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Thread: Need LImit help!!!!!!!!!!

  1. #1
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    Need LImit help!!!!!!!!!!

    find the lim as x approaches positive and negative infinity

    (3x^3 - x +1) / (x + 3)

    can you explain the answer
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  2. #2
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    Re: Need LImit help!!!!!!!!!!

    Quote Originally Posted by ubhutto View Post
    find the lim as x approaches positive and negative infinity

    (3x^3 - x +1) / (x + 3)

    can you explain the answer
    Start by dividing the top and bottom by x, to give

    $\displaystyle \displaystyle \begin{align*} \frac{3x^2 - 1 + \frac{1}{x}}{1 + \frac{3}{x}} \end{align*}$

    Then as you approach $\displaystyle \displaystyle \begin{align*} \infty \end{align*}$ the top goes to $\displaystyle \displaystyle \begin{align*} \infty \end{align*}$ and the bottom goes to 1, so the limit is $\displaystyle \displaystyle \begin{align*} \infty \end{align*}$.

    Try evaluating the limit as you approach $\displaystyle \displaystyle \begin{align*} -\infty \end{align*}$...
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  3. #3
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    Re: Need LImit help!!!!!!!!!!

    thanks

    can you also show me how to get the horizontal asymptote of the graph
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  4. #4
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    Re: Need LImit help!!!!!!!!!!

    Another way to do this to go ahead and divide: $\displaystyle \frac{3x^3- x+ 1}{x+1}= 3x^2- 3x+ 2- \frac{1}{x+1}$. As x goes to either + or - infinity, that last fraction goes to 0 so you only need to consider what happens to $\displaystyle 3x^2- 3x+ 2$ determines the limit.

    If you know what a "horizontal asymptote" is, you should realize that knowing the limit tells you everything you need to know about the horizontal asymptotes.
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