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Math Help - Finding this limit using Algebra

  1. #1
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    Finding this limit using Algebra

    Lim .... .... x-2
    x->-2 ....x^2-4

    I got 4 using algebra, but I did the thing where you put -1.999 instead of -2 into the function and got 1/2, and I tried an online limit calculator and got infinity...im so confused.
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  2. #2
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     \lim_{x \to 2}\frac{x-2}{x^2-4}

     \lim_{x \to 2}\frac{x-2}{(x-2)(x+2)}

     \lim_{x \to 2}\frac{1}{x+2} = \frac{1}{2+2}= \frac{1}{4}
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  3. #3
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    Quote Originally Posted by pickslides View Post
     \lim_{x \to 2}\frac{x-2}{x^2-4}

     \lim_{x \to 2}\frac{x-2}{(x-2)(x+2)}

     \lim_{x \to 2}\frac{1}{x+2} = \frac{1}{2+2}= \frac{1}{4}
    That would be right but the question is asking as x approaches -2
    Last edited by usmelikchees; August 30th 2009 at 08:10 PM.
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  4. #4
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    Apply Hospital's rule and you get \frac{-1}{4}:

    \lim_{x\rightarrow-2}\frac{x-2}{x^2 - 4} =<br />
\lim_{x\rightarrow-2}\frac{\frac{d}{dx}(x-2)}{\frac{d}{dx}(x^2 - 4)} =<br />
\lim_{x\rightarrow-2}\frac{1}{2x} =<br />
\frac{1}{2(-2)} = \frac{-1}{4}<br />
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  5. #5
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    Quote Originally Posted by eXist View Post
    Apply Hospital's rule and you get \frac{-1}{4}:

    \lim_{x\rightarrow-2}\frac{x-2}{x^2 - 4} =<br />
\lim_{x\rightarrow-2}\frac{\frac{d}{dx}(x-2)}{\frac{d}{dx}(x^2 - 4)} =<br />
\lim_{x\rightarrow-2}\frac{1}{2x} =<br />
\frac{1}{2(-2)} = \frac{-1}{4}<br />
    Ok thanks for the answer if its not too much trouble can i ask what the d/dx thing is?
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  6. #6
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    When you apply Hospital's rule you take the derivative of the top over the derivative of the bottom.

    That notation: \frac{d}{dx} just means that I'm taking the derivative of those two as if they were their own monomial, and ignoring the fact that one is actually in the denominator. Just a notation I use.
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  7. #7
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    Quote Originally Posted by eXist View Post
    When you apply Hospital's rule you take the derivative of the top over the derivative of the bottom.

    That notation: \frac{d}{dx} just means that I'm taking the derivative of those two as if they were their own monomial, and ignoring the fact that one is actually in the denominator. Just a notation I use.
    Ah yes thank you very very much.
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  8. #8
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    L'hospital's rule is.

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

    if \lim_{x\to a} g(x) = \lim_{x\to a} h(x) = 0

    then \lim_{x\to a} f(x)=\frac{g(x)}{h(x)} = \lim_{x\to a} f(x)=\frac{g'(x)}{h'(x)}

    Can L'hospital's rule be used here?
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  9. #9
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    Quote Originally Posted by pickslides View Post
    L'hospital's rule is.

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

    if \lim_{x\to a} g(x) = \lim_{x\to a} h(x) = 0

    then \lim_{x\to a} f(x)=\frac{g(x)}{h(x)} = \lim_{x\to a} f(x)=\frac{g'(x)}{h'(x)}

    Can L'hospital's rule be used here?
    Yes it can
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  10. #10
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    Quote Originally Posted by usmelikchees View Post
    Yes it can
    Actually, you can't, since the numerator doesn't approach either infinity or 0 when x\to -2

    In fact, the limit \lim_{x \to -2}\frac{x-2}{x^2-4} does not exist:

    \lim_{x \to -2}\frac{x-2}{x^2-4} = \lim_{x \to -2}\frac{1}{x+2}

    But,

    \lim_{x \to -2^+}\frac{1}{x+2} = +\infty
    And:  \lim_{x \to -2^-}\frac{1}{x+2} = -\infty

    Thus, the limit does not exist when x\to -2.
    Last edited by Defunkt; August 31st 2009 at 02:56 PM. Reason: typo
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  11. #11
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    Quote Originally Posted by eXist View Post
    Apply Hospital's rule and you get \frac{-1}{4}:

    \lim_{x\rightarrow-2}\frac{x-2}{x^2 - 4} =<br />
\lim_{x\rightarrow-2}\frac{\frac{d}{dx}(x-2)}{\frac{d}{dx}(x^2 - 4)} =<br />
\lim_{x\rightarrow-2}\frac{1}{2x} =<br />
\frac{1}{2(-2)} = \frac{-1}{4}<br />
    L'Hopital's rule is not applicable here as the numerator is not indeterminate at x=-2, the limit does not exist (it is not + or - infty as the sign as x approaches -2 from the left and right is different)

    CB
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