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Math Help - Evaluate the limit

  1. #1
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    Evaluate the limit



    the answer gave me 0/1 by dividing each term by t2
    but its wrongs



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  2. #2
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    Quote Originally Posted by wannous View Post


    the answer gave me 0/1 by dividing each term by t2
    but its wrongs



    Since you have a square root in the denominator, you will actually have to divide top and bottom by the SQUARE ROOT of the highest power.

    In other words, multiply top and bottom by \frac{1}{\sqrt{t^2}}.


    So \lim_{t \to \infty}\frac{2t - 9}{\sqrt{t^2 + 8t + 5}} = \lim_{t \to \infty}\frac{(2t - 9)\frac{1}{\sqrt{t^2}}}{\sqrt{t^2 + 8t + 5}\left(\frac{1}{\sqrt{t^2}}\right)}

     = \lim_{t \to \infty}\frac{2 - \frac{9}{t}}{\sqrt{1 + \frac{8}{t} + \frac{5}{t^2}}}

     = \frac{2}{\sqrt{1}}

     = 2.
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    clear explanation, thank you my good sir
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  4. #4
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    Quote Originally Posted by wannous View Post


    the answer gave me 0/1 by dividing each term by t2
    but its wrongs



    Hi wannous,

    use \sqrt{a}\sqrt{b}=\sqrt{ab}

    \sqrt{t^2+8t+5}=\sqrt{t^2(1+\frac{8}{t}+\frac{5}{t  ^2})}=\sqrt{t^2}\sqrt{1+\frac{8}{t}+\frac{5}{t^2}}

    =t\sqrt{1+\frac{8}{t}+\frac{5}{t^2}}

    As t approaches infinity, the fractions go to zero.
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