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
![{\displaystyle {\begin{aligned}&\lim _{x\in \mathbb {I} \to c}|g(x)|=\infty ,{\text{ then}}\\&\lim _{x\in \mathbb {I} \to c}{\frac {f(x)}{g(x)}}=\lim _{x\in \mathbb {I} \to c}{\frac {f'(x)}{g'(x)}},\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2c4958a9184869b14b23b95aba00555ec3f31901)
provided the limit on the right exists.
In the below proofs, I use the shorthands
![{\displaystyle f\equiv f(x),\ g\equiv g(x),\ f'\equiv f'(x),{\text{ and }}g'\equiv g'(x).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cca1e677fbecc112485a3a10f882a05e29fb41b6)
Also, I use the notation
to mean any open interval with endpoints
&
, with
nearer to
; i.e.,
![{\displaystyle a<b{\text{ if }}c{\text{ is a right endpoint, and }}b<a{\text{ if }}c{\text{ is a left endpoint}},\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c1a13c5c8723004d4b3931c70aaadffa2be703db)
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
:
![{\displaystyle \exists \xi '\in \mathbb {I'} :{\frac {f'(\xi ')}{g'(\xi ')}}={\frac {f-f(\xi )}{g-g(\xi )}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d119990cccd6615a3a66fcaf8c48485aa0c27af4)
In the limit, as
, this mean gradient becomes
![{\displaystyle \lim _{x\to c}{\left({\frac {f-f(\xi )}{g-g(\xi )}}={\frac {{\frac {f}{g}}-{\frac {f(\xi )}{g}}}{1-{\frac {g(\xi )}{g}}}}\right)}={\frac {\lim _{x\to c}{\frac {f}{g}}-\left(\lim _{x\to c}{\frac {f(\xi )}{g}}=0\right)}{1-\left(\lim _{x\to c}{\frac {g(\xi )}{g}}=0\right)}}=\lim _{x\to c}{\frac {f}{g}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b4b6acf0f8af25e2ec134aa3cad60c2442007abd)
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.
![{\displaystyle \therefore \exists \xi '\in \mathbb {I''} :{\frac {f'(\xi ')}{g'(\xi ')}}=\lim _{x\to c}{\frac {f}{g}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/047cd2dd2090737977959388887ea20a6f973b1b) | | (1) |
As (1) holds for all
,
![{\displaystyle \lim _{\xi \to c\,\therefore \xi '\to c}{\frac {f'(\xi ')}{g'(\xi ')}}=\lim _{x\to c}{\frac {f}{g}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6ce9dfcf3a8cf623642f539d7fa1838e6cd1109a)
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:
![{\displaystyle \exists L:\forall \epsilon >0,\,\exists \xi '\in \mathbb {I} :\{x\in \mathbb {I'} \}\Rightarrow \left\{\left|{\frac {f'}{g'}}-L\right|<\epsilon \right\}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/26cc150ccb4b00fababe7f87a2c916b5fcfd4cf0) | | (2) |
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:
![{\displaystyle \forall \epsilon >0,\,\exists \xi \in \mathbb {I'} \subset \mathbb {I} :\{x\in \mathbb {I''} \}\Rightarrow \left\{\max \left(\left|{\frac {f(\xi ')}{g}}\right|,\left|{\frac {g(\xi ')}{g}}\right|\right)<\epsilon \right\}\land \{g(\xi ')\neq g\};}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f25f44e32b95bc2cf52e6b18e055a611bc3513cb) | | (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
:
![{\displaystyle {\frac {f'(\xi '')}{g'(\xi '')}}={\frac {f-f(\xi ')}{g-g(\xi ')}},\ \xi ''\in \mathbb {I'''} \subset \mathbb {I'} ,\ x\in \mathbb {I''} .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0b03790f6adf71f162abec1f9d893fae5de5dd7f) | | (4) |
Since
, just as x is in (2),
![{\displaystyle \therefore \left|{\frac {f'(\xi '')}{g'(\xi '')}}-L\right\vert <\epsilon .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/197edb919a71414dabdba732ee1af81e324cc780)
Substituting from (4),
![{\displaystyle \therefore \left|{\frac {f-f(\xi ')}{g-g(\xi ')}}-L={\frac {\left({\frac {f}{g}}-L\right)-{\frac {f(\xi ')}{g}}+L{\frac {g(\xi ')}{g}}}{1-{\frac {g(\xi ')}{g}}}}\right\vert <\epsilon .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3019e12561c9ad1b2d1e5fc57d491f09820e5707)
Multiplying both sides by the absolute value of the denominator and using (3) (since
), repeatedly, together with the triangle inequality rule,
![{\displaystyle {\begin{aligned}&\therefore \left|\left({\frac {f}{g}}-L\right)-{\frac {f(\xi ')}{g}}+L{\frac {g(\xi ')}{g}}\right\vert <\left|1-{\frac {g(\xi ')}{g}}\right\vert \epsilon \leq \left(1+\left|{\frac {g(\xi ')}{g}}\right\vert \right)\epsilon <(1+\epsilon )\epsilon \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/361d387d86f92c160a5a254f857307a737a9ff02)
![{\displaystyle {\begin{aligned}\therefore \left|{\frac {f}{g}}-L\right\vert &<(1+\epsilon )\epsilon +\left|{\frac {f(\xi ')}{g}}\right\vert +|L|\left|{\frac {g(\xi ')}{g}}\right\vert \\&<(1+\epsilon )\epsilon +\epsilon +|L|\epsilon =(2+|L|+\epsilon )\epsilon \to 0{\text{ as }}\epsilon \to 0.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ba9c1d55c8033f7444fa1a989904cd317e1cc15a)
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.