Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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).

share|cite|improve this question
It seems to me that in order for you to prove the first part of the exercise, you only need to use L'Hopital's rule in combination with the Fundamental Theorem of Calculus to transform the limit involving the integrals to the limit involving the functions $f$ and $g$, which you already know by hypothesis that it is $0$. – Adrián Barquero Apr 9 '12 at 2:33
Sorry, Apostol hasn't covered L'Hopital's yet (next section), so it is supposed to be done by other methods. I'll edit to add that. – user23784 Apr 9 '12 at 2:36
Yes I just looked at my copy of Apostol's Calculus Volume I and indeed L'Hopital's rule is covered in the following section. By the way, you should specify in your question that you mean Apostol's Calculus because he has written several books so just saying Apostol chapter XX does not completely specify the book. – Adrián Barquero Apr 9 '12 at 2:47
@AdriánBarquero Thanks. Usually remember to do that, but slipped my mind this time. – user23784 Apr 9 '12 at 2:51
up vote 3 down vote accepted

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 $0$.

share|cite|improve this answer

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.

share|cite|improve this answer
Nice example! I tried for a while to prove the result, without success; now I know why... – Pedro Apr 9 '12 at 4:23

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.