Question 7.T.16: Let the functions f and g be differentiable on (a, b) and g′...

We shall prove the theorem in the case when

\underset{x→a^{+}}{\lim }  \frac{f^{′}(x)}{g^{′}(x)} =\ell

with \ell ∈ \mathbb{R}, and leave the cases when \ell = ±∞ as an exercise. Given ε > 0, there is a δ > 0 such that

\left|\frac{f^{′}(x)}{g^{′}(x)}  –  \ell\right| < ε  for all x ∈ (a, a + δ).        (7.13)

Choose any point c in (a, a + δ). Since f(x) → ∞ as x → a^{+}, there is a point d ∈ (a, c) (see Figure 7.8) such that

x ∈ (a, d) ⇒ f(x) > f(c).

Define the function h on (a, d) by

h(x) = \frac{1 − f(c)/f(x)}{1 − g(c)/g(x)},

and note that 1 − g(c)/g(x) ≠ 0 because g is injective on (a, b), and h(x) ≠ 0 because f(x) > f(c) on (a, d). Since f(c)/f(x) → 0 and g(c)/g(x) → 0 as x → a^{+}, it follows that \lim_{ x→a^{+}} h(x) = 1, so there is a t ∈ (a, d) such that

x ∈ (a, t) ⇒ |h(x) − 1| < ε^{′} = \min\left\{ε, 1/2\right\}

⇒ 1 − ε^{′} < h(x) < 1 + ε^{′}

⇒ \frac{1}{1|h(x)|} < \frac{1}{1 − ε^{′}} ≤ 2.        (7.14)

Let x ∈ (a, t). We shall complete the proof by showing that

\left|\frac{f(x)}{g(x)}  –  \ell\right| < kε

for some constant k. Observe that

\frac{f(x)}{g(x)} = \frac{f(x)}{g(x)} \frac{h(x)}{h(x)} = \frac{f(x)− f (c)}{g(x)− g(c)} \frac{1}{h(x)}.

If we apply Cauchy’s mean value theorem to f and g on [x, c], and use the above equation, we obtain, for some y ∈ (x, c),

\frac{f^{′}(y)}{g^{′}(y)} = \frac{f(x)− f (c)}{g(x)− g(c)} = \frac{f(x)}{g(x)}h(x)  for all x ∈ (a, d).

In view of (7.13) and (7.14), and with a < x < t < d < c < a + δ, we have

\left|\frac{f(x)}{g(x)}- \ell\right| = \left|\frac{f^{′}(y)}{g^{′}(y)} \frac{1}{h(x)} – \ell\right|

=\left|\frac{f^{′}(y)}{g^{′}(y)} – \ell h(x)\right| \frac{1}{\left|h(x)\right| }

≤ 2 \left(\left|\frac{f^{′}(y)}{g^{′}(y)} – \ell\right| + |\ell − \ell h(x)| \right)

≤ 2(ε + |\ell| ε)

≤ 2(1 + |\ell|)ε.