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Math Help - limit to infinity

  1. #1
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    limit to infinity

    Could someone please show me how to workout problems of these types:

    Find the limit. (If you need to use - or , enter -INFINITY or INFINITY.)


    Thanks in advance!
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  2. #2
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    Quote Originally Posted by ryan18 View Post
    Could someone please show me how to workout problems of these types:

    Find the limit. (If you need to use - or , enter -INFINITY or INFINITY.)


    Thanks in advance!
    Try multiplying by \frac{\sqrt{36x^2+x}+6x}{\sqrt{36x^2+x}+6x} and then dividing by x up and down (i think that it have to works)
    Last edited by felper; February 9th 2010 at 09:12 AM. Reason: Little mistake
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  3. #3
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    How did you get the \frac{\sqrt{36x^2+x}+36x^2}{\sqrt{36x^2+x}+36x^2}?
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  4. #4
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    Quote Originally Posted by ryan18 View Post
    How did you get the \frac{\sqrt{36x^2+x}+36x^2}{\sqrt{36x^2+x}+36x^2}?
    Oh, i did a little mistake. I've corrected it. Now that it's correct, with that factor, you can eliminate the root. See what happens:

    \lim_{x\to\infty}\sqrt{36x^2+x}-6x \cdot \frac{\sqrt{36x^2+x}+6x}{\sqrt{36x^2+x}+6x}=\lim_{  x\to\infty}\frac{x}{\sqrt{36x^2+x}+6x}=\lim_{x\to\  infty}\frac{1}{\sqrt{36+\frac{1}{x}}+6}

    Now you can conclude.
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  5. #5
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    How did you go from <br />
=\lim_{x\to\infty}\frac{x}{\sqrt{36x^2+x}+6x}=\lim  _{x\to\infty}\frac{1}{\sqrt{36+\frac{1}{x}}+6}<br />
    [/tex]?? It looks like in some places you multiplied out by \frac{1}{x} and other parts you multiplied out by \frac{1}{x^2}
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  6. #6
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    Quote Originally Posted by ryan18 View Post
    How did you go from <br />
=\lim_{x\to\infty}\frac{x}{\sqrt{36x^2+x}+6x}=\lim  _{x\to\infty}\frac{1}{\sqrt{36+\frac{1}{x}}+6}<br />
    [/tex]?? It looks like in some places you multiplied out by \frac{1}{x} and other parts you multiplied out by \frac{1}{x^2}
    I've multiplied by 1/x both sides. \frac{1}{x}(\sqrt{36x^2+x}+6x)=\frac{1}{x}(\sqrt{3  6x^2+x})+6=\frac{1}{\sqrt{x^2}}(\sqrt{36x^2+x})+6=  \sqrt{36+\frac{1}{x}}
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  7. #7
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    Ok I see now, so the denominator would look like \sqrt{36+\frac{1}{x}}+6 correct? and the \frac{1}{x} would tend to 0. Leaving \sqrt{36}+6=12? Making the final answer to be \frac{1}{12}??
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  8. #8
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    Yep
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  9. #9
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    Awesome thank you so much for putting up with my multiple questions!
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