Let $f,g$ be locally integrable on $[a,b)$. Show that if $$\lim_{x\to b}\frac{f(x)}{g(x)}=L$$ is finite and that $g$ is improperly integrable then $f$ is improperly integrable.
I'm quite lost as to how to approach this. I think the idea is that; as $x$ gets close to $b$, $f$ and $g$ are in a common ratio and kind of follow each other so if one converges the other must also. I'm having trouble using $$\lim_{x\to b}\frac{f(x)}{g(x)}=L$$ in any meaningful way. Anything to get me started would be helpful.
Thank you in advance for your time. Apologies for formatting.
The idea here is to show that $\int_a^x f(t)\;dt$ is one-sided Cauchy as $x\to b^-$ and therefore the one-sided limit must exist. We will use both the ratio limit and the improper integrability of $g$ to show this.
By the limit of ratios, for any $\epsilon_1>0$ there exists $\delta_1>0$ so that for $x \in (b-\delta_1,b)$, $$\left|\frac{f(x)}{g(x)}-L\right|<\epsilon_1$$ or equivalently: $$(L-\epsilon_1)g(x) < f(x) < (L+\epsilon_1)g(x) $$
By the Cauchy $\epsilon-\delta$ definition of one-sided limit (the improper integral $\lim_{x\to b^-}\int_a^x g(t)\;dt$): for any $\epsilon_2 > 0$ there exists $\delta_2 > 0$ so that for $x,y \in (b-\delta_2, b)$ $$\left|\int_a^x g(t)\;dt - \int_a^y g(t)\;dt\right| = \left|\int_y^x g(t)\;dt\right| < \epsilon_2$$
We then show that $f$ satisfies the above $\epsilon-\delta$ definition: Pick $\epsilon > 0$, let $\epsilon_1 \in (0, L)$ and $\epsilon_2 = \frac{\epsilon}{L+\epsilon_1}$. Pick $\delta=\min\{\delta_1,\delta_2\}$, then for $x,y \in (b-\delta, b)$,
\begin{align} (L-\epsilon_1)\underbrace{\int_y^x g(t)\;dt}_{>\;-\epsilon_2} < \int_y^x f(t)\;dt < (L+\epsilon_1)\underbrace{\int_y^x g(t)\;dt}_{<\;\epsilon_2} \end{align} using the definition, $\epsilon_2 = \frac{\epsilon}{L+\epsilon_1}$:
$$-\epsilon < -\underbrace{\frac{L-\epsilon_1}{L+\epsilon_1}}_{\in (0,1)} \epsilon =-(L-\epsilon_1)\epsilon_2 <\int_y^x f(t)\;dt < (L+\epsilon_1) \epsilon_2 = \epsilon $$
as this holds if we swap $x$ and $y$, then
$$\left|\int_y^x f(t)\;dt\right| = \left|\int_a^x f(t)\;dt -\int_a^y f(t)\;dt\right|<\epsilon$$ Hence the one-sided limit exists, i.e., $$\lim_{x\to b^-} \int_a^x f(t)\;dt \quad\text{exists.}$$