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Thread: 4 Limit problems.

  1. #1
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    4 Limit problems.

    Determine the limits of the following functions as $\displaystyle x\to\infty\$ , if they exist.

    $\displaystyle f(x)=4x^2-3x^3$.

    $\displaystyle f(x)=\frac{x^4 + 3x^2 - 2x}{2x^4 + 5}$

    $\displaystyle f(x)=\frac{xsin^2x}{x^2+5}$

    $\displaystyle
    f(x) =
    \begin{cases}
    0 & \mbox{if x is rational}\\
    \frac{1}{x}& \mbox{if x isn't rational}
    \end{cases}
    $


    Ive done quite a few of these now, but im stuck on these 4.

    edit: while doing my exersices i found a 5th problem that i wasn't sure about:

    $\displaystyle f(x)=\frac{\sqrt{x}-1}{x^2}$

    I think i got the limit as 0 using l'hopitals rule but it got a bit messy and im not sure if i did it right.
    Last edited by kbartlett; Sep 30th 2008 at 09:16 AM.
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by kbartlett View Post
    Determine the limits of the following functions as $\displaystyle x\to\infty\$ , if they exist.

    $\displaystyle f(x)=4x^2-3x^3$.

    $\displaystyle f(x)=\frac{x^4 + 3x^2 - 2x}{2x^4 + 5}$

    $\displaystyle f(x)=\frac{xsin^2x}{x^2+5}$

    $\displaystyle
    f(x) =
    \begin{cases}
    0 & \mbox{if x is rational}\\
    \frac{1}{x}& \mbox{if x isn't rational}
    \end{cases}
    $


    Ive done quite a few of these now, but im stuck on these 4.
    for the first one, try to factor $\displaystyle x^3$

    for the second one, multiply and divide $\displaystyle \frac{1}{x^4}$ or use l'hopital's rule

    third: i think, you have to multiply and divide $\displaystyle \frac{1}{x^2}$

    fourth, i think, it is obvious..
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  3. #3
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    For the 1st one, $\displaystyle 4x^2-3x^3=x^3(\frac{4}{x}-3)$, and as $\displaystyle x\to\infty$ it is obvious $\displaystyle x^3\to\infty$, and $\displaystyle \frac{4}{x}-3\to\--3$ so the limit is $\displaystyle -\infty$ is my method right?

    Using l'hopitals rule for the second i got the limit as 1/2.

    For the 4th one, i think the limit is 0 because $\displaystyle \frac{1}{x}\to\infty$ as $\displaystyle x\to\infty$ is that right?

    could any1 confirm if my answers are correct?
    Last edited by kbartlett; Sep 30th 2008 at 09:07 AM.
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  4. #4
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by kbartlett View Post
    For the 1st one, $\displaystyle 4x^2-3x^3=x^3(\frac{4}{x}-3)$, and as $\displaystyle x\to\infty$ it is obvious $\displaystyle x^3\to\infty$, and $\displaystyle \frac{4}{x}-3\to\--3$ so the limit is $\displaystyle -\infty$ is my method right?

    Using l'hopitals rule for the second i got the limit as 1/2.

    For the 4th one, i think the limit is 0 because $\displaystyle \frac{1}{x}\to\infty$ as $\displaystyle x\to\infty$ is that right?

    could any1 confirm if my answers are correct?
    yes.
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  5. #5
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    can any1 help me with 3 or 5?
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  6. #6
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    Talking

    still need help
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  7. #7
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    Quote Originally Posted by kbartlett View Post
    still need help
    For 3, you should use squeeze theorem.

    For 5: surely you can figure this out on your own?
    $\displaystyle \frac{\sqrt{x} - 1}{x^2} = \frac{\sqrt{x}}{x^2} - \frac{1}{x^2} = \sqrt{\frac{x}{x^4}} - \frac{1}{x^2} $
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  8. #8
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    For 3, i tried

    $\displaystyle \frac{xsin^2x}{x^2+5} <= \frac{x}{x^2+5}$

    but i couldn't think of anything to go the other side.

    Also thanks for the help with 5.
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  9. #9
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    $\displaystyle 0 \leq \sin^2{x} \leq 1$

    Try to manipulate it to get your original limit, then apply squeeze theorem.
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  10. #10
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    Since $\displaystyle 0 \leq sin^2{x} \leq 1$

    $\displaystyle \frac{-x}{x^2+5} \leq \frac{xsin^2x}{x^2+5} \leq \frac{x}{x^2+5}$

    Then using l'hopitals rule or otherwise if i show that $\displaystyle \frac{-x}{x^2+5} \to 0$ as $\displaystyle x \to \infty $ and $\displaystyle \frac{x}{x^2+5} \to 0 $ as $\displaystyle x \to \infty $ would that be right?
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  11. #11
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    Quote Originally Posted by kbartlett View Post
    Since $\displaystyle 0 \leq sin^2{x} \leq 1$

    $\displaystyle \frac{-x}{x^2+5} \leq \frac{xsin^2x}{x^2+5} \leq \frac{x}{x^2+5}$

    Then using l'hopitals rule or otherwise if i show that $\displaystyle \frac{-x}{x^2+5} \to 0$ as $\displaystyle x \to \infty $ and $\displaystyle \frac{x}{x^2+5} \to 0 $ as $\displaystyle x \to \infty $ would that be right?
    Excellent work, you are correct. But you don't really need to use L'H˘pital. Simply by dividing both sides of the fractions by the leading term of the denominator does the trick. For instance:

    $\displaystyle \frac{x}{x^2+5} \cdot \frac{\frac{1}{x^2}}{\frac{1}{x^2}} = \frac{\frac{1}{x}}{1+\frac{5}{x^2}}$
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  12. #12
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    Thankyou
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