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

&\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{align}$$ provided the limit on the right exists.

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.

As ($$) holds for all $$\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:

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:

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

Since $$\xi\in\mathbb{I'}$$, just as x'' is in ($$),
 * $$\therefore\left|\frac{f'(\xi)}{g'(\xi)}-L\right\vert<\epsilon.$$

Substituting from ($$),
 * $$\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 ($$) (since $$x\in\mathbb{I''}$$), repeatedly, together with the triangle inequality rule,
 * $$\begin{align}

&\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\le\left(1+\left|\frac{g(\xi')}{g}\right\vert\right)\epsilon<(1+\epsilon)\epsilon \end{align}$$
 * $$\begin{align}

\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\to0\text{ as }\epsilon\to0. \end{align}$$

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.