# Right limit of a composite function. Original functions limits are known.

1. Dec 21, 2011

### jens.w

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

If $$\lim_{x\rightarrow 0+}f(x)=A$$
and
$$\lim_{x\rightarrow 0-}f(x)=B$$ find

$$\lim_{x\rightarrow 0+}f(x^{3}-x)$$

2. Relevant equations

3. The attempt at a solution

I dont have one. Im dumbfounded. Mostly i have been trying to understand the meaning of the composite function. Im not sure if this is correct but

$$f(x^{3}-x) =f(x(x^{2}-1))= f(x)f(x^{2}-1)$$

Then i have at least isolated f(x), whose limit is known, but the other factor i dont know what to do about.
I've been thinking about the meaning of the limits for the original function f(x). Since A is not equal to B, f(x) is not even. f(x) could be odd, but we dont know that. I dont know how that would help me, its just something i thought of.

Last edited: Dec 21, 2011
2. Dec 22, 2011

### Staff: Mentor

No, it's not correct at all. It is not generally true that f(ab) = f(a)f(b).

Under certain conditions, though, you can reverse the order of taking a limit and evaluating a function. Your textbook probably has a theorem that deals with this.

3. Dec 22, 2011

### SammyS

Staff Emeritus
What is the sign of (x3-x), when 0 < x < 1 ?

4. Dec 22, 2011

### jens.w

Oh ok, darn. I just tested it with f(x) = Ax + B, wich was stupid.
I was trying to find a theorem about it in my textbook (Calculus by Adams) but failed. Any tips on books that can teach me more about composite functions?

Negative, since x3 < x there.

5. Dec 22, 2011

### jens.w

Am i wrong in assuming that if for instance

$$f(x)=Ax+B$$

then

$$f(x^{3}-x)=(Ax+B)^{3}-(Ax+B)$$ ?

6. Dec 22, 2011

### HallsofIvy

Staff Emeritus
Yes, you are wrong. You have the order of composition backwards. If $f(x)= x^3- x$ then $f(Ax+ B)= (Ax+B)^3- (Ax+B)$.

With f(x)= Ax+B, then $f(x^3- x)= A(x^3- x)+ B$.

In any case, this has nothing to do with the original question. You know That $x^3- x< 0$ for x between 0 and 1 and, of course, goes to 0 as x goes to 0. And you know that $\lim_{x\to 0^-}f(x)= B$.

7. Dec 23, 2011

### jens.w

Oh right, thats how compositions are structured! Darn i have forgotten alot.

Ok, so since we have x^3 - x in place of x, in the original function f(x), and x^3 - x < 0 when 0 < x < 1, we have created the situation

$$\lim_{x\rightarrow 0-}f(x)$$

Wich is B. The answer is B.

Did i understand it now?

Last edited: Dec 23, 2011
8. Dec 23, 2011

### gb7nash

That's correct.

What about the case $\lim_{x \to 0^+} f(x^3+x)$?

9. Dec 23, 2011

### jens.w

Ok, since $$x^{3}+x > 0$$ when $$0 < x < 1$$ and since $$x^{3}+x \rightarrow 0$$ when $$x \rightarrow 0$$ we should get

$$\lim_{x\rightarrow 0+}f(x^{3}+x) = A$$

Am i right?

10. Dec 23, 2011

### Dick

I'd agree with that.

11. Dec 23, 2011

### jens.w

Thank you very much Mark44, SammyS, HallsofIvy, gb7nash and Dick.