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Limit of exp -> exp (limit) ?

  1. Oct 25, 2014 #1
    1. The problem statement, all variables and given/known data

    Okay, so the problem is to find lim (x→∞) (e^x + x) ^ (1/x)

    I was given the solution in the assignment in which the first step was to take the natural log of the function, then exponentiate it.

    i.e.

    lim (x→∞) [exp ln( (e^x + x) ^ (1/x))]

    which I understand.

    However the next step, the solution equated this with:

    exp [ lim (x→∞) ln( (e^x + x) ^ (1/x))]

    which I became confused with. Are the two equal? Am I missing out on something?

    This assignment we're doing l'Hospital's Rule, if that's of any help...

    I probably won't be able to understand too abstract math, currently doing my first year in university...

    Thanks.
     
  2. jcsd
  3. Oct 25, 2014 #2

    pasmith

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    There is a theorem which states that if [itex]f[/itex] is continuous then [tex]
    \lim_{x \to a} f(g(x)) = f( \lim_{x \to a} g(x)).[/tex] (This implicitly asserts that either both limits exist and are equal, or neither of them exist.)
     
  4. Oct 25, 2014 #3
    Remember that before you can apply L'Hospital's rule you must convert the function into a 0/0 or ∞/∞ form.
     
  5. Oct 25, 2014 #4

    Ray Vickson

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    I would do it differently: (x+e^x)^(1/x) = (a b)^c = (a^c) (b^c), where a = e^x, b = 1+ x e^(-x), and c = 1/x.
     
  6. Oct 25, 2014 #5

    mfb

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    Staff: Mentor

    Then you still have to show x e^(-x) --> 0 for x->inf, but it is still an elegant solution as this limit was probably studied before.
     
  7. Oct 26, 2014 #6
    Ah. All make sense now. Is there anywhere I can read up on this rule? I don't think I was taught this, unless it's derivable from what I already know. Either that or I've been a horrible student. Haha..

    So it doesn't have to be a limit at infinity for this rule to hold, can be any limit right?

    Also didn't know this was considered calculus... Shows that I don't even know what I'm studying.
     
  8. Oct 26, 2014 #7

    Ray Vickson

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    The concept you want is "continuity", or "continuous functions". If your textbook does not have it, use Google. Basically, a function ##f(x)## is said to be continuous at ##x_0## if ##\lim_{x \to x_0} f(x) = f(x_0)##. Note that if ##f(x)## is continuous at ##x_0## and ##g(y)## is continuous at ##y_0 = f(x_0)##, then ##g(y) \to g(y_0)## as ##y \to y_0##, and ##y = f(x) \to y_0 = f(x_0)## as ##x \to x_0##, so we have ##g(f(x)) \to g(f(x_0))## as ##x \to x_0##. This just needs the continuity properties of ##f## and ##g## mentioned before.

    However, the less stringent result
    [tex] \lim_{x \to x_0} g(f(x)) = g(\lim_{x \to x_0} f(x))[/tex]
    requires weaker conditions: you need only have ##g(y)## continuous at ##y_0 = \lim_{x \to x_0} f(x)##; the function ##f(x)## need not be continuous (or even defined) at ##x = x_0##. Basically, that is the case you have, with ##x_0 = \infty##.

    After that (or, maybe before that) you need to worry about the question "are my functions continuous?""
     
    Last edited: Oct 26, 2014
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