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

&\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[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 \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}.






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[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, 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\}.






Given the continuity of f & g in \mathbb{I}\,, hence f and g being finite, and the monotone-increasing |g| in \mathbb{I}\, as xc (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\}\and\{g(\xi')\ne g\};






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''}.






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


Substituting from (4),


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,

&<(1+\epsilon)\epsilon+\epsilon+|L|\epsilon=(2+|L|+\epsilon)\epsilon\to0\text{ as }\epsilon\to0.

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.