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Math Help - Limit question

  1. #1
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    Limit question

    Determine Lim x->Infinity Sqrt[4 x^2 + 7 x - 2] - (2 x + 1)
    I know the answer, but i dont know the steps to get to it..
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  2. #2
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    Quote Originally Posted by Monster32432421 View Post
    Determine Lim x->Infinity Sqrt[4 x^2 + 7 x - 2] - (2 x + 1)
    I know the answer, but i dont know the steps to get to it..

    Try to rationalise the numerator .
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  3. #3
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    Quote Originally Posted by Monster32432421 View Post
    Determine Lim x->Infinity Sqrt[4 x^2 + 7 x - 2] - (2 x + 1)
    I know the answer, but i dont know the steps to get to it..

    Write \left[\sqrt{4x^2+7x-2}-(2x+1)\right]\,\frac{\sqrt{4x^2+7x-2}+(2x+1)}{\sqrt{4x^2+7x-2}+(2x+1)} =\frac{3x-3}{\sqrt{4x^2+7x-2}+(2x+1)} , and now multiply the right hand by \frac{1/x}{1/x} and do

    a little algebra + some arithmetic of limits.

    The limit is \frac{3}{4}=0.75

    Tonio
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  4. #4
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    also a little care , you may make mistakes very easily .
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  5. #5
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    Quote Originally Posted by tonio View Post
    Write \left[\sqrt{4x^2+7x-2}-(2x+1)\right]\,\frac{\sqrt{4x^2+7x-2}+(2x+1)}{\sqrt{4x^2+7x-2}+(2x+1)} =\frac{3x-3}{\sqrt{4x^2+7x-2}+(2x+1)} , and now multiply the right hand by \frac{1/x}{1/x} and do

    a little algebra + some arithmetic of limits.

    The limit is \frac{3}{4}=0.75

    Tonio
    thanks
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  6. #6
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    {\sqrt{4x^2+7x-2}\over x} = \sqrt{4x^2+7x-2\over x^2} = \sqrt{4 + \frac7x -\frac2{x^2}}
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