User:Rschwieb/Sandbox

Proof of l'Hôpital's rule
The following proof is due to, where a unified proof for the 0/0 and ±∞/±∞ indeterminate forms is given. Taylor notes that different proofs may be found in and.

Let f and g be functions satisfying the hypotheses in the General form section. Let $$\mathcal{I}$$ be the open interval in the hypothesis with endpoint c. Considering that $$g'(x)\neq 0$$ on this interval and g is continuous, $$\mathcal{I}$$ can be chosen smaller so that g is nonzero on $$\mathcal{I}$$.

For each x in the interval, define $$m(x)=\inf\frac{f'(\xi)}{g'(\xi)}$$ and $$M(x)=\sup\frac{f'(\xi)}{g'(\xi)}$$ as $$\xi$$ ranges over all values between x and c. (The symbols inf and sup denote the infimum and supremum.)

From the differentiability of f and g on $$\mathcal{I}$$, Cauchy's mean value theorem ensures that for any two distinct points x and y in $$\mathcal{I}$$ there exists a $$\xi $$ between x and y such that $$\frac{f(x)-f(y)}{g(x)-g(y)}=\frac{f'(\xi)}{g'(\xi)}$$. Consequently $$m(x)\leq \frac{f(x)-f(y)}{g(x)-g(y)} \leq M(x)$$ for all choices of distinct x and y in the interval. The value g(x)-g(y) is always nonzero for distinct x and y in the interval, for if it was not, the mean value theorem would imply the existence of a p between x and y such that g' (p)=0.

The definition of m(x) and M(x) will result in an extended real number, and so it is possible for them to take on the values ±∞. In the following two cases, m(x) and M(x) will establish bounds on the ratio f/g.

Case 1: $$\lim_{x\rightarrow c}f(x)=\lim_{x\rightarrow c}g(x)=0$$

For any x in the interval $$\mathcal{I}$$, and point y between x and c,
 * $$m(x)\leq\frac{f(x)-f(y)}{g(x)-g(y)}=\frac{\frac{f(x)}{g(x)}-\frac{f(y)}{g(x)}}{1-\frac{g(y)}{g(x)}}\leq M(x)$$

and therefore as y approaches c, $$\frac{f(y)}{g(x)}$$ and $$\frac{g(y)}{g(x)}$$ become zero, and so
 * $$m(x)\leq\frac{f(x)}{g(x)}\leq M(x)$$.

Case 2: $$\lim_{x\rightarrow c}|g(x)|=\infty$$

For any x in the interval $$\mathcal{I}$$, and point y between x and c,
 * $$m(x)\leq \frac{f(y)-f(x)}{g(y)-g(x)}=\frac{\frac{f(y)}{g(y)}-\frac{f(x)}{g(y)}}{1-\frac{g(x)}{g(y)}}\leq M(x)$$.

As y approaches c, both $$\frac{f(x)}{g(y)}$$ and $$\frac{g(x)}{g(y)}$$ become zero, and therefore
 * $$m(x)\leq \liminf_{x \rightarrow c} \frac{f(y)}{g(y)} \leq \limsup_{x \rightarrow c} \frac{f(y)}{g(y)}\leq M(x)$$

(For readers skeptical about the x 's under the limit superior and limit inferior, remember that y is always between x and c, and so as x approaches c, so will y. The limsup and liminf are necessary: we may not yet write "lim" since the existence of that limit has not yet been established.)

In both cases,


 * $$\lim_{x \rightarrow c}m(x)=\lim_{x \rightarrow c}M(x)=\lim_{x \rightarrow c}\frac{f'(x)}{g'(x)}=L$$.

By the Squeeze theorem, the limit exists and $$\lim_{x\rightarrow c}\frac{f(x)}{g(x)}=L$$. This is the result that was to be proven.