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Math Help - Limits of indeterminant form.

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    Limits of indeterminant form.

    Limits of indeterminant form.-20110106320.jpgPlease help with questions 3 and 6.
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    Quote Originally Posted by niazk90 View Post
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ID:	20372Please help with questions 3 and 6.
    For Q3 note that \displaystyle \frac{\sqrt{x^2 - x}}{x-9} = \frac{\sqrt{1 - \frac{1}{x}}}{1 - \frac{9}{x}}.


    For Q6 note that by multiplying by \displaystyle \frac{\sqrt{2x+3} + \sqrt{x - 1}}{\sqrt{2x+3} + \sqrt{x - 1}} the expression becomes \displaystyle \frac{x + 4}{\sqrt{2x+3} + \sqrt{x - 1}}.
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