Results 1 to 12 of 12

Math Help - 4 Limit problems.

  1. #1
    Junior Member
    Joined
    Mar 2008
    Posts
    49

    4 Limit problems.

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

    f(x)=4x^2-3x^3.

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

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

    <br />
 f(x) =<br />
 \begin{cases}<br />
0 & \mbox{if x is rational}\\<br />
\frac{1}{x}& \mbox{if x isn't rational}<br />
  \end{cases}<br />


    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:

    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; September 30th 2008 at 10:16 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor kalagota's Avatar
    Joined
    Oct 2007
    From
    Taguig City, Philippines
    Posts
    1,026
    Quote Originally Posted by kbartlett View Post
    Determine the limits of the following functions as x\to\infty\ , if they exist.

    f(x)=4x^2-3x^3.

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

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

    <br />
 f(x) =<br />
 \begin{cases}<br />
0 & \mbox{if x is rational}\\<br />
\frac{1}{x}& \mbox{if x isn't rational}<br />
  \end{cases}<br />


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

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

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

    fourth, i think, it is obvious..
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Mar 2008
    Posts
    49
    For the 1st one, 4x^2-3x^3=x^3(\frac{4}{x}-3), and as x\to\infty it is obvious x^3\to\infty, and \frac{4}{x}-3\to\--3 so the limit is -\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 \frac{1}{x}\to\infty as x\to\infty is that right?

    could any1 confirm if my answers are correct?
    Last edited by kbartlett; September 30th 2008 at 10:07 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor kalagota's Avatar
    Joined
    Oct 2007
    From
    Taguig City, Philippines
    Posts
    1,026
    Quote Originally Posted by kbartlett View Post
    For the 1st one, 4x^2-3x^3=x^3(\frac{4}{x}-3), and as x\to\infty it is obvious x^3\to\infty, and \frac{4}{x}-3\to\--3 so the limit is -\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 \frac{1}{x}\to\infty as x\to\infty is that right?

    could any1 confirm if my answers are correct?
    yes.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Mar 2008
    Posts
    49
    can any1 help me with 3 or 5?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Mar 2008
    Posts
    49

    Talking

    still need help
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    Joined
    Jun 2008
    Posts
    792
    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?
    \frac{\sqrt{x} - 1}{x^2} = \frac{\sqrt{x}}{x^2} - \frac{1}{x^2} = \sqrt{\frac{x}{x^4}} - \frac{1}{x^2}
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Mar 2008
    Posts
    49
    For 3, i tried

    \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.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member
    Joined
    Jun 2008
    Posts
    792
    0 \leq \sin^2{x} \leq 1

    Try to manipulate it to get your original limit, then apply squeeze theorem.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Junior Member
    Joined
    Mar 2008
    Posts
    49
    Since 0 \leq sin^2{x} \leq 1

    \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 \frac{-x}{x^2+5} \to 0 as  x \to \infty and \frac{x}{x^2+5} \to 0 as  x \to \infty would that be right?
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Super Member
    Joined
    Jun 2008
    Posts
    792
    Quote Originally Posted by kbartlett View Post
    Since 0 \leq sin^2{x} \leq 1

    \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 \frac{-x}{x^2+5} \to 0 as  x \to \infty and \frac{x}{x^2+5} \to 0 as  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:

    \frac{x}{x^2+5} \cdot \frac{\frac{1}{x^2}}{\frac{1}{x^2}} = \frac{\frac{1}{x}}{1+\frac{5}{x^2}}
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Junior Member
    Joined
    Mar 2008
    Posts
    49
    Thankyou
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Limit help (two problems)
    Posted in the Calculus Forum
    Replies: 5
    Last Post: September 19th 2011, 03:25 PM
  2. Limit problems
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: June 6th 2009, 05:04 AM
  3. Help with two limit problems...
    Posted in the Calculus Forum
    Replies: 18
    Last Post: October 14th 2008, 06:36 AM
  4. help with two limit problems please!
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 30th 2008, 01:57 AM
  5. Limit Problems
    Posted in the Calculus Forum
    Replies: 5
    Last Post: October 5th 2007, 04:09 AM

Search Tags


/mathhelpforum @mathhelpforum