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

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jens.w
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Homework Statement



If [tex]\lim_{x\rightarrow 0+}f(x)=A[/tex]
and
[tex]\lim_{x\rightarrow 0-}f(x)=B[/tex] find

[tex]\lim_{x\rightarrow 0+}f(x^{3}-x)[/tex]


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

[tex]f(x^{3}-x) =f(x(x^{2}-1))= f(x)f(x^{2}-1)[/tex]

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.
 
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jens.w said:

Homework Statement



If [tex]\lim_{x\rightarrow 0+}f(x)=A[/tex]
and
[tex]\lim_{x\rightarrow 0-}f(x)=B[/tex] find

[tex]\lim_{x\rightarrow 0+}f(x^{3}-x)[/tex]


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

[tex]f(x^{3}-x) =f(x(x^{2}-1))= f(x)f(x^{2}-1)[/tex]
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.
jens.w said:
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 said:
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?

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

Negative, since x3 < x there.
 
Am i wrong in assuming that if for instance

[tex]f(x)=Ax+B[/tex]

then

[tex]f(x^{3}-x)=(Ax+B)^{3}-(Ax+B)[/tex] ?
 
jens.w said:
Am i wrong in assuming that if for instance

[tex]f(x)=Ax+B[/tex]

then

[tex]f(x^{3}-x)=(Ax+B)^{3}-(Ax+B)[/tex] ?
Yes, you are wrong. You have the order of composition backwards. If [itex]f(x)= x^3- x[/itex] then [itex]f(Ax+ B)= (Ax+B)^3- (Ax+B)[/itex].

With f(x)= Ax+B, then [itex]f(x^3- x)= A(x^3- x)+ B[/itex].

In any case, this has nothing to do with the original question. You know That [itex]x^3- x< 0[/itex] for x between 0 and 1 and, of course, goes to 0 as x goes to 0. And you know that [itex]\lim_{x\to 0^-}f(x)= B[/itex].
 
Oh right, that's how compositions are structured! Darn i have forgotten a lot.

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

[tex]\lim_{x\rightarrow 0-}f(x)[/tex]

Wich is B. The answer is B.

Did i understand it now?
 
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Ok, since [tex]x^{3}+x > 0[/tex] when [tex]0 < x < 1[/tex] and since [tex]x^{3}+x \rightarrow 0[/tex] when [tex]x \rightarrow 0[/tex] we should get

[tex]\lim_{x\rightarrow 0+}f(x^{3}+x) = A[/tex]

Am i right?
 
jens.w said:
Ok, since [tex]x^{3}+x > 0[/tex] when [tex]0 < x < 1[/tex] and since [tex]x^{3}+x \rightarrow 0[/tex] when [tex]x \rightarrow 0[/tex] we should get

[tex]\lim_{x\rightarrow 0+}f(x^{3}+x) = A[/tex]

Am i right?

I'd agree with that.
 
Thank you very much Mark44, SammyS, HallsofIvy, gb7nash and Dick.