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Math Help - Could someone check my answers for limits please?

  1. #1
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    Could someone check my answers for limits please?

    Could someone check my calculus answers, please?

    1. Find lim x -> 3 (9-x^2)/(x^2-4x+3) I got -3/2

    2. Find lim x -> -1 (sqrt(5+x)-2)/1+x, I got 1/4 (Only the 5+x is under the square root, just to clarify.)

    If these are wrong, could someone please tell me the correct answer and how to get it so I know how?

    Thanks so much!
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  2. #2
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    Quote Originally Posted by anita123 View Post
    1. Find lim x -> 3 (9-x^2)/(x^2-4x+3) I got -3/2
    The rational function factors as:

    . . . . . \frac{(3\, -\, x)(3\, +\, x)}{(x\, -\, 3)(x\, +\, 1)}\, =\, \frac{-1(x\, -\, 3)(x\, +\, 3)}{(x\, -\, 3)(x\, +\, 1)}

    "Cancelling" gives us:

    . . . . . -\frac{x\, +\, 3}{x\, +\, 1}

    Evaluating at x = 3 results in:

    . . . . . -\frac{3\, +\, 3}{3\, +\, 1}\, =\, -\frac{6}{4}\, =\, -\frac{3}{2}

    So this looks good!

    Quote Originally Posted by anita123 View Post
    2. Find lim x -> -1 (sqrt(5+x)-2)/1+x, I got 1/4 (Only the 5+x is under the square root, just to clarify.)
    I will guess that all of "1 + x" is in the denominator, so the expression is as follows:

    . . . . . \frac{\sqrt{5\, +\, x}\, -\, 2}{1\, +\, x}

    When dealing with problematic expressions containing radicals, it can often be useful to multiply by the conjugate, so let's try that here:

    . . . . . \left(\frac{\sqrt{5\, +\, x}\, -\, 2}{1\, +\, x}\right)\left(\frac{\sqrt{5\, +\, x}\, +\, 2}{\sqrt{5\, +\, x}\, +\, 2}\right)

    . . . . . \frac{(5\, +\, x)\, -\, 4}{(1\, +\, x)\left(\sqrt{5\, +\, x}\, +\, 2\right)}

    . . . . . \frac{1\, +\, x}{(1\, +\, x)\left(\sqrt{5\, +\, x}\, +\, 2\right)}

    "Cancelling" the common factor and then evaluating at x = -1 leads to the result you got. Excellent!
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