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Math Help - limit to infinity

  1. #1
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    limit to infinity

    How do I do this?

    lim (square root(8x^3 + 5x + 10)) / (x^2)
    x->infinity

    Any help?
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  2. #2
    Member javax's Avatar
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    it goes to \infty
    the numerator is bigger than denominator
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  3. #3
    o_O
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    \lim_{x \to \infty} \frac{\sqrt{8x^3 + 5x + 10}}{x^2} \cdot \frac{\displaystyle {\color{red}\frac{1}{x^2}}}{\ \displaystyle{\color{red}\frac{1}{x^2}} \ } \ = \  \lim_{x \to \infty} \frac{\displaystyle \frac{\sqrt{8x^3 + 5x + 10}}{x^2}}{\displaystyle \frac{\ x^2 \ }{x^2}}

    = \lim_{x \to \infty} \frac{\displaystyle \frac{\sqrt{8x^3 + 5x + 10}}{\sqrt{x^4}}}{1} \ = \ \lim_{x \to \infty} \sqrt{\frac{8x^3 + 5x + 10}{x^4}} \ = \ \lim_{x \to \infty} \sqrt{\frac{8x^3}{x^4} + \frac{5x}{x^4} + \frac{10}{x^4}} \ = \ \hdots

    Quote Originally Posted by javax View Post
    it goes to \infty
    the numerator is bigger than denominator
    Not exactly. \sqrt{x^3} = x^{\frac{3}{2}} \ < \ x^2
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  4. #4
    Member javax's Avatar
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    Quote Originally Posted by o_O View Post
    \lim_{x \to \infty} \frac{\sqrt{8x^3 + 5x + 10}}{x^2} \cdot \frac{\displaystyle {\color{red}\frac{1}{x^2}}}{\ \displaystyle{\color{red}\frac{1}{x^2}} \ } \ = \ \lim_{x \to \infty} \frac{\displaystyle \frac{\sqrt{8x^3 + 5x + 10}}{x^2}}{\displaystyle \frac{\ x^2 \ }{x^2}}

    = \lim_{x \to \infty} \frac{\displaystyle \frac{\sqrt{8x^3 + 5x + 10}}{\sqrt{x^4}}}{1} \ = \ \lim_{x \to \infty} \sqrt{\frac{8x^3 + 5x + 10}{x^4}} \ = \ \lim_{x \to \infty} \sqrt{\frac{8x^3}{x^4} + \frac{5x}{x^4} + \frac{10}{x^4}} \ = \ \hdots



    Not exactly. \sqrt{x^3} = x^{\frac{3}{2}} \ < \ x^2
    opps, I didn't notice the root at all! Sorry
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  5. #5
    Super Member Showcase_22's Avatar
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    I tried using the sandwich rule but my final answer doesn't make much sense:



    The final bit doesn't make sense since I get that it convereges to a number between 0 and 0 that isn't 0. :s

    any helpers?
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  6. #6
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    Quote Originally Posted by Showcase_22 View Post
    I tried using the sandwich rule but my final answer doesn't make much sense:



    The final bit doesn't make sense since I get that it convereges to a number between 0 and 0 that isn't 0. :s

    any helpers?
    The sandwich theorem works fine.

    If f(x) < g(x) < h(x) and  \lim_{x \rightarrow a} f(x) = \lim_{x \rightarrow a} h(x) = L then  \lim_{x \rightarrow a} g(x) = L.
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  7. #7
    Super Member Showcase_22's Avatar
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    Oh right. It just didn't look right when I got to the end.

    I think that's the first time i've ever used the sandwich rule. It's nice to know I was doing it correctly!
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