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Math Help - infinite limits

  1. #1
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    infinite limits

    I'm still trying to grasp the idea of limits as a whole, and infinite limits, individually. I really don't know where to start when it comes to multiplying the numerator and denom by a certain power of 1/x to talk myself through the problem... Here are a few of my homework problems, any addition info to show me how you work through the problem would be greatly appreciated.


    1. lim as x approaches infinity of x + x^3 + x^5/1 - x^2 + x^4 =


    2. lim as t approaches negative infinity of x^2 + 2/t^3 + t^2 -1


    3. lim as x approaches infinity of x + 2/ sq rt (9x^2 + 1)


    4. lim as x approaches infinity of [(sq rt(9x^2 + x)) - 3x]
    *(9x^2 + x) is the only thing supposed to be under the radical.
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  2. #2
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    Hello, obsmith08!

    The rule I've taught my students is:
    Divide numerator and denominator by the highest power of the variable in the denominator.


    1)\;\;\lim_{x\to\infty}\frac{x + x^3 + x^5}{1 - x^2 + x^4}

    Divide top and bottom by x^4\!:\;\;\frac{\dfrac{x}{x^4} + \dfrac{x^3}{x^4} + \dfrac{x^5}{x^4}}{\dfrac{1}{x^4} - \dfrac{x^2}{x^4} + \dfrac{x^4}{x^4}} . = \;\frac{\dfrac{1}{x^3} + \dfrac{1}{x} + x}{\dfrac{1}{x^4} - \dfrac{1}{x^2} + 1}

    . . Then: . \lim_{x\to\infty} \frac{\frac{1}{x^3} + \frac{1}{x} + x}{\frac{1}{x^4} - \frac{1}{x^2} + 1} \;=\;\frac{0+0+\infty}{0-0+1} \;=\;\infty




    2)\;\;\lim_{x\to-\infty}\frac{x^2 + 2}{x^3 + x^2 -1}

    Divide top and bottom by x^3\!:\;\;\frac{\dfrac{x^2}{x^3} + \dfrac{2}{x^3}}{\dfrac{x^3}{x^3} + \dfrac{x}{x^3} - \dfrac{1}{x^3}} \;=\;<br />
\frac{\dfrac{1}{x} + \dfrac{2}{x^3}}{1 + \dfrac{1}{x} - \dfrac{1}{x^3}}
    .Then: . \lim_{x\to-\infty}\frac{\dfrac{1}{x} + \dfrac{2}{x^3}}{1 + \dfrac{1}{x} - \dfrac{1}{x^3}}\;=\;\frac{0+0}{1+0-0} \;=\;0




    3)\;\;\lim_{x\to\infty}\frac{x + 2}{\sqrt{9x^2 + 1}}

    \text{Divide top and bottom by }x\!:\;\;\frac{\dfrac{x+2}{x}}{\dfrac{\sqrt{9x^2+1  }}{x}} \;=\;\frac{\dfrac{x}{x} + \dfrac{2}{x}} {\dfrac{\sqrt{9x^2+1}}{\sqrt{x^2}}} \;=\;\frac{1+\dfrac{2}{x}} {\sqrt{\dfrac{9x^2}{x^2} + \dfrac{1}{x^2}}} . =\;\frac{1 + \dfrac{2}{x}} {\sqrt{9 + \dfrac{1}{x^2}}}
    . . Then: . \lim_{x\to\infty}\frac{1 + \dfrac{2}{x}} {\sqrt{9 + \dfrac{1}{x^2}}} \;=\;\frac{1+0}{\sqrt{9+0}} \;=\;\frac{1}{3}




    4)\;\;\lim_{x\to\infty}\left[\sqrt{9x^2 + x} - 3x\right]
    This one requires a different technique.


    Multiply top and bottom by the conjugate:

    . . \frac{\sqrt{9x^2+x} - 3x}{1}\cdot\frac{\sqrt{9x^2+x} + 3x}{\sqrt{9x^2+x} + 3x} \;=\;\frac{(9x^2+x) - 9x^2}{\sqrt{9x^2+x} + 3x} \;=\;\frac{x}{\sqrt{9x^2+x}+3x}


    \text{Divide top and bottom by }x\!:\;\;\frac{\dfrac{x}{x}} {\dfrac{\sqrt{9x^2+x}}{\sqrt{x^2}} + \dfrac{3x}{x}} \;=\;\frac{1} {\sqrt{\dfrac{9x^2}{x^2}+\dfrac{x}{x^2}} + 3} \;=\;\frac{1}{\sqrt{9+\dfrac{1}{x}} + 3}

    . . Then: . \lim_{x\to\infty}\frac{1}{\sqrt{9+\frac{1}{x}} + 3} \;=\;\frac{1}{\sqrt{9+0} + 3} \;=\;\frac{1}{3+3} \;=\;\frac{1}{6}

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