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Proof of L'Hopital's Rule


Date: 12/23/98 at 01:21:34
From: James Williams
Subject: Proof of L'hopital's rule

Can you show me a proof of L'Hopital's Rule?

Thanks


Date: 12/23/98 at 10:27:11
From: Doctor Anthony
Subject: Re: Proof of L'hopital's rule

To prove L'Hopital's Rule (sometimes spelled L'Hospital's Rule), we 
use the Taylor expansion:

   f(a+h) = f(a) + hf'(a) + terms in h^2 and higher
   g(a+h) = g(a) + hg'(a) + terms in h^2 and higher

So:

         f(a+h)     f(a)+h*f'(a)
    Lt   ------ ->  ------------
   h->0  g(a+h)     g(a)+h*g'(a)

so with f(a) = g(a) = 0 we get:

         f(a+h)     h*f'(a)       f'(a)
    Lt   ------- ->  -------  ->  ------
   h->0  g(a+h)     h*g'(a)       g'(a)

We can use l'Hopital's also if f'(a) -> infinity and g'(a) -> infinity:

   f(a)      infinity           1/g(a)       0
   ---- ->   --------     so   --------  -> ---
   g(a)      infinity           1/f(a)       0

and applying l'Hopital's to this latter expression, we get:

    f(a)       -g'(a)/[g(a)]^2        g'(a)*[f(a)]^2
   ------  ->  ----------------  ->  ----------------
    g(a)       -f'(a)/[f(a)]^2        f'(a)*[g(a)]^2

                         
and cross-multiplying:

    f'(a)        f(a)
   -------  ->  ------
    g'(a)        g(a)

Therefore whether we have 0/0 or infinity/infinity we can use 
l'Hopital's rule.
 
- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/   


Date: 12/23/98 at 11:43:47
From: Doctor Rob
Subject: Re: Proof of L'hopital's rule

Thanks for writing to Ask Dr. Math!

This is a rather complicated business. First of all, there are several 
forms of the Rule. I will state and prove one of them, and try to 
indicate how other forms are corollaries of this one.

======================================================================
Theorem (L'Ho^pital's Rule):

Let f(x) and g(x) be differentiable on the interval a <= x < b, with 
g'(x) nonzero throughout.  If:

  (i)  lim  f(x) = 0  and  lim  g(x) = 0
      x->b-               x->b-

or if

  (ii)  lim  f(x) = infinity  and  lim  g(x) = infinity
       x->b-                      x->b-

and if:

    lim  f'(x)/g'(x) = L
   x->b-

then:

    lim  f(x)/g(x) = L
   x->b-

Note: b may be any real number or infinity, and L may be any real 
number or infinity.

======================================================================

Proof:  We restrict our attention to the case where b and L are real
numbers.

Assume first hypothesis (i) is satisfied. We may set f(b) = 0 and
g(b) = 0, and then f(x) and g(x) are continuous on the closed interval
a <= x <= b. Applying Cauchy's Mean Value Theorem (see below), we see
that for any x with a < x < b, there is a point x0 with x < x0 < b 
such that:

   f'(x0)/g'(x0) = [f(b)-f(x)]/[g(b)-g(x)]
                 = f(x)/g(x)

Since:

    lim   f'(x0)/g'(x0) = L
   x0->b-

given any epsilon > 0, we may choose delta > 0 so that:

   |f(x)/g(x) - L| = |f'(x0)/g'(x0) - L| < epsilon

whenever b - delta(n) < x < b.  This proves that:

    lim  f(x)/g(x) = L
   x->b-

This is the desired conclusion for hypothesis (i).
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 
Now assume hypothesis (ii). Given epsilon > 0, we first choose 
delta > 0 so that, whenever b - delta < x1 < b, then:

   |f'(x1)/g'(x1) - L| < epsilon

Set x0 = b - delta, and take any points x with x0 < x < b. By Cauchy's
Mean Value Theorem (see below), there is a choice of x1 with 
x0 < x1 < b, such that:

   [f(x)-f(x0)]/[g(x)-g(x0)] = f'(x1)/g'(x1)

Thus:

   |[f(x)-f(x0)]/[g(x)-g(x0)] - L| < epsilon

Now, defining:

   h(x) = [1-f(x0)/f(x)]/[1-g(x0)/g(x)]

we can rewrite this as:

   |h(x)*f(x)/g(x) - L| < epsilon

which is valid for all x with x0 < x < b. Now by using hypothesis 
(ii), it follows that:

    lim  h(x) = 1
   x->b-

Choose x1 with x0 < x1 < b so that |h(x)-1| < epsilon and h(x) > 1/2,
whenever x1 < x < b.  For such values of x, we have:

   |h(x)*[f(x)/g(x)-L]| = |h(x)*f(x)/g(x) - L*h(x)|
                        <= |h(x)*f(x)/g(x) - L| + |L*[1-h(x)]|
                        < epsilon + |L|*epsilon

and:

   |f(x)/g(x) - L| < (1+|L|)*epsilon/h(x) < 2*(1+|L|)*epsilon

This proves that:

    lim  f(x)/g(x) = L
   x->b-

Q.E.D.
======================================================================

Now the proof above depends in two places on the following.

Cauchy's Mean Value Theorem:

Suppose that two functions f(x) and g(x) are continuous in the closed
interval a <= x <= b and differentiable in the open interval a < x < b. 
Suppose further that g'(x) is nonzero for x in a < x < b. Then there 
exists at least one number c with a < c < b such that:

   [f(b)-f(a)]/[g(b)-g(a)] = f'(c)/g'(c)
----------------------------------------------------------------------
Proof: Let:

   h(x) = [g(b)-g(a)]*[f(x)-f(a)] - [f(b)-f(a)]*[g(x)-g(a)]

Then clearly h(a) = h(b) = 0. Since h is continuous in [a,b] and
differentiable in (a,b), we can apply Rolle's Theorem, which tells us
that there exists at least one number c in (a,b) such that h'(c) = 0.
Then:

   h'(c) = [g(b)-g(a)]*f'(c) - [f(b)-f(a)]*g'(c) = 0

Rearranging this equation, and using the fact that g'(c) is nonzero,
we get:

   [f(b)-f(a)]/[g(b)-g(a)] = f'(c)/g'(c)

Q.E.D.

======================================================================

Now let's talk about variations on L'Ho^pital's Rule.

The first is to see that if L = +infinity or -infinity, the same ideas
work to give a proof, although we have to modify the parts dealing 
with epsilons and deltas to accommodate the definition of an infinite 
limit. Instead of |F(x) - L| < epsilon, you need |F(x)| > 1/epsilon, 
and so on.

The second is to see that if we replace the limits with x->b- with ones 
with x->b+, and replace the condition a <= x < b with b < x <= a, the 
conclusion still holds. This is a corollary of the stated theorem,
obtained by substituting y = 2*b - x, and applying the theorem.

The third is to see that if we replace the one-sided limits by two-
sided ones, the conclusion still holds, as a corollary of the theorem 
and the immediately preceding paragraph.

The fourth is to see that if we take b = +infinity or -infinity, the
conclusion of the theorem still holds. We prove this by substituting
x = 1/y, and then as x -> +infinity, y -> 0+, and as x -> -infinity,
y -> 0-.  Also, dx/dy = -1/y^2, and by the chain rule:

   f'(x) = d/dy[f(1/y)] = f'(1/y)*d/dy[1/y] = f'(1/y)*(-1/y^2)

and similarly for g'(x). Thus:

    lim      f'(x)/g'(x) =  lim  [f'(1/y)*(-1/y^2)]/[g'(1/y)*(-1/y^2)]
   x->+infinity             y->0+
                         =  lim  f'(1/y)/g'(1/y)
                            y->0+
                         =  lim  f(1/y)/g(1/y)
                            y->0+

by the theorem, so:

       lim     f'(x)/g'(x) =     lim     f(x)/g(x)
   x->+infinity              x->+infinity

which was to be shown. The case x -> -infinity is handled in the same
way.

I know this is long and complicated, but at least I warned you!

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Analysis
High School Calculus

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