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Math Help - Sum and product of limits

  1. #1
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    Sum and product of limits

    Evaluate the limit or explain why it does not exist.

    1.) (1/x - 1/2 )(1/(x-2)) as x approaches 2

    2.) 2x/(x+4) + 8/(x+4) as x approaches -4

    I concluded that both 1 and 2 do not exist. *crosses fingers*
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  2. #2
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    How did you conclude that?

    #1: \lim_{x \to 2} \left(\frac{1}{x} - \frac{1}{2}\right)\left(\frac{1}{x-4}\right) = \lim_{x \to 2} \left(\frac{1}{x(x-4)} - \frac{1}{2(x-4)}\right) = \hdots

    #2: \lim_{x \to -4} \left(\frac{2x}{x+4} + \frac{8}{x+4}\right) = \lim_{x \to -4} \left(\frac{2x+8}{x+4}\right) = \hdots

    Factor out the numerator and what cancels ?
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  3. #3
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    Quote Originally Posted by NotEinstein View Post
    Evaluate the limit or explain why it does not exist.

    1.) (1/x - 1/2 )(1/(x-4)) as x approaches 2

    2.) 2x/(x+4) + 8/(x+4) as x approaches -4

    I concluded that both 1 and 2 do not exist. *crosses fingers*
    No on both counts.

    Obviously \frac{1}{x} - \frac{1}{2} \rightarrow 0 as x --> 2 ....

    The second one can be re-witten as \frac{2x + 8}{x+4} = \frac{2(x+4)}{x+4} ....

    Knowledge of basic algebra and its application is essential when studying limits.
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  4. #4
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    Quote Originally Posted by o_O View Post
    How did you conclude that?

    #1: \lim_{x \to 2} \left(\frac{1}{x} - \frac{1}{2}\right)\left(\frac{1}{x-4}\right) = \lim_{x \to 2} \left(\frac{1}{x(x-4)} - \frac{1}{2(x-4)}\right) = \hdots

    #2: \lim_{x \to -4} \left(\frac{2x}{x+4} + \frac{8}{x+4}\right) = \lim_{x \to -4} \left(\frac{2x+8}{x+4}\right) = \hdots

    Factor out the numerator and what cancels ?
    for #1 it was actually x-2 in the denominator of the 2nd function. i edited it, but i will look at your work and go from there
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    Quote Originally Posted by NotEinstein View Post
    for #1 it was actually x-2 in the denominator of the 2nd function. i edited it, but i will look at your work and go from there
    Then some simple algebra (get under a common denominator etc.) will re-arrange it into \frac{2-x}{2x(x - 2)} = ....
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  6. #6
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    Quote Originally Posted by mr fantastic View Post
    Then some simple algebra (get under a common denominator etc.) will re-arrange it into \frac{2-x}{2x(x - 2)} = ....
    Is there any way to justify this using the product limit law? first function being f(x) 2nd being g(x)? Or is that implied already?
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  7. #7
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    Quote Originally Posted by mr fantastic View Post
    Then some simple algebra (get under a common denominator etc.) will re-arrange it into \frac{2-x}{2x(x - 2)} = ....
    Quote Originally Posted by NotEinstein View Post
    Is there any way to justify this using the product limit law? first function being f(x) 2nd being g(x)? Or is that implied already?
    Actually I was expecting that you'd cancel the common factor and then take the limit .....
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