Editing L'Hôpital's rule (section)

===General proof===
The following proof is due to {{harvtxt|Taylor|1952}}, where a unified proof for the <math display="inline">\frac{0}{0}</math> and <math display="inline">\frac{\pm \infty}{\pm \infty}
</math> indeterminate forms is given. Taylor notes that different proofs may be found in {{harvtxt|Lettenmeyer|1936}} and {{harvtxt|Wazewski|1949}}.

Let ''f'' and ''g'' be functions satisfying the hypotheses in the [[L'hopital's rule#General form|General form]] section. Let <math>\mathcal{I}</math> be the open interval in the hypothesis with endpoint ''c''. Considering that <math>g'(x)\ne 0</math> on this interval and ''g'' is continuous, <math>\mathcal{I}</math> can be chosen smaller so that ''g'' is nonzero on <math>\mathcal{I}</math>.{{efn|1=Since ''g' '' is nonzero and ''g'' is continuous on the interval, it is impossible for ''g'' to be zero more than once on the interval. If it had two zeros, the [[mean value theorem]] would assert the existence of a point ''p'' in the interval between the zeros such that ''g' ''(''p'')&nbsp;=&nbsp;0. So either ''g'' is already nonzero on the interval, or else the interval can be reduced in size so as not to contain the single zero of ''g''.}}

For each ''x'' in the interval, define <math>m(x)=\inf\frac{f'(t)}{g'(t)}</math> and <math>M(x)=\sup\frac{f'(t)}{g'(t)}</math> as <math>t</math> 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 <math>\mathcal{I}</math>, [[Cauchy's mean value theorem]] ensures that for any two distinct points ''x'' and ''y'' in <math>\mathcal{I}</math> there exists a <math>\xi</math> between ''x'' and ''y'' such that <math>\frac{f(x)-f(y)}{g(x)-g(y)}=\frac{f'(\xi)}{g'(\xi)}</math>. Consequently, <math>m(x)\leq \frac{f(x)-f(y)}{g(x)-g(y)} \leq M(x)</math> 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 {{sfrac|''f''|''g''}}.

'''Case 1:''' <math>\lim_{x\to c}f(x)=\lim_{x\to c}g(x)=0</math>

For any ''x'' in the interval <math>\mathcal{I}</math>, and point ''y'' between ''x'' and ''c'',
:<math>m(x)\le \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)}}\le M(x)</math>
and therefore as ''y'' approaches ''c'', <math>\frac{f(y)}{g(x)}</math> and <math>\frac{g(y)}{g(x)}</math> become zero, and so
:<math>m(x)\leq\frac{f(x)}{g(x)}\leq M(x).</math>

'''Case 2:''' <math>\lim_{x\to c}|g(x)|=\infty</math>

For every ''x'' in the interval <math>\mathcal{I}</math>, define <math>S_x=\{y\mid y \text{ is between } x \text{ and } c\}</math>. For every point ''y'' between ''x'' and ''c'',
: <math>m(x)\le \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)}} \le M(x).</math>

As ''y'' approaches ''c'', both <math>\frac{f(x)}{g(y)}</math> and <math>\frac{g(x)}{g(y)}</math> become zero, and therefore
: <math>m(x)\le \liminf_{y\in S_x} \frac{f(y)}{g(y)} \le \limsup_{y\in S_x} \frac{f(y)}{g(y)} \le M(x).</math>

The [[limit superior]] and [[limit inferior]] are necessary since the existence of the limit of {{sfrac|''f''|''g''}} has not yet been established.

It is also the case that 
:<math>\lim_{x\to c}m(x)=\lim_{x\to c}M(x)=\lim_{x\to c}\frac{f'(x)}{g'(x)}=L.</math>
{{efn|
The limits <math>\lim_{x\to c} m(x)</math> and <math>\lim_{x\to c} M(x)</math> both exist as they feature nondecreasing and nonincreasing functions of ''x'', respectively.

Consider a sequence <math>x_i \to c</math>. Then <math>\lim_i m(x_i) \le \lim_i \frac{f'(x_i)}{g'(x_i)} \le \lim_i M(x_i)</math>, as the inequality holds for each ''i''; this yields the inequalities <math>\lim_{x\to c}m(x) \le \lim_{x\to c}\frac{f'(x)}{g'(x)} \le \lim_{x\to c}M(x)</math>

The next step is to show <math>\lim_{x\to c}M(x) \le \lim_{x\to c}\frac{f'(x)}{g'(x)}</math>. Fix a sequence of numbers <math>\varepsilon_i > 0</math> such that <math> \lim_i \varepsilon_i = 0</math>, and a sequence <math>x_i\to c </math>. For each ''i'', choose <math>x_i < y_i < c</math> such that <math>\frac{f'(y_i)}{g'(y_i)} + \varepsilon_i \ge \sup_{x_i < \xi < c}\frac{f'(\xi)}{g'(\xi)}</math>, by the definition of <math>\sup</math>. Thus
<math display="block">\begin{align}
\lim_i M(x_i) &\leq \lim_i \frac{f'(y_i)}{g'(y_i)} + \varepsilon_i \\
&= \lim_i \frac{f'(y_i)}{g'(y_i)} + \lim_i \varepsilon_i \\
&= \lim_i \frac{f'(y_i)}{g'(y_i)}
\end{align}</math> as desired.

The argument that <math>\lim_{x\to c}m(x) \ge \lim_{x\to c} \frac{f'(x)}{g'(x)}</math> is similar.
}}
and
:<math>\lim_{x\to c}\left(\liminf_{y\in S_x}\frac{f(y)}{g(y)}\right)=\liminf_{x\to c}\frac{f(x)}{g(x)}</math> and <math>\lim_{x\to c}\left(\limsup_{y\in S_x} \frac{f(y)}{g(y)}\right)=\limsup_{x\to c}\frac{f(x)}{g(x)}. </math>

In case 1, the [[squeeze theorem]] establishes that <math>\lim_{x\to c}\frac{f(x)}{g(x)}</math> exists and is equal to ''L''. In the case 2, and the squeeze theorem again asserts that <math>\liminf_{x\to c}\frac{f(x)}{g(x)}=\limsup_{x\to c}\frac{f(x)}{g(x)}=L</math>, and so the limit <math>\lim_{x\to c}\frac{f(x)}{g(x)}</math> exists and is equal to ''L''. This is the result that was to be proven.

In case 2 the assumption that ''f''(''x'') diverges to infinity was not used within the proof. This means that if |''g''(''x'')| diverges to infinity as ''x'' approaches ''c'' and both ''f'' and ''g'' satisfy the hypotheses of L'Hôpital's rule, then no additional assumption is needed about the limit of ''f''(''x''): It could even be the case that the limit of ''f''(''x'') does not exist. In this case, L'Hopital's theorem is actually a consequence of Cesàro–Stolz.<ref>{{cite web|url= http://www.imomath.com/index.php?options=686|website=IMOmath|publisher=[[International Mathematical Olympiad]]|title=L'Hopital's Theorem}}</ref>

In the case when |''g''(''x'')| diverges to infinity as ''x'' approaches ''c'' and ''f''(''x'') converges to a finite limit at ''c'', then L'Hôpital's rule would be applicable, but not absolutely necessary, since basic limit calculus will show that the limit of ''f''(''x'')/''g''(''x'') as ''x'' approaches ''c'' must be zero.