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Is this proposition true?

  1. Dec 2, 2004 #1

    quasar987

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    Is the following true? Is so, under what conditions, and what are, roughly, the arguments used to prove it?

    f,g two functions of superiorly unbounded domain and such that for x > N, g(x) is continuous and f(g(x)) is continuous.

    [tex]\lim_{x \rightarrow \infty} f(g(x)) = f(\lim_{x \rightarrow \infty} g(x))[/tex]

    I'm trying to show what

    [tex]\lim_{x \rightarrow \infty} ln \left(\frac{x-1}{x+1} \right)=0[/tex]

    where ln is the natural logarithm (I think some people use the notation log for that). And without that "theorem", I don't see how to do it. :confused:
     
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  3. Dec 2, 2004 #2

    matt grime

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    log is continuous. (x-1)/(x+1) tends to 1 as x tends to infinity. you see how to put that together? (don't even know what superiorly bounded means....)
     
  4. Dec 2, 2004 #3

    quasar987

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    superiorly unbounded: Is that not a word? I meant to say the domain has no upper bound.

    I see how to put that together if the proposition I stated is true.

    I guess you're saying it is, then.
     
  5. Dec 2, 2004 #4

    matt grime

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    I have no desire to look at your proposition since:

    a function is continuous at a point x if lim as y tends to x of f(y) is f(x). So your result is a consequnce of the fact that log is continuous at 1, that's all.
     
  6. Dec 2, 2004 #5

    quasar987

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    So you're saying that if

    [tex]\lim_{x \rightarrow \infty} f(x) = L[/tex]

    And if g(x) is continuous at L, then

    [tex]\lim_{x \rightarrow \infty} g(f(x)) = \lim_{x \rightarrow L} g(x)[/tex]

    is that right?

    I can see from the definition of continuity how it's true for x --> x0 an accumulation point of the domain of f and if f(x0) is an accumulation point of the domain of g but for x --> infinity, I wasn't sure.
     
  7. Dec 2, 2004 #6

    shmoe

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    g doesn't have to be continuous at L for your last post to be true-it's just a statement of the limit of g, not how this limit relates to g(L).

    You need continuity of g at L if you want to assert that

    [tex]\lim_{x \rightarrow \infty} f(x) = L[/tex]

    implies

    [tex]\lim_{x \rightarrow \infty} g(f(x)) = \lim_{x \rightarrow L} g(x)=g(L)[/tex]

    which is what you're using for this problem.
     
  8. Dec 2, 2004 #7

    quasar987

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    Ok, so for my problem, since log is continuous at 1, all I have to do is prove that the function [itex]f(x) = (x-1)/(x+1)[/itex] as x approches infinity is 1. I have never done that. So if someone could check and see if these steps are correct, I'd be very grateful.

    We want to show that [itex]\forall \epsilon > 0[/itex], there is an M such that for x element of the domain of f (i.e element of the real), x > M implies

    [tex]\left |\frac{x-1}{x+1}-1 \right| < \epsilon[/tex]

    We see that

    [tex]\left|\frac{x-1}{x+1}-1 \right| = \left|\frac{x-1}{x+1}-\frac{x+1}{x+1} \right| = \left|\frac{x-1-x-1}{x+1} \right| = \left|\frac{-2}{x+1} \right| = \left|\frac{2}{x+1} \right| = \frac{2}{\left|{x+1} \right|}[/tex]

    and

    [tex]\frac{2}{\left|{x+1} \right|} < \epsilon \Leftrightarrow |x+1|>\frac{2}{\epsilon}[/tex]

    But [itex]|x+1| \leq |x|+1[/itex]. Therefor, if we can show that

    [tex]|x|+1>\frac{2}{\epsilon} \Leftrightarrow |x|>\frac{2}{\epsilon}-1 \ \forall \epsilon>0[/tex]

    we will have won. We can suppose without loss of generality that [itex]M \geq 0 \Rightarrow x>0 \Rightarrow |x| = x[/itex]. So that choosing M = max{0, [itex]\frac{2}{\epsilon}-1[/itex]} does the trick.
     
  9. Dec 2, 2004 #8

    shmoe

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    You hit a snag when you applied the triangle inequality. Knowing [itex]|x+1| \leq |x|+1[/itex] will not help you get a lower bound for [itex]|x+1|[/itex]. Try looking at [itex]|x+1|\geq |x|-|1|[/itex].
     
  10. Dec 2, 2004 #9

    quasar987

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    Right! I did that part too mechanically. Thanks!
     
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