# User:Toolnut

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 ${\displaystyle \mathbb {I} }$, 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}}}

provided the limit on the right exists.

In the below proofs, I use the shorthands

${\displaystyle f\equiv f(x),\ g\equiv g(x),\ f'\equiv f'(x),{\text{ and }}g'\equiv g'(x).}$

Also, I use the notation ${\displaystyle (a,b)\,}$ to mean any open interval with endpoints ${\displaystyle a\,}$ & ${\displaystyle b\,}$, with ${\displaystyle b\,}$ nearer to ${\displaystyle c\,}$; i.e.,

${\displaystyle a

and ${\displaystyle \to c}$ implies a one-sided approach from within ${\displaystyle \mathbb {I} }$.

### Proof 1

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 ${\displaystyle \xi \in \mathbb {I} \,}$. We may apply Cauchy's mean value theorem to the finite interval ${\displaystyle \mathbb {I'} =(\xi ,x)\subset \mathbb {I} \,}$:

${\displaystyle \exists \xi '\in \mathbb {I'} :{\frac {f'(\xi ')}{g'(\xi ')}}={\frac {f-f(\xi )}{g-g(\xi )}}.}$

In the limit, as ${\displaystyle x\to c}$, this mean gradient becomes

${\displaystyle \lim _{x\to c}{\left({\frac {f-f(\xi )}{g-g(\xi )}}={\frac {{\frac {f}{g}}-{\frac {f(\xi )}{g}}}{1-{\frac {g(\xi )}{g}}}}\right)}={\frac {\lim _{x\to c}{\frac {f}{g}}-\left(\lim _{x\to c}{\frac {f(\xi )}{g}}=0\right)}{1-\left(\lim _{x\to c}{\frac {g(\xi )}{g}}=0\right)}}=\lim _{x\to c}{\frac {f}{g}},}$

provided that f & g do not blow up in the open interval ${\displaystyle \mathbb {I''} =(\xi ,c)}$, i.e. f(ξ) & g(ξ) are finite, for all choices of ${\displaystyle \xi \in \mathbb {I} }$, which is true because their individual differentiabilities guarantee their continuities in that interval.

${\displaystyle \therefore \exists \xi '\in \mathbb {I''} :{\frac {f'(\xi ')}{g'(\xi ')}}=\lim _{x\to c}{\frac {f}{g}}.}$

(1)

As (1) holds for all ${\displaystyle \xi \in \mathbb {I} \,}$,

${\displaystyle \lim _{\xi \to c\,\therefore \xi '\to c}{\frac {f'(\xi ')}{g'(\xi ')}}=\lim _{x\to c}{\frac {f}{g}}.}$

### Proof 2

Let L be the second limit (given to exist), assumed finite. Due to the continuity of f' & g' , plus the fact that g'≠0, ${\displaystyle f'/g'\,}$ is continuous in ${\displaystyle \mathbb {I} }$. Defining ${\displaystyle \mathbb {I'} =(\xi ',c)\subset \mathbb {I} \,}$, the existence of the limit is expressed as follows:

${\displaystyle \exists L:\forall \epsilon >0,\,\exists \xi '\in \mathbb {I} :\{x\in \mathbb {I'} \}\Rightarrow \left\{\left|{\frac {f'}{g'}}-L\right|<\epsilon \right\}.}$

(2)

Given the continuity of f & g in ${\displaystyle \mathbb {I} \,}$, hence f and g being finite, and the monotone-increasing |g| in ${\displaystyle \mathbb {I} \,}$ as xc (because it is given that g'≠0 and g≠0), and defining ${\displaystyle \mathbb {I''} =(\xi ,c)\subset \mathbb {I'} \,}$, then ${\displaystyle x\in \mathbb {I''} }$ 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:

${\displaystyle \forall \epsilon >0,\,\exists \xi \in \mathbb {I'} \subset \mathbb {I} :\{x\in \mathbb {I''} \}\Rightarrow \left\{\max \left(\left|{\frac {f(\xi ')}{g}}\right|,\left|{\frac {g(\xi ')}{g}}\right|\right)<\epsilon \right\}\land \{g(\xi ')\neq g\};}$

(3)

Now, given the differentiabilities of f & g and that g'≠0, everywhere in ${\displaystyle \mathbb {I} \,}$, and, from (3), that g(ξ' )≠g for ${\displaystyle x\in \mathbb {I''} \,}$, we may apply Cauchy's MVT to the finite interval ${\displaystyle \mathbb {I'''} =(\xi ',x)\subset \mathbb {I'} \,}$:

${\displaystyle {\frac {f'(\xi '')}{g'(\xi '')}}={\frac {f-f(\xi ')}{g-g(\xi ')}},\ \xi ''\in \mathbb {I'''} \subset \mathbb {I'} ,\ x\in \mathbb {I''} .}$

(4)

Since ${\displaystyle \xi ''\in \mathbb {I'} }$, just as x is in (2),

${\displaystyle \therefore \left|{\frac {f'(\xi '')}{g'(\xi '')}}-L\right\vert <\epsilon .}$

Substituting from (4),

${\displaystyle \therefore \left|{\frac {f-f(\xi ')}{g-g(\xi ')}}-L={\frac {\left({\frac {f}{g}}-L\right)-{\frac {f(\xi ')}{g}}+L{\frac {g(\xi ')}{g}}}{1-{\frac {g(\xi ')}{g}}}}\right\vert <\epsilon .}$

Multiplying both sides by the absolute value of the denominator and using (3) (since ${\displaystyle x\in \mathbb {I''} }$), 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}}}
{\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}}}

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.