Apostol Calculus, Volume I, Chapter 7.11, Problem 31 (p. 291), $o$-notation I am working on the following problem from Apostol, Calculus, Volume I:

Given two functions $f$ and $g$ with derivatives in some interval
  containing $0$, where $g$ is positive.  Assume also $f(x) = o(g(x))$
  as $x \to 0$.  Prove or disprove each of the following statements:
(a)$\displaystyle{\int_0^x f(t) dt = o\left(\int_0^x g(t) dt \right) \text{ as } x \to 0},$
(b)$f'(x) = o(g'(x)) \text{ as } x \to 0.$

This is from a section of exercises following the introduction of $o$-notation, and that seems to be the source of my problems.  I'm just confused about how to think about and deal with o-notation in these proofs.
I know,
$$f(x) = o(g(x)) \text{ as } x \to 0,$$
means that
$$ \lim_{x \to 0} \dfrac{f(x)}{g(x)} = 0 \text{ as } x \to 0.$$
In part (a) this translates to
$$ \lim_{x \to 0} \dfrac{\int_0^x f(t) dt}{\int_0^x g(t) dt} = 0 \text{ as } x \to 0$$
It seems like this should be true as, in some sense $f(x) = o(g(x))$ as $x \to 0$ implies that $g(x)$ is much larger than $f(x)$ as $x \to 0$.  Hence, I would think the integral of $g$ is also much larger.  Obviously that is not remotely close to rigorous, and may well be wrong.  Is this, morally, a reasonable way to think about the situation?
If someone can let me know how to approach this, and really a good way to think about $o$-notation in this context, I would appreciate it.  The more explicit the better, as I feel like my whole use of $o$-notation could best be described as "fuzzy" at the moment.  Thanks.
Edited: The book has not yet covered L'Hopital's rule, so it intends another approach.  If someone can approach it without using L'Hopital that would be helpful (as I'd like to learn what Apostol is intending to teach me).
 A: What you wanted to do intuitively may be formalized as follows. We know that $\displaystyle\lim_{x \rightarrow 0} \frac{f(x)}{g(x)} = 0$. This means that for every $\epsilon > 0$ there is a $\delta > 0$ such that $|x| < \delta$ implies $\left|\frac{f(x)}{g(x)}\right| < \epsilon$, that is, $|f(x)| < \epsilon |g(x)|$. Then we can say:
$\left|\frac{\displaystyle\int_{0}^{x} f(t) dt}{\displaystyle\int_{0}^{x} g(t) dt}\right| \leq \frac{\displaystyle\int_{0}^{x} |f(t)| dt}{\left|\displaystyle\int_{0}^{x} g(t) dt\right|} < \frac{\displaystyle\epsilon \displaystyle\int_{0}^{x} |g(t)| dt}{\left|\displaystyle\int_{0}^{x} g(t) dt\right|}$. Since $g$ is positive, the top and bottom cancel out, leaving $\left|\frac{\displaystyle\int_{0}^{x} f(t) dt}{\displaystyle\int_{0}^{x} g(t) dt}\right| < \epsilon$; therefore it tends to $0$.
A: For (b) consider $$f(x) := x^2 \sin\left(\frac{1}{x}\right)$$ for $x\neq 0$, and $f(0) :=  0$. It can be shown that $f$ is differentiable on $\mathbb{R}$ with $f'(0)=0$ and $$f'(x) = 2x\sin\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right)$$ for $x \neq 0$.
We have that $f(x) = o(x)$, but $f'(x) \neq o(1)$ since the limit
$$
\lim_{x \to 0}\left(2x\sin\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right)\right)
$$
does not exist.
