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Math Help - Limits

  1. #1
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    Limits

    Consider f (x) = (-7x^3 + 2)/(10x^3 + 3x + 2)

    a) Construct a table of values to determing lim f(x) x to infinity

    b) Use algebra to determin lim f(x)
    x to infinity


    c) Sketch the graph and indicate on the graph the behavior described by the limit.
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  2. #2
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    part a. plug and chug.


    For part b, get rid of everything but the highest powers in the num. and den.

    You are left with: \frac{-7x^{3}}{10x^{3}}

    Now, you can see what the limit is, can't you?.


    part c. You have a graphing calculator?.
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  3. #3
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    This is my 18th Post!!!
    Attached Thumbnails Attached Thumbnails Limits-picture3.gif  
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  4. #4
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    Quote Originally Posted by galactus
    part a. plug and chug.


    For part b, get rid of everything but the highest powers in the num. and den.

    You are left with: \frac{-7x^{3}}{10x^{3}}

    Now, you can see what the limit is, can't you?.


    part c. You have a graphing calculator?.
    Umm am not really sure I understand what you meant y plug and chug.
    Can I plug in any value that I like? Arent I supposed to follow a sequence or something??

    For part B how can I just get rid of everything but the highest powers?
    L'Hospital Rule??
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  5. #5
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    Just plug in higher and higher values and watch it approach the limit.

    \frac{-7(1)^{3}+2}{10(1)^{3}+3(1)+2}=\frac{-1}{3}

    \frac{-7(100)^3+2}{10(100)^{3}+3(100)+2}=\frac{-3499999}{5000151}=-.699978660644

    \frac{-7(1000)^{3}+2}{10(1000)^{3}+3(1000)+2}=\frac{-3499999999}{5000001501}=-.69999978966

    See, it's approaching -.7\;\ or\;\ \frac{-7}{10}

    Now, make it official with part b. Rational functions, especially with polynomials, are one of the easiest things in which to take a limit.

    For part b, no, just erase them. In a rational function, if the limit approaches infinity, it is unaffected if you discard everything but the highest powers. Think about it. See?.

    Suppose I had \lim_{x\to\infty}\frac{3x+5}{6x-8}

    If x gets huge, towards infinity, what are the 5 and 6 going to amount to?. Nothing. So, get rid of them and take the limit of 3x/6x. Which is 3/6=1/2.

    Yours works the same.

    \frac{-7x^{3}}{10x^{3}}
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  6. #6
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    Quote Originally Posted by askmemath
    For part B how can I just get rid of everything but the highest powers?
    L'Hospital Rule??
    You have,
    \frac{-7x^3 + 2}{10x^3 + 3x + 2}
    Divide numerator and denominator by x^3,
    \frac{-7+\frac{2}{x^3}}{10+\frac{3}{x^2}+\frac{2}{x^3}} \to  -\frac{7}{10} as, x\to\infty
    Becuase,
    \frac{k}{x^n}\to 0
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