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Math Help - last question of the night..thank goodness

  1. #1
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    last question of the night..thank goodness

    after hours of trying to make sense of this practice test, im at my last question...another doozy i have no idea on (you may have seen my plethora of other ones that i could not beat)

    Use the
    M - (delta)definition of limit at infinity to prove the following limits:
    lim x>infinity... 3x^2/(5x^2 +1)=3/5
    lim x> 1+...1/(x^2 -1)= infinity
    Last edited by mr fantastic; December 17th 2009 at 04:04 AM. Reason: Changed microscopic fontsize to default so that mortals have a fighting chance of reading this post.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by twostep08 View Post
    after hours of trying to make sense of this practice test, im at my last question...another doozy i have no idea on (you may have seen my plethora of other ones that i could not beat)


    Use the
    M - (delta)definition of limit at infinity to prove the following limits:

    lim x>infinity... 3x^2/(5x^2 +1)=3/5

    lim x> 1+...1/(x^2 -1)= infinity




    Do you know any theorems that you can use? Otherwise, note that \left|\frac{3x^2}{5x^2+1}-\frac{3}{5}\right|=\left|\frac{3}{25x^2+5}\right|\  leqslant\frac{3}{x^2}. So choosing x such that x>\frac{1}{\sqrt{\frac{3}{\varepsilon}}} ensures that \left|\frac{3x^2}{5x^2+1}\right|\leqslant\frac{3}{  x^2}<\frac{3}{\sqrt{\frac{3}{\varepsilon}^2}}=\var  epsilon..................and done.
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