structural:reformatting, structural:summarization
General proof

The following proof is due to Taylor (1952), where a unified proof for the 0 0 {\textstyle {\frac {0}{0}}} and ± ∞ ± ∞ {\textstyle {\frac {\pm \infty }{\pm \infty }}} indeterminate forms is given. Taylor notes that different proofs may be found in Lettenmeyer (1936) and Wazewski (1949).

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

For each x in the interval, define m (x) = inf f ′ (t) g ′ (t) {\displaystyle m(x)=\inf {\frac {f'(t)}{g'(t)}}} and M (x) = sup f ′ (t) g ′ (t) {\displaystyle M(x)=\sup {\frac {f'(t)}{g'(t)}}} as t {\displaystyle t} 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 I {\displaystyle {\mathcal {I}}}, Cauchy's mean value theorem ensures that for any two distinct points x and y in I {\displaystyle {\mathcal {I}}} there exists a ξ {\displaystyle \xi } between x and y such that f (x) − f (y) g (x) − g (y) = f ′ (ξ) g ′ (ξ) {\displaystyle {\frac {f(x)-f(y)}{g(x)-g(y)}}={\frac {f'(\xi)}{g'(\xi)}}}. Consequently, m (x) ≤ f (x) − f (y) g (x) − g (y) ≤ M (x) {\displaystyle 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 → c f (x) = lim x → c g (x) = 0 {\displaystyle \lim _{x\to c}f(x)=\lim _{x\to c}g(x)=0}

For any x in the interval I {\displaystyle {\mathcal {I}}}, and point y between x and c,

and therefore as y approaches c, f (y) g (x) {\displaystyle {\frac {f(y)}{g(x)}}} and g (y) g (x) {\displaystyle {\frac {g(y)}{g(x)}}} become zero, and so

Case 2: lim x → c | g (x) | = ∞ {\displaystyle \lim _{x\to c}|g(x)|=\infty }

For every x in the interval I {\displaystyle {\mathcal {I}}}, define S x = { y ∣ y is between x and c } {\displaystyle S_{x}=\{y\mid y{\text{ is between }}x{\text{ and }}c\}}. For every point y between x and c,

As y approaches c, both f (x) g (y) {\displaystyle {\frac {f(x)}{g(y)}}} and g (x) g (y) {\displaystyle {\frac {g(x)}{g(y)}}} become zero, and therefore

The limit superior and limit inferior are necessary since the existence of the limit of f / g has not yet been established.

It is also the case that

and

In case 1, the squeeze theorem establishes that lim x → c f (x) g (x) {\displaystyle \lim _{x\to c}{\frac {f(x)}{g(x)}}} exists and is equal to L. In the case 2, and the squeeze theorem again asserts that lim inf x → c f (x) g (x) = lim sup x → c f (x) g (x) = L {\displaystyle \liminf _{x\to c}{\frac {f(x)}{g(x)}}=\limsup _{x\to c}{\frac {f(x)}{g(x)}}=L}, and so the limit lim x → c f (x) g (x) {\displaystyle \lim _{x\to c}{\frac {f(x)}{g(x)}}} 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.

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.