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Math Help - question regarding limit to infinity

  1. #1
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    question regarding limit to infinity

    Hi my question is this:
    If two functions, say f and g are defined on an open set (a,\infty), a\in\mathbb{R} where
    \lim_{x \to \infty}f(x)=L and \lim_{x \to \infty}g(x)=+\infty,
    how can I show that \lim_{x\to\infty}(f\circ g)(x)=L
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by dannyboycurtis View Post
    Hi my question is this:
    If two functions, say f and g are defined on an open set (a,\infty), a\in\mathbb{R} where
    \lim_{x \to \infty}f(x)=L and \lim_{x \to \infty}g(x)=+\infty,
    how can I show that \lim_{x\to\infty}(f\circ g)(x)=L
    Since g(x)\to\infty as x\to\infty, then \lim_{x\to\infty}f(g(x)) = \lim_{g(x)\to\infty}f(g(x)), which is L by definition.

    (See tonio's post below for a full-blown \epsilon-proof.)
    Last edited by redsoxfan325; October 19th 2009 at 07:22 PM.
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  3. #3
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    Quote Originally Posted by dannyboycurtis View Post
    Hi my question is this:
    If two functions, say f and g are defined on an open set (a,\infty), a\in\mathbb{R} where
    \lim_{x \to \infty}f(x)=L and \lim_{x \to \infty}g(x)=+\infty,
    how can I show that \lim_{x\to\infty}(f\circ g)(x)=L

    Choose any e > 0 ==> there exists a real number M s.t. |f(x) - L| < e whenever x > R.

    Since g(x) --> oo when x --> oo there exists a number T s.t. g(x) > R whenever x > T ==> for ANY x > T you get

    |f(g(x)) - L| < e , and this means f(g(x)) --> L when x --> oo

    Tonio
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  4. #4
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by tonio View Post
    Choose any e > 0 ==> there exists a real number M s.t. |f(x) - L| < e whenever x > R.
    You mean there exists a real number R s.t...
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  5. #5
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    Or he means "whenever x> M".
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