L’Hospitals Rule

Theorem 3.3.1 (Cauchy Mean Value Theorem)

If f and g be functions that are continuous on [a,b] and differentiable on (a,b). Then there exists at least one point c \in (a,b) such that

[f(b) - f(a)]g^\prime(c) = [g(b)-g(a)]f^\prime(c).

Proof

Let

h(x) = [f(b) - f(a)]g(x) = [g(b)-g(a)]f(x) \quad \text{for all $x \in [a,b]$}.

Then h is continuous on [a,b] and differentiable on (a,b). Furthermore

h(a) = f(b)g(a) - g(b)f(a) = h(b)

Thus by the Mean Value Theorem there exists c \in (a,b) such that h^\prime(c) = 0. That is,

h^\prime(c) = [f(b)-f(a)]g^\prime(c) - [g(b)-g(a)]f^\prime(c) = 0.

Theorem 3.3.2 (L’Hospitals Rule)

Let f and g be differentiable on (a,\infty). Suppose that \lim_{x \to \infty}f(x) = \lim_{x \to \infty}g(x) = \infty and that g^\prime(x) \not = 0 for x \in (a,\infty). If

\lim_{x \to \infty} \frac{f^\prime(x)}{g^\prime(x)} = L, \quad \text{where $L \in \mathbb{R}$},

then

\lim_{x\to\infty} \frac{f(x)}{g(x)} = L.

Proof

Given \varepsilon > 0, there exists an N_1 > a such that x > N_1 implies that

\left| \frac{f^\prime(x)}{g^\prime(x)} - L \right| < \frac{\varepsilon}{2}.

Since \lim_{x\to\infty} f(x) = \lim_{x\to\infty} g(x) = \infty, there exists an N_2 > N_1 such that x > N_2 implies that f(x) > 0, and g(x) > 0.

Furthermore, there exists N_3 > N_2 such that x > N_3 implies that f(x) > f(N_2) and g(x) > g(N_2). For any point x > N_3, the Cauchy mean value theorem implies that there exists some point c in (N_2,x) such that

\frac{f^\prime(c)}{g^\prime(c)} = \frac{f(x) - f(N_2)}{g(x) - g(N_2)} = \frac{f(x)}{g(x)} \frac{1 - f(N_2)/f(x)}{1 - g(N_2)/g(x)}.

But then for x > N_3 we have

\frac{f(x)}{g(x)} = \frac{f^\prime(c)}{g^\prime(c)} F(x) \quad \text{where } F(x) = \frac{1 - g(N_2)/g(x)}{1-f(N_2)/f(x)}.