Results 1 to 4 of 4

Math Help - Evaluating limit

  1. #1
    Junior Member
    Joined
    Jan 2009
    Posts
    66

    Evaluating limit

    Did l work out this problem correctly ? I thought there was a step which l did wrongly. Can you pliz help

    BTW the question and attempt to question is on attached image
    Attached Thumbnails Attached Thumbnails Evaluating limit-img_0002.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member General's Avatar
    Joined
    Jan 2010
    From
    Kuwait
    Posts
    562
    Quote Originally Posted by nyasha View Post
    Did l work out this problem correctly ? I thought there was a step which l did wrongly. Can you pliz help

    BTW the question and attempt to question is on attached image
    You have a mistake in the 2nd line.
    How in earth did you get x^{\frac{3}{2}} ?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member General's Avatar
    Joined
    Jan 2010
    From
    Kuwait
    Posts
    562
    Quote Originally Posted by nyasha View Post
    Did l work out this problem correctly ? I thought there was a step which l did wrongly. Can you pliz help

    BTW the question and attempt to question is on attached image
    You have a mistake in the 2nd line.
    How in earth did you get x^{\frac{3}{2}} ?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,408
    Thanks
    1328
    I was wondering why he would write x^3 as \left(x^{3/2}right)^2 myself but I notice it disappears in the next line!

    Nyasha, Yes, what you have done is correct. Although, in fact, it was not necessary to be that general. Just showing that
    \lim_{x\to 0}\frac{x^3}{3x^3}= \frac{1}{3}
    where the limit is taken along the horizontal line y= -2, and
    \lim_{y\to -2}\frac{4(y+ 2)^3}{-(y+2)^2}= 0
    where the limit is taken along the vertical line x= 0, are different is sufficient to show that the limit does not exist.

    If you were doing that thinking that if you got a result that did NOT depend on m that would tell you that the limit did exist and would give you the limit, you are mistaken. For one thing, the vertical line, x= constant, cannot be written in the form y= mx so you would not have proven that the limit along all straight lines is the same. And even if you did that, it would not be enough. There exist examples where the limit along any straight line gives the same thing but limits taken along curves gives a different result- in those cases, the limit itself still does not exist.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Need help evaluating the limit.
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 13th 2010, 06:18 PM
  2. Evaluating a limit
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 2nd 2010, 01:01 PM
  3. evaluating a limit
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 12th 2009, 05:01 PM
  4. evaluating a limit
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: October 17th 2009, 09:54 PM
  5. Evaluating A Limit
    Posted in the Calculus Forum
    Replies: 3
    Last Post: July 1st 2009, 10:43 PM

Search Tags


/mathhelpforum @mathhelpforum