Results 1 to 2 of 2

Math Help - need help with three conceptual calculus problems!

  1. #1
    Member
    Joined
    May 2007
    Posts
    150

    need help with three conceptual calculus problems!

    ok so 1.

    true or false (and provide reasoning, not a full length proof) that
    lim as x -> a of f(x)*g(x) can not exist
    if lim x->a f(x) and lim x->a g(x) both exist
    so its kind of a reasoning to show that the product limit law always works if both limits exist?


    2. In the formal definition of limits: lim x->c f(x) = L, if for every E>0 there is a d>0 such that IF 0<|x-c|<d THEN |f(x)-L|<E

    can you switch the inequalities after the IF and THEN, and does the definition still hold?


    3. TRUE or FALSE

    f(x) = 1 if x is rational
    = 2 if x is irrational

    the limit as X-->0 exists for f(x)

    im not sure about this one cause of whether the right and left hand limits meet and if so, at 1 or 2??!?!


    thanks! please do asap
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by stones44 View Post
    true or false (and provide reasoning, not a full length proof) that
    lim as x -> a of f(x)*g(x) can not exist
    if lim x->a f(x) and lim x->a g(x) both exist
    so its kind of a reasoning to show that the product limit law always works if both limits exist?
    If \lim_{x\to a}f(x) and \lim_{x\to a}g(x) exist then \lim_{x\to a}f(x)g(x) exists. That is a theorem.

    2. In the formal definition of limits: lim x->c f(x) = L, if for every E>0 there is a d>0 such that IF 0<|x-c|<d THEN |f(x)-L|<E

    can you switch the inequalities after the IF and THEN, and does the definition still hold?
    No

    3. TRUE or FALSE

    f(x) = 1 if x is rational
    = 2 if x is irrational

    the limit as X-->0 exists for f(x)

    im not sure about this one cause of whether the right and left hand limits meet and if so, at 1 or 2??!?!
    Assume that \lim_{x\to 0}f(x) = L (that is, assume it exists).


    If L < 1 then pick \epsilon > 0 so that L + \epsilon < 1.
    This means there is \delta > 0 so that if 0 < |x| < \delta \implies |f(x) - L| < \epsilon.
    Therefore L - \epsilon < f(x) < L + \epsilon < 1 for 0 < |x| < \delta.
    But this is a contradiction since f(x) \not < 1.

    If L > 2 then pick \epsilon > 0 so that L - \epsilon > 2.
    Repeating a similar argument will show f(x) > 2 and this is a contradiction too.

    Therefore 1\leq L \leq 2 if it exists. Using a similar argument as above we can show L\not= 1, L\not = 2. This forces 1 < L < 2. But then this means there is \epsilon > 0 so that 1< L - \epsilon and L+\epsilon < 2.
    Thus, there is \delta > 0 so that 0 < |x| < \delta \implies |f(x) - L| < \epsilon.
    Thus, 1 < L - \epsilon < f(x) < L + \epsilon < 2.
    This is a contradiction beceause f does not take values strictly between 1 and 2.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: July 27th 2011, 10:58 AM
  2. Replies: 1
    Last Post: March 5th 2011, 10:41 PM
  3. Two conceptual problems with binomial distributions
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: February 7th 2011, 12:23 PM
  4. Replies: 4
    Last Post: August 14th 2010, 06:44 AM
  5. Replies: 2
    Last Post: June 25th 2010, 10:41 PM

Search Tags


/mathhelpforum @mathhelpforum