Results 1 to 7 of 7

Math Help - limit proof

  1. #1
    Member
    Joined
    Nov 2006
    Posts
    142

    limit proof

    I have that the limit of f(x) as x goes to c exists and that the limit of g(x) as x goes to c does not exist.

    How do I prove that the limit of h(x) as x goes to c does not exist (where h = f + g)?

    What can you say about the limit of g using epsilon notation to reach a contradiction?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Sep 2008
    Posts
    34
    lim g(x) = lim h(x) - lim f(x)
    because the "limit" operation distributes over addition/subtraction. <-- Are you allowed to assume this in your problem or do you essentially have to prove this?
    If RHS is finite then LHS is finite. QED.
    Last edited by hwhelper; September 20th 2008 at 01:25 PM. Reason: Making sure assumptions are OK
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Nov 2006
    Posts
    142
    I guess that does work, but I don't want to use that.

    abs(g(x) - L) >= epsilon I think, but g doesn't technically have a limit, so what do you say?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by PvtBillPilgrim View Post
    I guess that does work, but I don't want to use that.

    abs(g(x) - L) >= epsilon I think, but g doesn't technically have a limit, so what do you say?
    i guess that could work.

    you have for all \epsilon > 0 and all those \delta's and stuff that

    |f(x) - F| < \frac {\epsilon}2 but |g(x) - G| \ge \frac {\epsilon}2 = \frac {\epsilon}2 + \zeta, where \zeta > 0

    where F and G are the limits of f(x) and g(x) respectfully.

    had \lim_{x \to c}h(x) existed, we would have |h(x) - H| < \epsilon, where H = F + G is the limit of h(x)

    but, |h(x) - H| = |[f(x) + g(x)] - [F + G]| = |[f(x) - F] + [g(x) - G]|  \le |f(x) - F| + |g(x) - G| = \frac {\epsilon}2 + \bigg( \frac {\epsilon}2 + \zeta \bigg) >  \epsilon

    i don't really like this though. hwhelper's way is perhaps better. maybe you can fix it up
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Nov 2006
    Posts
    142
    it's weird because at the end you get that

    abs(f - F) < epsilon
    and abs(g- G) >= epsilon

    does this imply that abs(h - (F+G)) >= epsilon?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by PvtBillPilgrim View Post
    it's weird because at the end you get that

    abs(f - F) < epsilon
    and abs(g- G) >= epsilon

    does this imply that abs(h - (F+G)) >= epsilon?
    yeah, the inequalities turned in opposing directions does make things unclear. that's why i liked hwhelper's solution better. it is more straight forward. an epsilon proof is more subtle, i'll have to think about it more
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by PvtBillPilgrim View Post
    I have that the limit of f(x) as x goes to c exists and that the limit of g(x) as x goes to c does not exist.

    How do I prove that the limit of h(x) as x goes to c does not exist (where h = f + g)?

    What can you say about the limit of g using epsilon notation to reach a contradiction?
    If the limit of (f+g) existed then the limit of [(f+g)-f] = g would have existed.
    A contradiction.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Proof of limit
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 14th 2011, 01:37 PM
  2. [SOLVED] limit proof
    Posted in the Differential Geometry Forum
    Replies: 18
    Last Post: December 23rd 2010, 09:48 PM
  3. Limit Proof
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: June 29th 2010, 06:30 PM
  4. Another limit proof
    Posted in the Calculus Forum
    Replies: 4
    Last Post: October 11th 2008, 12:38 PM
  5. Limit Proof
    Posted in the Calculus Forum
    Replies: 6
    Last Post: October 9th 2007, 01:28 PM

Search Tags


/mathhelpforum @mathhelpforum