Results 1 to 6 of 6

Math Help - proof on limits

  1. #1
    Junior Member
    Joined
    Dec 2009
    Posts
    30

    Cool proof on limits

    Problem
    Let f:R-->R be a differentiable function. Prove that if lim x-->infinity f(x) and lim x-->infinity f'(x) both exist and are finite, then lim x--> infinity f'(x) = 0.



    Please help and assist in which def or theorem i should look at.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member mohammadfawaz's Avatar
    Joined
    Feb 2010
    From
    Lebanon - Beirut
    Posts
    100
    You should notice that for some \epsilon>0, |f(x)-f(t)|<\epsilon whenever x,t>X for some X. Now look at the definition of derivative: f'(x) = \lim_{t->x}{\frac{f(t)-f(x)}{t-x)}}. This number exists by differentiability of f.
    Now, for another given \epsilon>0, you need to prove that |f'(x)|<\epsilon using above relations.

    Hope this helps
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Dec 2009
    Posts
    30
    i don't see what you're getting at?

    i don't see how you end up at lim x-->infinity f'(x) =0
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member mohammadfawaz's Avatar
    Joined
    Feb 2010
    From
    Lebanon - Beirut
    Posts
    100
    Forget what I said.
    I have another idea in mind.
    Use MVT on the interval [x,x+1]:
    we have: f(x+1)-f(x) = f'(c)(x+1-x) where c belongs to (x,x+1). Hence, f(x+1)-f(x) = f'(c).
    When x goes to infinity, x+1 goes to infinity and so does c. So, both f(x) and f(x+1) go to the limit L. Therefore, f'(c) goes to L-L = 0.
    Hope this helps
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Dec 2009
    Posts
    30
    thank you. this one makes more sense and now i see where you are coming from with the MVT.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member Pinkk's Avatar
    Joined
    Mar 2009
    From
    Uptown Manhattan, NY, USA
    Posts
    419
    You can also use L'Hospital's Rule; since both those limits are finite we can write \lim_{x\to \infty}[f(x)+f'(x)] = L. Now let a(x) = f(x)e^{x} and let b(x)= e^{x}; observe that \frac{a(x)}{b(x)} = f(x) since e^{x} \neq 0 for all x\in \mathbb{R}. Since \lim_{x\to \infty}\frac{a'(x)}{b'(x)} = L and \lim_{x\to \infty}|b(x)| = +\infty, we can apply L'Hospital's Rule, and so \lim_{x\to \infty}\frac{a(x)}{b(x)} = \lim_{x\to \infty}f(x) = L. It then follows that \lim_{x\to \infty}f'(x) = 0.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Limits proof
    Posted in the Calculus Forum
    Replies: 4
    Last Post: September 20th 2009, 11:45 PM
  2. Limits Proof
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 9th 2009, 11:49 PM
  3. Proof with limits
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 30th 2008, 12:15 PM
  4. proof of 2 limits
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 25th 2007, 09:55 AM
  5. limits/proof
    Posted in the Calculus Forum
    Replies: 4
    Last Post: April 8th 2007, 02:44 PM

Search Tags


/mathhelpforum @mathhelpforum