Finding One-Sided Limits Using a Cubic Function

[ahmadmz]

Homework Statement

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

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

Homework Equations

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?

 

[Mark44]

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?

 

[ahmadmz]

g(x) approaches 0 from both sides.

 

[Mark44]

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

 

[ahmadmz]

Yes g(x) --> 0+

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

 

[Mark44]

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

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

 

[ahmadmz]

It approaches A?

 

[HallsofIvy]

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.

 

[ahmadmz]

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.

 

[HallsofIvy]

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.