Results 1 to 4 of 4

Math Help - Problem with precise definition of the limit

  1. #1
    Junior Member
    Joined
    Jan 2010
    Posts
    49

    Problem with precise definition of the limit

    Question :

    let h(x)=x^2, x<2
    = 3, x=2
    =2, x>2

    Show that
    lim(x-->2)h(x)does not equal to 4

    my teacher give the answer :

    for 2<x<2+delta => h(x) = 2 => abs(h(x)-4)=2

    Thus for epsilon < 2, abs(h(x)-4) >= epsilon whenever 2<x<2+delta no matter how small we choose
    delta > 0 => lim(x-->2)h(x)does not equal to 4


    i cannot understand the second line of the explanation. why we have to consider epsilon < 2

    why cannot consider this : abs(h(x)-4)=2 < epsilon thus,  epsilon should > 2 right?

    can anyone explain it to me? i really could'nt understand this!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    14,973
    Thanks
    1121
    In order that the limit be 4, you must be able to make |f(x)- 4|< \epsilon for any \epsilon> 0. Your teacher was pointing out that the difficulty (the fact that you can't make it " <\epsilon" happens if you take \epsilon< 2.

    The fact that it is true for some \epsilon ( \epsilon> 2) doesn't matter. It has to be true for all \epsilon.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jan 2010
    Posts
    49
    Quote Originally Posted by HallsofIvy View Post
    In order that the limit be 4, you must be able to make |f(x)- 4|< \epsilon for any \epsilon> 0. Your teacher was pointing out that the difficulty (the fact that you can't make it " <\epsilon" happens if you take \epsilon< 2.

    The fact that it is true for some \epsilon ( \epsilon> 2) doesn't matter. It has to be true for all \epsilon.

    can you give some example, i.e. replace the epsilon and delta with a number so that i can 'see' my teacher's explanation. still confuse...

    what is the relationship between a delta and epsilon in this case? i can't see the relationship... wat the delta have to do with the epsilon value in this problem???
    anyone???
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jan 2010
    Posts
    49
    In order to show that lim x->x_0 f(x) does not equal to L, you have to show that

    for every \delta > 0 , there exist \epsilon > 0

    thus |x - x_0| < \delta and |f(x) - L|>\epsilon

    so in the above case

    |x-2|<\delta ==> 2-\delta<x<2+\delta

    and thus 2<x<2+\delta which is from the graph

    ==> h(x) = 2 ==> |h(x) - L| = |2-4| = |-2| = 2 >\epsilon

    if we take \epsilon = 0.0001,thus |h(x) - L|> \epsilon (since \epsilon > 0

    thus, it shows that lim(x-->2)h(x)does not equal to 4
    Last edited by bobey; February 23rd 2010 at 01:30 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Precise definition of a limit question
    Posted in the Calculus Forum
    Replies: 4
    Last Post: September 20th 2011, 03:10 PM
  2. Precise Def of a Limit Problem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: July 19th 2010, 04:38 PM
  3. Precise definition of a limit.
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 27th 2010, 04:35 PM
  4. Precise Definition of a Limit Problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 26th 2009, 11:56 AM
  5. Precise definition of a limit
    Posted in the Calculus Forum
    Replies: 3
    Last Post: July 2nd 2009, 09:42 PM

Search Tags


/mathhelpforum @mathhelpforum