Results 1 to 4 of 4

Math Help - L'Hopital Rule

  1. #1
    Member
    Joined
    Jun 2008
    Posts
    170

    L'Hopital Rule

    Prove the following: Let  a<b be real numbers, let  f: [a,b] \to \bold{R} and  g: [a,b] \to \bold{R} be functions which are differentiable on  [a,b] . Suppose that  f(a) = g(a) = 0 , that  g' is nonzero on  [a,b] , and  \lim_{x \to a: x \in (a,b]} \frac{f'(x)}{g'(x)} exists and equals  L . Then  g(x) \neq 0 for all  x \in (a,b] , and  \lim_{x \to a: x \in (a,b]} \frac{f(x)}{g(x)} exists and equals  L .

    Suppose that  g(x) = 0 for some  x \in (a,b] . Since  g(a) = 0 , then by Rolle's theorem we can find a  y \in (a,b] such that  g'(y) = 0 . But this contradicts the fact that  g' is nonzero on  [a,b] . Hence  g(x) \neq 0 for all  x \in (a,b] .

    Now to show that  \lim_{x \to a: x \in (a,b]} \frac{f'(x)}{g'(x)} exists and equals  L , we can show that  \lim_{n \to \infty} \frac{f(x_{n})}{g(x_{n})} = L for any sequence  (x_{n})_{n=1}^{\infty} which converges to  x . How would I show this? I am guessing that I have to create a new function, and somehow manipulate it so that  \frac{f(x)}{g(x)} = \frac{f'(x)}{g'(x)} . And this is equivalent to showing that  \frac{f(x_{n})}{g(x_{n})} = \frac{f'(x_{n})}{g'(x_{n})} .
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member wingless's Avatar
    Joined
    Dec 2007
    From
    Istanbul
    Posts
    585
    The mean value theorem says that there is an x in the interval (a,b) such that g'(x) [f(b)-f(a)] = f'(x)[g(b)-g(a)].

    Now, we know that g' is nonzero (you assumed it), and if g(a) \neq g(b), we can write \frac{f'(x)}{g'(x)} = \frac{f(b)-f(a)}{g(b)-g(a)}.

    As we know f(a) = g(a) = 0,

    \frac{f'(x)}{g'(x)} = \frac{f(b)}{g(b)}

    If we take the limit of both sides, then x and b will approach the same point.

    \lim_{x\to x_0}\frac{f'(x)}{g'(x)} = \lim_{b\to b_0}\frac{f(b)}{g(b)} = \lim_{x\to x_0}\frac{f(x)}{g(x)}

    \lim_{x\to x_0}\frac{f(x)}{g(x)} = \lim_{x\to x_0}\frac{f'(x)}{g'(x)}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jun 2008
    Posts
    170
    You could use Rolle's Theorem also right? In terms of sequences you can define a function  z_{n}: [a, x_{n}] \to \bold{R} by  z_{n}(x) =  f(x)g(x_{n}) - g(x)f(x_{n}) . So this is continuous on the interval, and is zero at the endpoints. And  z'_{n}(x) = f'(x)g(x_{n})- g'(x)f(x_{n}) . Using Rolle's Theorem we get  \frac{f(x_{n})}{g(x_{n})} = \frac{f'(y_{n})}{g'(y_{n})} .  x_n \to a as  n \to \infty and so  y_n \in (a, x_n) \to a as  n \to \infty by Squeeze Theorem.

    And so  \frac{f'(x_{n})}{g'(x_{n})} \to L, \frac{f'(y_{n})}{g'(y_{n})} \to L \implies \frac{f(x_{n})}{g(x_{n})} \to L .
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member wingless's Avatar
    Joined
    Dec 2007
    From
    Istanbul
    Posts
    585
    Quote Originally Posted by particlejohn View Post
    You could use Rolle's Theorem also right? In terms of sequences you can define a function  z_{n}: [a, x_{n}] \to \bold{R} by  z_{n}(x) =  f(x)g(x_{n}) - g(x)f(x_{n}) . So this is continuous on the interval, and is zero at the endpoints. And  z'_{n}(x) = f'(x)g(x_{n})- g'(x)f(x_{n}) . Using Rolle's Theorem we get  \frac{f(x_{n})}{g(x_{n})} = \frac{f'(y_{n})}{g'(y_{n})} .  x_n \to a as  n \to \infty and so  y_n \in (a, x_n) \to a as  n \to \infty by Squeeze Theorem.

    And so  \frac{f'(x_{n})}{g'(x_{n})} \to L, \frac{f'(y_{n})}{g'(y_{n})} \to L \implies \frac{f(x_{n})}{g(x_{n})} \to L .
    But what you did here was proving Mean Value Theorem using Rolle's Theorem =)
    (except you used sequences instead of function. see here)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. L' Hopital's Rule
    Posted in the Calculus Forum
    Replies: 3
    Last Post: February 26th 2010, 11:08 AM
  2. L'Hopital's Rule
    Posted in the Calculus Forum
    Replies: 4
    Last Post: July 28th 2009, 05:40 PM
  3. líHopitalís Rule Help..
    Posted in the Calculus Forum
    Replies: 5
    Last Post: March 27th 2009, 11:06 PM
  4. L'hopital Rule qn
    Posted in the Calculus Forum
    Replies: 2
    Last Post: August 26th 2008, 10:36 AM
  5. L'Hopital's Rule
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 29th 2008, 10:02 AM

Search Tags


/mathhelpforum @mathhelpforum