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Math Help - Limits

  1. #1
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    Limits

    How could I show by means of an example that lim x ->a [f(x) + g(x)] may exist even though neither lim x->a f(x) nor lim x->a g(x) exists?
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    Quote Originally Posted by colerelm1 View Post
    How could I show by means of an example that lim x ->a [f(x) + g(x)] may exist even though neither lim x->a f(x) nor lim x->a g(x) exists?
    Hint let \displaystyle f(x)=\frac{1}{x-a} and let g(x)=-f(x)

    So f(x)+g(x)=0 but...
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  3. #3
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    Quote Originally Posted by colerelm1 View Post
    How could I show by means of an example that lim x ->a [f(x) + g(x)] may exist even though neither lim x->a f(x) nor lim x->a g(x) exists?
    If it is an example you seek,

     \mathop {\lim} \limits_{ x \to 3}  ( \frac{1}{x-3} - \frac{6}{x^2-9})

    Where a = 3 and f(x) is the left function and g(x) is the right function. We can see that neither of these limits exist by themselves! But, we can combine them,

      \mathop {\lim} \limits_{ x \to 3}   ( \frac{1}{x-3} - \frac{6}{x^2-9}) = \mathop {\lim} \limits_{ x \to 3}   ( \frac{1}{x-3} - \frac{6}{(x-3)(x+3)}) = \mathop {\lim} \limits_{ x \to 3}   \frac{ x+3 - 6}{(x-3)(x+3)} =   \mathop {\lim} \limits_{ x \to 3}   \frac{ x- 3}{(x-3)(x+3)} = \frac{1}{6}

    If you want to actually prove this result for all cases then that would probably be a bit more trickseee (as golunm would say)
    Last edited by AllanCuz; February 16th 2011 at 01:03 PM.
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