Results 1 to 12 of 12

Math Help - L'Hopital's Rule help please

  1. #1
    Newbie
    Joined
    Feb 2011
    Posts
    15

    L'Hopital's Rule help please

    I need help with this question. I think I'm supposed to use L'Hopital's Rule, but I'm not sure if I'm doing it correctly.

    lim as x -->0 (2x^6+6x^3)/(4x^5+3x^3)

    I got 0 at first, but then I got 2. I'm just so unsure! Any help at all is greatly appreciated!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,831
    Thanks
    1602
    If you're going to use L'Hospital's Rule

    \displaystyle \lim_{x \to 0}\frac{2x^6 + 6x^3}{4x^5 + 3x^3} = \lim_{x \to 0}\frac{2x^3 + 6}{4x^2 + 3}

    \displaystyle = \lim_{x \to 0}\frac{6x^2}{8x} by L'Hospital's Rule

    \displaystyle = \lim_{x \to 0}\frac{12x}{8} by L'Hospital's Rule

    \displaystyle = 0.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Jun 2009
    From
    Africa
    Posts
    641
    How did you get 2 ??
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Feb 2011
    Posts
    15
    I applied L'Hopital's rule like 4 times then stopped. I knew that would be wrong :/

    But how did you figure out this part?

    lim as x-->0= 2x^3+ 6/ 4x^2+3

    I thought, you just went straight to finding the derivative after the original was found to be indeterminate? Please tell me how that was found.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Jun 2009
    From
    Africa
    Posts
    641
    simplify first =)
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    I think the idea is

    \displaystyle\frac{2x^6+6x^3}{4x^5+3x^3}=\left[\frac{x^3}{x^3}\right]\frac{2x^3+6}{4x^2+3}

    and the limit as x approaches zero is 2.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Feb 2011
    Posts
    15
    Ohhh! Okay! But then why do we stop simplifying at 2x^3/ 4x^2 ? Can't we keep going? Like get rid of all the x's? Please forgive me if this comes off as stupid, I just wanna get it down once and for all lol :/
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member
    Joined
    Jun 2009
    From
    Africa
    Posts
    641
    or,
    \frac{2x^6+6x^3}{4x^5+3x^3}=\frac{x}{2}-\frac{3x-12}{8x^2+6}
    which has 2 as limit when x approaches 0.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Feb 2011
    Posts
    15
    Okay Raoh, thanks for that but I don't follow :/ I'm not good at manipulating math at all!

    I see you and Archie Meade agree that the answer should be 2? Is 0 out of the question then? lol This seems to be a controversial question :P
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,831
    Thanks
    1602
    I stupidly applied L'Hospital's Rule when it wasn't \displaystyle \frac{0}{0} ><

    Serves me right for doing maths at 6am without any sleep :P
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Newbie
    Joined
    Feb 2011
    Posts
    15
    Wow thanks guys!Good thing that's cleared up! It would've took me a while to figure this out -___- Hopefully one day I can do math like you guys lol

    Thanks again!
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by kikiya View Post
    Ohhh! Okay! But then why do we stop simplifying at 2x^3/ 4x^2 ? Can't we keep going? Like get rid of all the x's? Please forgive me if this comes off as stupid, I just wanna get it down once and for all lol :/
    At

    \displaystyle\frac{2x^3+6}{4x^2+3}

    We do get rid of the x's !!
    Now it's safe to set the remaining ones all to zero as we are beyond getting 0/0
    which was the indeterminate form.

    As you were asked to apply L'Hopital's Rule...
    Initially, we have 0/0, so...

    Differentiate numerator and denominator

    \displaystyle\frac{12x^5+18x^2}{20x^4+9x^2}

    Still 0/0, so differentiate again...

    \displaystyle\frac{60x^4+36x}{80x^3+18x}

    Still 0/0, so differentiate again..

    \displaystyle\frac{240x^3+36}{240x^2+18}

    It's no longer indeterminate, especially there is no potential for zero in the denominator,
    so now it's safe to set the remaining x's to zero to evaluate the limit.
    Last edited by Archie Meade; February 25th 2011 at 02:15 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. l'Hopital's rule
    Posted in the Calculus Forum
    Replies: 10
    Last Post: February 23rd 2011, 12:11 AM
  2. L'Hopital's Rule
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 1st 2010, 07:44 AM
  3. L'Hopital's Rule
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 11th 2009, 11:39 PM
  4. L Hopital Rule
    Posted in the Calculus Forum
    Replies: 8
    Last Post: May 3rd 2009, 10:40 AM
  5. l'Hopital's Rule
    Posted in the Calculus Forum
    Replies: 10
    Last Post: October 21st 2007, 03:52 PM

Search Tags


/mathhelpforum @mathhelpforum