instead.

10. Nov 6, 2015

### haruspex

Yes, that works, but it is not the chain limit rule as such, which would be too blunt an instrument here.
Glad you figured out this is what Andrew meant.