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

[jens.w]

Homework Statement

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)

Homework Equations

The Attempt at a Solution

I don't have one. I am dumbfounded. Mostly i have been trying to understand the meaning of the composite function. I am 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 don't 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 don't know that. I don't know how that would help me, its just something i thought of.

 

[Mark44]

Homework Statement

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)

Homework Equations

The Attempt at a Solution

I don't have one. I am dumbfounded. Mostly i have been trying to understand the meaning of the composite function. I am not sure if this is correct but

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

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.

Then i have at least isolated f(x), whose limit is known, but the other factor i don't 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 don't know that. I don't know how that would help me, its just something i thought of.

 

[SammyS]

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

 

[jens.w]

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.

Oh ok, darn. I just tested it with f(x) = Ax + B, which 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?

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

Negative, since x³ < x there.

 

[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) ?

 

[HallsofIvy]

Am i wrong in assuming that if for instance

f(x)=Ax+B

then

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

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&lt; 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.

 

[jens.w]

Oh right, that's 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?

 

[gb7nash]

That's correct.

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

 

[jens.w]

Ok, since x^{3}+x &gt; 0 when 0 &lt; x &lt; 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?

 

[Dick]

Ok, since x^{3}+x &gt; 0 when 0 &lt; x &lt; 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?

I'd agree with that.

 

[jens.w]

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