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 $\mathbb {I}$, with c (an extended real number) at one extremity, and

${\begin{aligned}&\lim _{x\in \mathbb {I} \to c}|g(x)|=\infty ,{\text{ then}}\\&\lim _{x\in \mathbb {I} \to c}{\frac {f(x)}{g(x)}}=\lim _{x\in \mathbb {I} \to c}{\frac {f'(x)}{g'(x)}},\end{aligned}}$

provided the limit on the right exists.

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In the below proofs, I use the shorthands

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

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

$a<b{\text{ if }}c{\text{ is a right endpoint, and }}b<a{\text{ if }}c{\text{ is a left endpoint}},\,$

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

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

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

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

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

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

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

(1)

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

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

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, $f'/g'\,$ is continuous in $\mathbb {I}$. Defining $\mathbb {I'} =(\xi ',c)\subset \mathbb {I} \,$, the existence of the limit is expressed as follows:

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

(2)

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

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

(3)

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

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

(4)

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

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

Substituting from (4),

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

Multiplying both sides by the absolute value of the denominator and using (3) (since $x\in \mathbb {I''}$), repeatedly, together with the triangle inequality rule,

${\begin{aligned}&\therefore \left|\left({\frac {f}{g}}-L\right)-{\frac {f(\xi ')}{g}}+L{\frac {g(\xi ')}{g}}\right\vert <\left|1-{\frac {g(\xi ')}{g}}\right\vert \epsilon \leq \left(1+\left|{\frac {g(\xi ')}{g}}\right\vert \right)\epsilon <(1+\epsilon )\epsilon \end{aligned}}$

${\begin{aligned}\therefore \left|{\frac {f}{g}}-L\right\vert &<(1+\epsilon )\epsilon +\left|{\frac {f(\xi ')}{g}}\right\vert +|L|\left|{\frac {g(\xi ')}{g}}\right\vert \\&<(1+\epsilon )\epsilon +\epsilon +|L|\epsilon =(2+|L|+\epsilon )\epsilon \to 0{\text{ as }}\epsilon \to 0.\end{aligned}}$

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.