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Math Help - Evaluate the limit

  1. #1
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    Evaluate the limit

    \\ \begin{array}{c}\lim\\t\rightarrow0\end{array} \ \  \frac{t+\frac{a}{t}}{t+\frac{b}{t}} \\ \\ \\ \begin{array}{c}\lim\\t\rightarrow0\end{array} \ \  \frac{t^{2}+at}{t^{2}+bt} \\ \\ \\ \begin{array}{c}\lim\\t\rightarrow0\end{array} \ \  \frac{t(t+a)}{t(t+b)}

    The book gives the answer as \frac{a}{b} but I can't figure out how to get to that solution. Up there is everything I have thought of and I'm stuck because it still has a zero in the denominator.

    Can anyone give me a hint on what to do (no need to type any LaTeX code)

    ----------------------
    Edit: I can't believe I made a mistake simply multiplying (\frac{a}{t})(t) \& (\frac{b}{t})(t)
    Last edited by Fenixx09; September 1st 2011 at 08:02 PM. Reason: my mistake
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  2. #2
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    Re: Evaluate the limit

    Quote Originally Posted by Fenixx09 View Post
    \\ \begin{array}{c}\lim\\t\rightarrow0\end{array} \ \  \frac{t+\frac{a}{t}}{t+\frac{b}{t}} \\ \\ \\ \begin{array}{c}\lim\\t\rightarrow0\end{array} \ \  \frac{t^{2}+at}{t^{2}+bt} \\ \\ \\ \begin{array}{c}\lim\\t\rightarrow0\end{array} \ \  \frac{t(t+a)}{t(t+b)}

    The book gives the answer as \frac{a}{b} but I can't figure out how to get to that solution. Up there is everything I have thought of and I'm stuck because it still has a zero in the denominator.

    Can anyone give me a hint on what to do (no need to type any LaTeX code)

    ----------------------
    Edit: I can't believe I made a mistake simply multiplying (\frac{a}{t})(t) \& (\frac{b}{t})(t)
    When you multiply top and bottom by \displaystyle t you should actually get \displaystyle \lim_{t \to 0}\frac{t^2 + a}{t^2 + b}.

    So you have

    \displaystyle \begin{align*}\lim_{t \to 0} \frac{t^2 + a}{t^2 + b} &= \lim_{t \to 0}\frac{t^2 + b + a - b}{t^2 + b} \\ &= \lim_{t \to 0} \left(\frac{t^2 + b}{t^2 + b} + \frac{a - b}{t^2 + b}\right) \\ &= \lim_{t \to 0} \left(1 + \frac{a - b}{t^2 + b}\right) \\ &= 1 + \frac{a - b}{b} \\ &= 1 + \frac{a}{b} - \frac{b}{b} \\ &= 1 + \frac{a}{b} - 1 \\ &= \frac{a}{b}  \end{align*}
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  3. #3
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    Re: Evaluate the limit

    Quote Originally Posted by Fenixx09 View Post
    \\ \begin{array}{c}\lim\\t\rightarrow0\end{array} \ \  \frac{t+\frac{a}{t}}{t+\frac{b}{t}} \\ \\ \\ \begin{array}{c}\lim\\t\rightarrow0\end{array} \ \  \frac{t^{2}+at}{t^{2}+bt} \\ \\ \\ \begin{array}{c}\lim\\t\rightarrow0\end{array} \ \  \frac{t(t+a)}{t(t+b)}

    The book gives the answer as \frac{a}{b} but I can't figure out how to get to that solution. Up there is everything I have thought of and I'm stuck because it still has a zero in the denominator.

    Can anyone give me a hint on what to do (no need to type any LaTeX code)

    ----------------------
    Edit: I can't believe I made a mistake simply multiplying (\frac{a}{t})(t) \& (\frac{b}{t})(t)
    \lim_{t\to 0}  \frac{t(t+a)}{t(t+b)} = \lim_{t\to 0} \frac{(t+a)}{(t+b)}=\frac{a}{b}

    CB
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  4. #4
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    Re: Evaluate the limit

    Quote Originally Posted by CaptainBlack View Post
    \lim_{t\to 0}  \frac{t(t+a)}{t(t+b)} = \lim_{t\to 0} \frac{(t+a)}{(t+b)}=\frac{a}{b}

    CB
    That is true CB, but \displaystyle \frac{t + \frac{a}{t}}{t + \frac{b}{t}} \neq \frac{t^2 + at}{t^2 + bt}...
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