# One sided limit question.

1. Jun 24, 2009

1. The problem statement, all variables and given/known data

If lim (x->0+) f(x) = A and lim (x->0-) f(x) = B

Find lim(x->0+) f(x^3-x)

2. Relevant equations

3. The attempt at a solution

I'm not sure how to do this. We know the right and left hand limits at x, how is it possible to find the right hand limit at x^3-x?

2. Jun 24, 2009

### Staff: Mentor

If you haven't already done so, take a look at the graph of g(x) = x3 -x = x(x2 - 1). As x --> 0+, what does g(x) approach? As x --> 0-, what does g(x) approach?

3. Jun 24, 2009

g(x) approaches 0 from both sides.

4. Jun 24, 2009

### Staff: Mentor

As x --> 0+, g(x) --> 0+, right?

5. Jun 24, 2009

Yes g(x) --> 0+

I don't know what i'm missing :/
That makes it lim (x->0+) f(0) ?

6. Jun 24, 2009

### Staff: Mentor

You're trying to find
$$\lim_{x \rightarrow 0^+} f(g(x))$$

As y -->0+, f(y) approaches which value, A or B?

7. Jun 24, 2009

It approaches A?

8. Jun 25, 2009

### HallsofIvy

Staff Emeritus
Let u= x3- x. u is a polynomial in x and all polynomials are continuous so, as x goes to 0, u goes to 0. BUT if x= 0.001, x3= 0.000000001 so x3- x= 0.000000001-0.001= -0.000999999. u is NEGATIVE for x between 0 and 1. If x= -.001, x3= -0.000000001 so x3- x= -0.000000001+0.001= 0.000999999. u is POSITIVE for x< 0.

Last edited: Jun 25, 2009
9. Jun 25, 2009

Thank you! Now i get it :)
So as x-->0+ f(x^3-x) has the limit B and as x-->0- it has the limit A.

All i needed to do was take some numbers and see what happens..
Is there a way to show this with symbols instead of words like this? Just wondering.

10. Jun 25, 2009

### HallsofIvy

Staff Emeritus
Well, you could note that x3- x= x(x- 1). If x< 0 both of those are negative so the x3- x> 0 for all x< 0 and if 0< x< 1, x is positive while x- 1 is negative so x3-x< 0 for 0< x< 1.