The Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

Explanation of L'Hopital's Rule

Date: 02/06/2001 at 23:34:26
From: Samuel Ward
Subject: L'Hospital's Rule

In certain cases, L'Hopital's Rule connects the limit of a quotient 
(f/g) to the limit of the quotient of the derivatives (f'/g'). This 
is true when f and g go to 0 or infinity at the point where the limit 
is taken. 

I understand how to use this rule, and I somewhat understand the 
proof, but I still do not understand why this happens. Can you help?
Please also try to describe explicitly how to think of the roles of 
the limit, the derivative, and the quotient.

Date: 02/07/2001 at 20:55:05
From: Doctor Fenton
Subject: Re: L'Hospital's Rule

Hi Samuel,

Thanks for writing to Dr. Math. You've posed a very good question. One
way you can think of this is to use the idea of derivative: a function
f(x) is differentiable at x=a if f(x) is very close to its tangent 
line y = f'(a)*(x-a) + f(a) near x = a. Specifically,

     f(x) = f(a) + f'(a)*(x-a) + E1(x)

where E1(x) is an error term which goes to 0 as x goes to a. In fact, 
E1(x) must approach 0 so fast that 

     lim  ----- = 0
     x->a  x-a


    E1(x)   f(x)-f(a)
    ----- = --------- - f'(a)   
     x-a       x-a

and we know from the definition of derivative that this quantity has 
the limit 0 at a.

Similarly, if g is differentiable at x = a,

     g(x) = g(a) + g'(a)*(x-a) + E2(x)

where E2(x) is another error term which goes to 0 as x->a. If you're 
computing the limit of f(x)/g(x) as x->a and if g(a) is not equal to 
0, then as x->a, the numerator becomes indistinguishable from f(a) 
and the denominator from g(a), so the limit is

     lim    f(x)   f(a)
     x->a   ---- = ----
            g(x)   g(a)

If both f(a) and g(a) are 0, then we must use the tangent 
approximations to say that

     f(x)   f(a) + f'(a)*(x-a) + E1(x)  
     ---- = --------------------------
     g(x)   g(a) + g'(a)*(x-a) + E2(x) 

            f'(a)*(x-a) + E1(x)  
          = ---------------------
            g'(a)*(x-a) + E2(x)  

            f'(a) + [E1(x)/(x-a)]  
          = ---------------------
            g'(a) + [E2(x)/(x-a)]

and we have seen that the second term becomes negligible as x->a.

In other words, when both function values approach 0 as x->a, the 
ratio of the function values just reduces to the ratio of the slopes 
of the tangents, because both functions are very close to their 
tangent lines.

Does this clarify the situation? If you still have questions, please 
write again.

- Doctor Fenton, The Math Forum   
Associated Topics:
High School Calculus

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.