User:Toolnut

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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[edit]

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 .

Proof 1[edit]

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.

 

 

 

 

(1)

As (1) holds for all ,

Proof 2[edit]

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:

 

 

 

 

(2)

Given the continuity of f & g in , hence f and g being finite, and the monotone-increasing |g| in as xc (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:

 

 

 

 

(3)

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 :

 

 

 

 

(4)

Since , just as x is in (2),

Substituting from (4),

Multiplying both sides by the absolute value of the denominator and using (3) (since ), repeatedly, together with the triangle inequality rule,

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.