Jeremy E. Riley, an electrical engineer, alumnus of the University of Utah. Love to contribute to science and math. Hoping to someday be a college professor and write books.
My Proof of one of L'Hôpital's Rules
Given two differentiable functions f & g of x, with g'(x)≠0 and g(x)≠0, in a finite or infinite open interval , with c (an extended real number) at one extremity, and
provided the limit on the right exists.
In the below proofs, I use the shorthands
Also, I use the notation to mean any open interval with endpoints & , with nearer to ; i.e.,
and implies a one-sided approach from within .
In this proof, all variables and functions may take on the values of the extended real number system. A limit is considered to "exist" when it has a definite value, including one of -∞ or +∞, but not a range of values.
Define the variable . We may apply Cauchy's mean value theorem to the finite interval :
In the limit, as , this mean gradient becomes
provided that f & g do not blow up in the open interval , i.e. f(ξ) & g(ξ) are finite, for all choices of , which is true because their individual differentiabilities guarantee their continuities in that interval.
As (1) holds for all ,
Let L be the second limit (given to exist), assumed finite. Due to the continuity of f' & g' , plus the fact that g'≠0, is continuous in . Defining , the existence of the limit is expressed as follows:
Given the continuity of f & g in , hence f and g being finite, and the monotone-increasing |g| in as x→c (because it is given that g'≠0 and g≠0), and defining , then is closer to c than is some ξ, itself chosen to be closer to c than ξ' to make |g| large enough to satisfy the following:
Now, given the differentiabilities of f & g and that g'≠0, everywhere in , and, from (3), that g(ξ' )≠g for , we may apply Cauchy's MVT to the finite interval :
Since , just as x is in (2),
Substituting from (4),
This proves the theorem for finite limits, including zero. The case where the limit is ∞ can be reduced to one that is 0, by swopping the roles of the functions f & g, and the proof is complete.