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Limits? changing and transformation

  1. Nov 11, 2012 #1
    okay, so I'm at the most elementary stage of learning limits and there are things which leave me baffled at times, namely two.
    1. lim (x -> a) f(x) = lim (x+k -> a+k) f(x)

    how? the physical reason behind this?

    2. the theorem to evaluate limits of the form --- 1^infinity
    if f(x)=g(x)=0 (lim: x->a)
    that is,
    lim (x -> a) [1+f(x)]^{1/g(x)} =
    e^{lim x -> a. f(x)/g(x))
     
  2. jcsd
  3. Nov 11, 2012 #2

    LeonhardEuler

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    Gold Member

    For 1, I don't know how technically you learned the definition of the limit, but you know the idea of the limit is that [itex]\lim_{x\rightarrow a}f(x)=L[/itex] means that f(x) is very close to L as long as x is close enough to a. "x is close enough to a" is literally the exact same condition as "x+k is close enough to a+k".

    For 2, you know that one definition of e is
    [tex]e=\lim_{x\rightarrow 0}(1+x)^{1/x}[/tex]
    Well,
    [tex](1+f(x))^{1/g(x)} = (1+f(x))^{\frac{f(x)}{f(x)g(x)}}=[(1+f(x))^{1/f(x)}]^{\frac{f(x)}{g(x)}}[/tex]
    If you call f(x) "t", you will see that as t->0, you get the limit to be e inside. You have to do a bit more proving to show that everything goes through as you would expect, but at least you see why it makes sens to expect that result.
     
  4. Nov 12, 2012 #3
    yes, for 1 I get that x is close to a, so x+k is close to a+k, but the doubt I have is how the result Is same?
     
  5. Nov 12, 2012 #4

    MarneMath

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    Education Advisor

    One way to see the first property is let y = x + k, and thus f(x) = f(y-k). From there it is pretty straight forward. I'm a bit to lazy to find a formal proof, but I imagine the argument would follow that line. To be honest, I never came across that property, so I'm not entirely sure if it's always true, but it seems right.
     
  6. Nov 12, 2012 #5

    HallsofIvy

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    [itex]\lim_{x\to a}f(x)= L[/itex] means that "'Given any [itex]\epsilon> 0[/itex], there exist [itex]\delta> 0[/itex] such that if [itex]|x- a|< \delta[/itex], then [itex]|f(x)- L|< \epsilon[/itex]".

    And it should be easy to see that we can replace x by x+ k and a by a+ k to get exactly the same result: |(x+k)- (a+ k)|= |x- a|.

    Be careful, this does NOT say that the limit "at x= a+ k" is the same as the limit "at x= a". We are still dealing with |f(x)- L|, NOT with |f(x+k)- L|.
     
  7. Nov 13, 2012 #6
    thank you! now it's clear. we just express a expression in some other way by changing the limit but the meaning and result remains the same
     
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