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Math Help - Limit problem: adding fractions with variables in the denom.

  1. #1
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    Limit problem: adding fractions with variables in the denom.

    \lim_{x \to 1}\frac{1}{x-1}+\frac{1}{x^2-3x+2}

    \lim_{x \to 1}\frac{1}{x-1}+\frac{1}{(x-1)(x-2)}

    This is where I get lost. Do I multiply (x-1) to both sides?
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  2. #2
    MHF Contributor chisigma's Avatar
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    Is...

    \frac{1}{x-1} + \frac{1}{(x-1)\cdot (x-2)} = \frac{x-2+1}{(x-1)\cdot (x-2)} = \frac{1}{x-2}

    ... so that...

    Kind regards

    \chi \sigma
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  3. #3
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    Quote Originally Posted by Monkee View Post
    \lim_{x \to 1}\frac{1}{x-1}+\frac{1}{x^2-3x+2}

    \lim_{x \to 1}\frac{1}{x-1}+\frac{1}{(x-1)(x-2)}

    This is where I get lost. Do I multiply (x-1) to both sides?
    You need a common denominator.

    So multiply the first fraction by \frac{x - 2}{x - 2}.


    Then it becomes:

    \lim_{x \to 1}\frac{x - 2}{(x - 1)(x - 2)} + \frac{1}{(x - 1)(x - 2)}

     = \lim_{x \to 1}\frac{x - 2 + 1}{(x - 1)(x - 2)}

     = \lim_{x \to 1}\frac{x - 1}{(x - 1)(x - 2)}

     = \lim_{x \to 1}\frac{1}{x - 2}.

    You should be able to go from here.
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  4. #4
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    Great!

    \lim_{x \to 1}\frac{1}{x - 2}

    = \frac{1}{1 - 2}

    = \frac{1}{-1}

    = -1

    But why do we use x-2 as a common denominator?
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  5. #5
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    Quote Originally Posted by Monkee View Post
    Great!

    \lim_{x \to 1}\frac{1}{x - 2}

    = \frac{1}{1 - 2}

    = \frac{1}{-1}

    = -1

    But why do we use x-2 as a common denominator?
    Because in order for you to add the fractions together, you need to have the denominators the same; thus you multiply both fractions as necessary to obtain a common denominator.
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  6. #6
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    Also, x - 2 is not the common denominator.

    (x - 1)(x - 2) IS.
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