Question 7.T.15: Let the functions f and g be differentiable on [a, ∞) and su...
Let the functions f and g be differentiable on [a, ∞) and suppose
\underset{x→∞}{\lim } f(x) = \underset{x→∞}{\lim } g(x) = 0,
g^{′}(x) ≠ 0 for all x > a.
If \underset{x→∞}{\lim } \frac {f^{′}(x)} {g^{′}(x)} exists in \bar{\mathbb{R}}, then
\underset{x→∞}{\lim } \frac {f(x)} {g(x)} = \underset{x→∞}{\lim } \frac {f^{′}(x)} {g^{′}(x)}.
Clearly, we can take a > 0. Define the functions F and G on (0, 1 / a] by
F(x) = f \left(\frac{1}{x}\right) , G(x) = g\left(\frac{1}{x}\right).
This means
f(x) = F\left(\frac{1}{x}\right), g(x) = G\left(\frac{1}{x}\right) , x ∈ [a, ∞),
and so the limit we have to evaluate is
\underset{x→∞}{\lim } \frac{F(1/x)}{G(1/x)}.
But since x → ∞ if and only if t = 1/x → 0^{+}, this is the same as the limit
\underset{t→0^{+}}{\lim } \frac{F(t)}{G(t)}.
By the chain rule,
F^{′}(t) = f^{′} \left(\frac{1}{t}\right)· \left(− \frac{1}{t^{2}}\right),
G^{′}(t) = G^{′} \left(\frac{1}{t}\right)· \left(− \frac{1}{t^{2}}\right).
Hence
\underset{t→0^{+}}{\lim } \frac{F(t)}{G(t)} = \underset{t→0^{+}}{\lim } \frac{f^{′}(1/t)}{g^{′}(1/t)} = \underset{x→∞}{\lim } \frac {f^{′}(x)} {g^{′}(x)} .
To complete the proof, we need to show that
\underset{t→0^{+}}{\lim } \frac{F(t)}{G(t)} = \underset{t→0^{+}}{\lim } \frac{F^{′}(t)}{G^{′}(t)} (7.12)
The functions F and G are differentiable on (0, 1 / a), and they may be extended as continuous functions to [0, 1 / a) by the definitions
F(0) = \underset{t→0^{+}}{\lim } F(t) = \underset{x→∞}{\lim } f(x) = 0,
G(0) = \underset{t→0^{+}}{\lim } G(t) = \underset{x→∞}{\lim } g(x) = 0,
This allows us to apply Theorem 7.14 and obtain (7.12).