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Math Help - Limit question

  1. #1
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    Limit question

    So, the equation is:

    \lim (\sqrt{9x^2 + x} -3x)
    x\rightarrow\infty

    Well, I first tried solving in the way i solved equations that didn't contain sqrt, dividing all the equation by the biggest power of x, in this case, dividing by X, getting

    \lim (\sqrt{9 + 1/x} -3)
    x\rightarrow\infty

    And, by limit laws i would get 3-3 = 0, however, according to my book (which I trust more than myself) the answer is 1/6, so I tried doing in a different way:

    \lim (\sqrt{9x^2 + x} -3x)*(\sqrt{9x^2 + x} +3x)/(\sqrt{9x^2 + x} +3x)
    x\rightarrow\infty

    and now getting :
     1/(\sqrt{9 + 1/x} + 3), which is, 1/6.

    So, my question is, why I couldn't do it the first way? Isn't it supposed to be the same thing? I mean, clearly not but, what rule am I breaking by doing it in the first way?

    Thanks in advance
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  2. #2
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    Quote Originally Posted by Capes View Post
    So, the equation is:

    \lim (\sqrt{9x^2 + x} -3x)
    x\rightarrow\infty

    Well, I first tried solving in the way i solved equations that didn't contain sqrt, dividing all the equation by the biggest power of x, in this case, dividing by X, getting

    \lim (\sqrt{9 + 1/x} -3)
    x\rightarrow\infty

    And, by limit laws i would get 3-3 = 0, however, according to my book (which I trust more than myself) the answer is 1/6, so I tried doing in a different way:

    \lim (\sqrt{9x^2 + x} -3x)*(\sqrt{9x^2 + x} +3x)/(\sqrt{9x^2 + x} +3x)
    x\rightarrow\infty

    and now getting :
     1/(\sqrt{9 + 1/x} + 3), which is, 1/6.

    So, my question is, why I couldn't do it the first way? Isn't it supposed to be the same thing? I mean, clearly not but, what rule am I breaking by doing it in the first way?

    Thanks in advance
    You are treating it like it is a rational function.

    \displaystyle \lim_{x \to \infty}\sqrt{9x^2+x}-3x=\lim_{x \to \infty}|x|\sqrt{9+\frac{1}{x}}-3x

    Since x > 0

    \displaystyle \lim_{x \to \infty}x[\sqrt{9+\frac{1}{x}}-3]

    This is an indeterminate form infinity times 0 just as the first was infinity minus infinity!
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  3. #3
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    Quote Originally Posted by TheEmptySet View Post
    You are treating it like it is a rational function.

    \displaystyle \lim_{x \to \infty}\sqrt{9x^2+x}-3x=\lim_{x \to \infty}|x|\sqrt{9+\frac{1}{x}}-3x

    Since x > 0

    \displaystyle \lim_{x \to \infty}x[\sqrt{9+\frac{1}{x}}-3]

    This is an indeterminate form infinity times 0 just as the first was infinity minus infinity!
    Sorry, can you please explain in a simpler way? I have re-read what you wrote but im still lost here, what do you mean by "This is an indeterminate form infinity times 0 just as the first was infinity minus infinity!", for example?
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  4. #4
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    Quote Originally Posted by Capes View Post
    Sorry, can you please explain in a simpler way? I have re-read what you wrote but im still lost here, what do you mean by "This is an indeterminate form infinity times 0 just as the first was infinity minus infinity!", for example?
    There are many indeterminate froms Indeterminate form - Wikipedia, the free encyclopedia

    The main point is that the ordinary rules of arthimetic break down in these cases

    A simpler version of what you have above could be

    \displaystyle \lim_{x \to \infty}2x-x=\infty-\infty the question is what is infinity minus infintiy? is it zero? infinity? or maybe 12?

    The point is we don't know. When you multiplied by the conjugate you used algegra to change it into a from that we can compute. Later in you calucus class you will learn more about these forms and a theorem called L'H˘pital's rule - Wikipedia, the free encyclopedia
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  5. #5
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    thanks for being patient and helping me as always, man!
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