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Math Help - Theorem on the composition of limits?

  1. #1
    Member mfetch22's Avatar
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    Theorem on the composition of limits?

    Is there a theorem on the composition of limits? What I mean is, is there some equation relating the limits of two functions with the limit of the composition of a function? And what about 'nested' limits? I know the following theorems about limits:

    \lim_{x \to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a}g(x)

    and

    \lim_{x \to a} [f(x)g(x)] = [\lim_{x \to a} f(x)] [\lim_{x \to a}g(x)]

    and

    IF \lim_{x \to a} f(x) \neq 0 then

    \lim_{x \to a} \frac{1}{f(x)} = \frac{1}{\lim_{x \to a} f(x)}

    But what about something that would give a simplification of the following limits:

    \lim_{x \to a} f(g(x)) \;\;\; \mathrm{or} \;\;\; \lim_{x \to a} g(f(x))

    (Assuming that we know \lim_{x \to a} f(x) and \lim_{x \to a} g(x), and assuming these limits exist)

    Is there anything we can say about the limit of the composition of these functions? And what about nested limits? Is there anything we can say about an expression like:

    \lim_{x \to a} [ \lim_{x \to a} f(x)]

    ? Is there any "rule" about nested limits that can be used to help in the manipulation of such an expression?

    Thanks in advance
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  2. #2
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    If you looks at this \lim_{x \to a} [ \lim_{x \to a} f(x)], it's kind of trivial. Once you find the inner limit, it'll be just a number or there won't be a limit. The outer limit is meaningless in this case.

    However, there is the following

     \lim_{y \to y_o} [ \lim_{x \to x_o} f(x,y)] =  \lim_{x \to x_o} [ \lim_{y \to y_o} f(x,y)]

    For this you need uniform convergence.
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  3. #3
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    Quote Originally Posted by mfetch22 View Post
    Is there a theorem on the composition of limits? What I mean is, is there some equation relating the limits of two functions with the limit of the composition of a function? And what about 'nested' limits? I know the following theorems about limits:

    \lim_{x \to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a}g(x)

    and

    \lim_{x \to a} [f(x)g(x)] = [\lim_{x \to a} f(x)] [\lim_{x \to a}g(x)]

    and

    IF \lim_{x \to a} f(x) \neq 0 then

    \lim_{x \to a} \frac{1}{f(x)} = \frac{1}{\lim_{x \to a} f(x)}
    Be very careful with these "rules". They pretty much only apply when the limits on the right hand side exist and are finite. If you forget about that, you will get the wrong answers to questions later on.
    But what about something that would give a simplification of the following limits:

    \lim_{x \to a} f(g(x)) \;\;\; \mathrm{or} \;\;\; \lim_{x \to a} g(f(x))

    (Assuming that we know \lim_{x \to a} f(x) and \lim_{x \to a} g(x), and assuming these limits exist)
    Well, If f is continuous at \lim_{x \to a} g(x), then \lim_{x \to a} f(g(x))=f(\lim_{x \to a} g(x)), and similarly for the other thing.
    Is there anything we can say about the limit of the composition of these functions? And what about nested limits? Is there anything we can say about an expression like:

    \lim_{x \to a} [ \lim_{x \to a} f(x)]

    ? Is there any "rule" about nested limits that can be used to help in the manipulation of such an expression?
    See what Vlasev said (and if you haven't heard of uniform convergence, don't worry about it).
    Thanks in advance
    No problem.
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  4. #4
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    Would you say that for any functions f(x) and g(x)
    \lim_{x \to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a}g(x)
    is wrong because it is only true when limits exist and are finite?
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  5. #5
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    Quote Originally Posted by infinitepersepctives View Post
    Would you say that for any functions f(x) and g(x)
    \lim_{x \to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a}g(x)
    is wrong because it is only true when limits exist and are finite?
    The usual statement includes "if all limits exist".
    But the statement without that is false if, for example, f(x)= 1/(x-a) and g(x)= -1/(x-a). In that case, f(x)+ g(x) exists and its limit, as x goes to a, is 0 while the limits on the right do not exist.
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  6. #6
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    Very helpful, thank you!
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