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Math Help - Help with limits to infinity

  1. #1
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    Help with limits to infinity

    I'm in a beginning calculus class after not touching math of any sort for 4 years. I've forgotten a lot of the more complicated algebra, and I'm struggling with simplifying limits right now. I'm doing my homework right now and coming across a lot of problems I'm getting wrong and I can't understand why, so I am asking for help.

    first problem:

    1.) lim  x\to \infty                   \frac {\sqrt{9x^6 - x}}{x^3 + 1}

    To work this one out, I multiplied the top and bottom by \sqrt{9x^6 - x},
    so I get
    \frac{9x^6-x}{(x^3+1)(\sqrt{9x^6-x})}

    From there, I divided everything by x^3 since that's the highest x value in the denominator in order to get rid of some of the numbers. I wind up with:
    \frac{9x^3-0}{ (1+0)(\sqrt{9-0} }
    Which is \frac{9x^3}{3 } which equals
    3x^3

    The back of my book however says this answer should just be 3. Why is the x^3 cancelled?
    ---------------------


    Here's another I got wrong:
    lim x\to \infty \sqrt {9x^2 + x} - 3x

    All I could think to do was to divide everything by X but I don't even think you're allowed to do that without a denominator (are you?). Anyway when I do that I get
    \sqrt{9+0} - 3 = 0

    Apparently the answer is supposed to be 1/6, how do you get this?
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  2. #2
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    \lim_{x\to\infty}\frac{\sqrt{9x^6}}{x^3}=3

    When looking at a fraction of a limit to infinity, you need to only look at the highest power terms of the denominator and numerator.
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  3. #3
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    For the second one, multiply and divide by the conjugate.
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  4. #4
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    Quote Originally Posted by ftmichael View Post
    I'm in a beginning calculus class after not touching math of any sort for 4 years. I've forgotten a lot of the more complicated algebra, and I'm struggling with simplifying limits right now. I'm doing my homework right now and coming across a lot of problems I'm getting wrong and I can't understand why, so I am asking for help.

    first problem:

    1.) lim  x\to \infty                   \frac {\sqrt{9x^6 - x}}{x^3 + 1}

    To work this one out, I multiplied the top and bottom by \sqrt{9x^6 - x},
    so I get
    \frac{9x^6-x}{(x^3+1)(\sqrt{9x^6-x})}

    From there, I divided everything by x^3 since that's the highest x value in the denominator in order to get rid of some of the numbers. I wind up with:
    \frac{9x^3-0}{ (1+0)(\sqrt{9-0} }
    Which is \frac{9x^3}{3 } which equals
    3x^3

    The back of my book however says this answer should just be 3. Why is the x^3 cancelled?
    L'Hospital's Rule works nicely in both cases. If you have an indeterminate form \displaystyle \frac{0}{0} or \displaystyle \frac{\infty}{\infty}, then

    \displaystyle \lim_{x \to c}\frac{f(x)}{g(x)} = \lim_{x \to c}\frac{f'(x)}{g'(x)}

    The first is already in \displaystyle \frac{\infty}{\infty} form, and the second can be made into that form using this transformation...

    \displaystyle \begin{align*} \sqrt{9x^2 + x} - 3x &= \ln{\left(e^{\sqrt{9x^2 + x} - 3x}\right)} \\ &= \ln{\left(\frac{e^{\sqrt{9x^2 + x}}}{e^{3x}}\right)}\end{align*}

    Recall that because of the continuity of the logarithm, \displaystyle \lim_{x \to c}\ln{[f(x)]} = \ln{\left[\lim_{x \to c}f(x)\right]}, and the "stuff" inside the logarithm is now of the form \displaystyle \frac{\infty}{\infty}. So you can use L'Hospital's Rule there too
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  5. #5
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    Quote Originally Posted by Prove It View Post
    L'Hospital's Rule works nicely in both cases. If you have an indeterminate form \displaystyle \frac{0}{0} or \displaystyle \frac{\infty}{\infty}, then

    \displaystyle \lim_{x \to c}\frac{f(x)}{g(x)} = \lim_{x \to c}\frac{f'(x)}{g'(x)}

    The first is already in \displaystyle \frac{\infty}{\infty} form, and the second can be made into that form using this transformation...

    \displaystyle \begin{align*} \sqrt{9x^2 + x} - 3x &= \ln{\left(e^{\sqrt{9x^2 + x} - 3x}\right)} \\ &= \ln{\left(\frac{e^{\sqrt{9x^2 + x}}}{e^{3x}}\right)}\end{align*}

    Recall that because of the continuity of the logarithm, \displaystyle \lim_{x \to c}\ln{[f(x)]} = \ln{\left[\lim_{x \to c}f(x)\right]}, and the "stuff" inside the logarithm is now of the form \displaystyle \frac{\infty}{\infty}. So you can use L'Hospital's Rule there too
    Thanks for all that help, but I haven't learned this rule yet and I don't feel comfortable enough trying stuff outside of what I've already been taught. Thanks for all that effort though!

    DWsmith:

    Thanks for the first reply, I realized i made a stupid mistake on the first problem i posted and after working that out I got 3. Conjugating first was all I needed for the second problem as well.

    Another question, How do you find horizontal and vertical asymptotes? I know for horizontal you have to make the limit of a function + or - infinity, and I'd assume for vertical it's the limit of a function when x approaches 0? I can provide a practice problem if it would be easier to grasp what im saying.
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  6. #6
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    Quote Originally Posted by ftmichael View Post
    Thanks for all that help, but I haven't learned this rule yet and I don't feel comfortable enough trying stuff outside of what I've already been taught. Thanks for all that effort though!

    DWsmith:

    Thanks for the first reply, I realized i made a stupid mistake on the first problem i posted and after working that out I got 3. Conjugating first was all I needed for the second problem as well.

    Another question, How do you find horizontal and vertical asymptotes? I know for horizontal you have to make the limit of a function + or - infinity, and I'd assume for vertical it's the limit of a function when x approaches 0? I can provide a practice problem if it would be easier to grasp what im saying.
    Vertical asymptote occur when the denominator is zero.
    Horizontal--you need to check the limits at negative and positive infinity.
    Slant--when the highest degree of the numerator is greater than the highest degree of the denominator. To solve, you do long division and disregard the remainder.
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  7. #7
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by ftmichael View Post
    I'm in a beginning calculus class after not touching math of any sort for 4 years. I've forgotten a lot of the more complicated algebra, and I'm struggling with simplifying limits right now. I'm doing my homework right now and coming across a lot of problems I'm getting wrong and I can't understand why, so I am asking for help.

    first problem:

    1.) lim  x\to \infty                   \frac {\sqrt{9x^6 - x}}{x^3 + 1}

    To work this one out, I multiplied the top and bottom by \sqrt{9x^6 - x},
    so I get
    \frac{9x^6-x}{(x^3+1)(\sqrt{9x^6-x})}

    From there, I divided everything by x^3 since that's the highest x value in the denominator in order to get rid of some of the numbers. I wind up with:
    \frac{9x^3-0}{ (1+0)(\sqrt{9-0} }
    Which is \frac{9x^3}{3 } which equals
    3x^3

    The back of my book however says this answer should just be 3. Why is the x^3 cancelled?
    ---------------------


    Here's another I got wrong:
    lim x\to \infty \sqrt {9x^2 + x} - 3x

    All I could think to do was to divide everything by X but I don't even think you're allowed to do that without a denominator (are you?). Anyway when I do that I get
    \sqrt{9+0} - 3 = 0

    Apparently the answer is supposed to be 1/6, how do you get this?

    \lim_{x\to\infty}\frac{\sqrt{9x^6-x}}{x^3+1}

    I will prove by definition that the above limit equals to 3.

    (\forall \varepsilon >0)(\exists\Delta \in\mathbb{R})(\forall x>\Delta ):|f(x)-L|<\varepsilon

    Where f(x)=\frac{\sqrt{9x^6-x}}{x^3+1} and L=3.

    Proof:

    Let be \varepsilon >0 given.

    |\frac{\sqrt{9x^6-x}}{x^3+1}-1|=|\frac{\sqrt{9x^6-x}-3x^3-3}{x^3+1}|\overset{x\geq \sqrt[5]{\frac{1}{9}}}{<}{|\frac{3x^3-3x^3-3}{x^3+1}|}<|\frac{3}{x^3}|<\varepsilon

    |\frac{3}{x^3}|<\varepsilon

    or:

    3<x^3\varepsilon

    \frac{3}{\varepsilon }<x^3

    or:

    x>\sqrt[3]{\frac{3}{\varepsilon }}

    We set \sqrt[3]{\frac{3}{\varepsilon }} be our delta.

    \Delta =\sqrt[3]{\frac{3}{\varepsilon }}

    Hence,


    (\forall \varepsilon >0)(\exists \sqrt[3]{\frac{3}{\varepsilon }} \in\mathbb{R})(\forall x>\sqrt[3]{\frac{3}{\varepsilon }}):|f(x)-3|<\varepsilon
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  8. #8
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by dwsmith View Post
    For the second one, multiply and divide by the conjugate.
    Lets do it!

    \lim_{x\to\infty}\sqrt{9x^2+x}-3x=?

    \sqrt{9x^2+x}-3x=\frac{(\sqrt{9x^2+x}-3x)(\sqrt{9x^2+x}+3x)}{\sqrt{9x^2+x}+3x}=\frac{9x^  2+x-9x^2}{\sqrt{9x^2+x}+3x}=

     \frac{\frac{x}{x}}{\frac{\sqrt{9x^2+x}+3x}{x}}=\fr  ac{1}{\sqrt{\frac{9x^2+x}{x^2}}+\frac{3x}{x}}=\fra  c{1}{\sqrt{9+\frac{1}{x}}+3}

    THERE IS STRANGE LATEX ERROR THAT I CAN'T FIND!

    Now,


    \lim_{x\to\infty}\sqrt{9x^2+x}-3x=\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  }
    Last edited by Also sprach Zarathustra; June 12th 2011 at 01:06 AM.
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