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Thread: 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 $\displaystyle \frac{1}{\sqrt{t^2}}$.


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

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

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

    $\displaystyle = 2$.
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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 $\displaystyle \sqrt{a}\sqrt{b}=\sqrt{ab}$

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

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

    As t approaches infinity, the fractions go to zero.
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