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Math Help - two limit problems

  1. #1
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    two limit problems

    please, Im having trouble finding these as x goes to infinity

    sqrt(2x+1)/x+4 limit is 0 and sqrt(x^2 + 2x) - x limit is 1

    for the first I take both sides by 1/x but end up getting sqrt(2)/1 and for the second im unsure. i have the answers because of the back of the book but I dont know how to get them
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  2. #2
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    \frac{\sqrt{2x+1}}{x+4} =

    \frac{\frac{\sqrt{2x+1}}{x}}{\frac{x+4}{x}} =

    \frac{\frac{\sqrt{2x+1}}{\sqrt{x^2}}}{\frac{x+4}{x  }} =

    \frac{\sqrt{\frac{2}{x}+\frac{1}{x^2}}}{1 + \frac{4}{x}}

    now take the limit.


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

    \frac{(x^2 + 2x) - x^2}{\sqrt{x^2 + 2x} + x} =

    \frac{2x}{\sqrt{x^2 + 2x} + x} =

    \frac{2}{\sqrt{1 + \frac{2}{x}} + 1}

    now take the limit.
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  3. #3
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    thank you that helped me a lot
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  4. #4
    Senior Member mollymcf2009's Avatar
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    Skeeter,

    Could you use l'Hopital's Rule for the first problem instead? I tried it and got the same answer.
    Thanks!
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  5. #5
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    yes, L'Hopital will work on that problem.

    however, consider this limit ...

    \lim_{x \to \infty} \frac{x}{\sqrt{x^2+1}}

    it's easy to see the limit is 1, but try using L'Hopital and see what happens.
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  6. #6
    Senior Member mollymcf2009's Avatar
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    The x^2 and the square root in the denominator is what causes problems using l"Hopital for this one right?
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