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Math Help - Limits

  1. #1
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    Limits

    Evaluate.

    1. \lim_{x \to \--2} \frac{x + 4}{x + 3} = 2

    2. \lim_{x \to \--2} \frac{x^2 + x - 2}{x^2 + 5x + 6} = -3

    3. \lim_{x \to \-2} \frac {\sqrt {(x-1)} - 1}{(x - 2)} = 0.5

    4. \lim_{x \to \-1} \frac{x - 1}{x^2 - 1} = 0.5

    Are my answers correct?
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Macleef View Post
    Evaluate.

    1. \lim_{x \to \--2} \frac{x + 4}{x + 3} = 2

    2. \lim_{x \to \--2} \frac{x^2 + x - 2}{x^2 + 5x + 6} = -3

    3. \lim_{x \to \-2} \frac {\sqrt {(x-1)} - 1}{(x - 2)} = 0.5

    4. \lim_{x \to \-1} \frac{x - 1}{x^2 - 1} = 0.5

    Are my answers correct?
    1.yes
    2.yes
    3.yes
    4.yes
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  3. #3
    Super Member Aryth's Avatar
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    1. \lim_{x\to -2}\frac{x + 4}{x + 3}

    = \frac{-2 + 4}{-2 + 3} = \frac{2}{1} = 2

    2. \lim_{x \to \--2} \frac{x^2 + x - 2}{x^2 + 5x + 6}

    = \lim_{x \to \--2} \frac{2x + 1}{2x + 5}

    = \frac{-4 + 1}{-4 + 5} = \frac{-3}{1} = -3

    3. \lim_{x\to 2}\frac {\sqrt {(x-1)} - 1}{(x - 2)}

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

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

    = \frac{1}{2\sqrt{(2 - 1)}}

    = \frac{1}{2}

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

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

    = \frac{1}{2}

    So, yes to all.
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