Compute the quotient and product of functions with given properties

Let $f$ and $g$ be twice-differentiable functions such that

$f(0) = \frac{2}{g(0)}, \qquad f'(0) = 2g'(0) = 4g(0), \qquad g''(0) = 5f''(0) = 6f(0) = 3.$

If $h(x) = \frac{f(x)}{g(x)}$ , compute $h'(0)$ .
If $k(x) = f(x) g(x) \sin x$ , compute $k'(0)$ .
Compute $\lim_{x \to 0} \frac{g'(x)}{f'(x)}$ .

Using the rule for the derivative of a quotient we have,
$h(x) = \frac{f(x)}{g(x)} \quad \implies \quad h'(x) = \frac{f'(x) g(x) - f(x) g'(x)}{(g(x))^2}.$

Substituting the given expressions for $f(0), g(0), f'(0)$ , and $g'(0)$ we have,

$\begin{align*} h'(0) &= \frac{f'(0) g(0) - f(0) g'(0)}{(g(0))^2} \\ &= \frac{4 (g(0))^2 - f(0)g'(0)}{(g(0))^2} & (f'(0) = 4g(0))\\ &= \frac{4 (g(0))^2 - 4}{(g(0))^2} &(g'(0) = 2 g(0) \text{ and } f(0) = \frac{2}{g(0)}) \\ &= 4 - \frac{4}{(g(0))^2} \\ &= 4 - \frac{4}{\frac{4}{(f(0))^2}} & (g(0) = \frac{2}{f(0)}) \\ &= 4 - \frac{1}{4} & (6 f(0) = 3 \implies f(0) = \frac{1}{2}) \\ &= \frac{15}{4}. \end{align*}$
Using the product rule (more precisely, the product rule for more than two terms that we proved in this exercise) we have,
$k'(x) = f'(x) g(x) \sin x + f(x) g'(x) \sin x + f(x) g(x) \cos x.$

Then, we evaluate at 0,

$\begin{align*} k'(0) &= f'(0) g(0) \sin 0 + f(0) g'(0) \sin 0 + f(0) g(0) \cos 0 \\ &= f(0) g(0) \\ &= 2 \end{align*}$

since we are given $f(0) = \frac{2}{g(0)}$ .
Since we are given that $f$ and $g$ are twice differentiable functions, we know that $f'$ and $g'$ are differentiable; hence, $f'$ and $g'$ are continuous since differentiability implies continuity (from Section 4.4 of Apostol). From the definition of continuity of a function at a point we know this means,
$\lim_{x \to 0} f'(x) = f'(0) \qquad \text{and} \qquad \lim_{x \to 0} g'(x) = g'(0).$

Furthermore, $f'(0) \neq 0$ since $f'(0) = 4g(0)$ and $g(0) = \frac{2}{f(0)} = \frac{2}{1/2} = 4$ . So we can then compute,

$\begin{align*} \lim_{x \to 0} \frac{g'(x)}{f'(x)} &= \frac{\lim_{x \to 0} g'(x)}{\lim_{x \to 0} f'(x)} \\ &= \frac{f'(0)}{g'(0)} \\ & = \frac{1}{2} \end{align*}$

since we are given $f'(0) = 2g'(0)$ .