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Math Help - proof of 2 limits

  1. #1
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    Question proof of 2 limits

    hi im stuck with this question for a while. its a real analysis topic... i was wondering if anyone can help me by
    the question:
    f is defined on (0,infinity) to R.

    i want to prove that lim x->infinity f(x) =L if and only if lim x->0 f(1/x)=L ..the x->0 is a positive so its right limit.

    my solution so far

    i decided to prove each one seperately and let y =f(x),
    but the thing is i was getting confused with the definitions and what to do next when using the definitions.

    so for the first equation

    lim x->infinity f(x) =L

    let epsilon>0, N is in natural numbers, y>=N => ! y- L ! < epsilon

    this doesnt seem quite right and i wasnt to sure what to do next.

    so for the second limit

    lim x->0 f(1/x)=L ..the x->0 is a positive so its right limit.

    let epsilon>0

    tilda = 1/epsilon

    next 0<x- 0 <1/epsilon

    this gives ! f(1/x) - L ! < epsilon

    since y= f(x) where the x has to be replaced then ! f(x)- L!< epsilon ...hence the quesiton is proved...

    can some one check this and help me correct it please thanks
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  2. #2
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    Quote Originally Posted by dopi View Post
    hi im stuck with this question for a while. its a real analysis topic... i was wondering if anyone can help me by
    the question:
    f is defined on (0,infinity) to R.

    i want to prove that lim x->infinity f(x) =L if and only if lim x->0 f(1/x)=L ..the x->0 is a positive so its right limit.
    You are missing something, it should be \lim_{x\to 0^+} f\left( \frac{1}{x} \right) = L.

    Let 0,+\infty)\mapsto \mathbb{R}" alt="f0,+\infty)\mapsto \mathbb{R}" /> and define 0,+\infty)\mapsto \mathbb{R}" alt="g0,+\infty)\mapsto \mathbb{R}" /> as g(x) = f\left( \frac{1}{x} \right).


    I will use the sequencial approach to limit because this approach is the nicest. Let x_n be a sequence in (0,\infty) converging to 0, i.e. \lim \ x_n = 0. Then, we know (this is an easy theorem, but try to prove it) \lim \ \frac{1}{x_n} = +\infty. Thus, \frac{1}{x_n} is a sequence in (0,\infty) approaching +\infty. Thus, this means \lim f\left( \frac{1}{x_n} \right) = L. This implies then that \lim g(x_n) = \lim f\left( \frac{1}{x_n} \right) = L. Q.E.D.


    The theorem I used was let x_n be a sequence of positive real numbers which is convergent to zero. Then \lim \frac{1}{x_n} = +\infty. This theorem goes both ways.
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  3. #3
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    Given \lim _{x \to \infty } f(x) = L means that:
    \left( {\varepsilon  > 0} \right)\left( {\exists B > 0} \right)\left[ {x > B \Rightarrow \left| {f(x) - L} \right| < \varepsilon } \right].

    Select \left( {\delta  > 0} \right)\left[ {\delta  < \frac{1}{B}} \right]. Now if 0 < x < \delta  \Rightarrow B < \frac{1}{x} \Rightarrow \left| {f\left( {\frac{1}{x}} \right) - L} \right| < \varepsilon .
    Or \lim _{x \to 0^ +  } f\left( {\frac{1}{x}} \right) = L.
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