On proving that a limit involving an integral and part of its integrand tends to zero. Suppose $f$ is continuous, $F(x)=\int_{a}^{x}{f(t)dt}$ is bounded on $[a,b)$ and $\int_{a}^{b}{f(t)dt}$ converges as an improper integral. Suppose also that $g>0$, $g'$ is non-negative and locally integrable on $[a,b)$ and that $\lim_{x\to b^{-}}{g(x)}=\infty$
Prove:
$\lim_{x\to b^{-}}{\frac{1}{g(x)}\int_{a}^{x}{f(t)g(t)dt}}=0$
I´ve been trying prove this result that was part of an exercise in a real analysis PDF.
Can´t seem to get very far proving it so I am not sure if there is some theorem or lemma I´m not thinking of that will make it obvious or that I´ve been given insufficient information. I´ve tried integrating by parts which gives me: $\frac{1}{g(x)}\Big(F_0(t)g(t)\Big|_{a}^{x} -\int_{a}^{x}{F_0(t)g'(t)dt}\Big)$ and letting $F_0(x)$ be $\int_{x}^{b}{f(t)dt}$.
It seemed to help since: $\frac{1}{g(x)}\Big(F_0(t)g(t)\Big|_{a}^{x}\Big)\to 0$ as $x\to b^{-}$, but I can´t seem to get anywhere on the second integral.
I´m asking if anyone can help me prove it or explaing why it can´t be done.
 A: You are almost there. It remains to show that the growth of $g$ is offset by the vanishing of $F_0$ at $b$, so that
$$\Biggl\lvert\,\int_a^x F_0(t)g'(t)\,dt\,\Biggr\rvert$$
grows slower than $g$ near $b$. To exploit the vanishing of $F_0$, split the interval into two parts. One part where $\lvert F_0(t)\rvert$ may be large, but that stays at a safe distance from $b$, and the remaining part close to $b$ where $\lvert F_0(t)\rvert$ is small.
Thus, let $\varepsilon > 0$ be given, and pick $c \in (a,b)$ such that $\lvert F_0(t)\rvert \leqslant \varepsilon$ for $c \leqslant t \leqslant b$. For all $x \in (c,b)$ we have
\begin{align}
\Biggl\lvert\, \int_a^x F_0(t)g'(t)\,dt\,\Biggr\rvert &\leqslant \Biggl\lvert\, \int_a^c F_0(t)g'(t)\,dt\,\Biggr\rvert + \int_c^x \lvert F_0(t)\rvert\,g'(t)\,dt \\
&\leqslant \Biggl\lvert\, \int_a^c F_0(t)g'(t)\,dt\,\Biggr\rvert +  \varepsilon \bigl(g(x) - g(c)\bigr)\,.
\end{align}
From this it follows that
$$\limsup_{x \to b^-} \frac{1}{g(x)}\Biggl\lvert\, \int_a^x F_0(t)g'(t)\,dt\,\Biggr\rvert \leqslant \varepsilon \tag{$\ast$}$$
since $g(x) \to +\infty$ and the integral as well as $g(c)$ are constants (for every fixed $c$).
But $(\ast)$ holds for every $\varepsilon > 0$, hence
$$\limsup_{x \to b^-} \frac{1}{g(x)}\Biggl\lvert\, \int_a^x F_0(t)g'(t)\,dt\,\Biggr\rvert  \leqslant 0\,.$$
Since the expression on the left hand side is nonnegative, this means the limit exists and is $0$.
