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

In my text book, they state the following:

$$\begin{align*}f(x) &= (\frac{1}{x} + \frac{1}{2}) (x-\frac{1}{2}x^2+\frac{1}{3}x^3+O(x^4))-1& ,x \rightarrow 0\\&= 1-\frac{1}{2}x+\frac{1}{3}x^2+\frac{1}{2}x-\frac{1}{4}x^3+O(x^3)-1& ,x \rightarrow 0 \end{align*}$$

However, when I calculate this, I get $1-\frac{1}{2}x+\frac{1}{3}x^2+\frac{1}{2}x-\frac{1}{4}x^3+O(x^3)+\frac{O(x^4)}{2}-1$. That $O(x^4)$ part disappears I guess, due to the big O notation. However, I cannot figure out why.

Furthermore, a few pages later, they say that $\lim_{x\rightarrow 0} O(x) = 0$. Which I do not really understand, since $O(x)$ defines a set of functions, no?

share|cite|improve this question
up vote 5 down vote accepted

For your first question, both $O(x^3)$ and $O(x^4)$ are error terms as $x$ approaches zero. Since $x^4$ goes to zero faster than $x^3$ as $x$ goes to zero, the larger error, $O(x^3)$ will subsume the smaller $O(x^4)$.

For your second question, you're correct in interpreting $O(x)$ as a set of functions. In this context, $O(x)$ is the set of all functions $f(x)$ for which $\mid f(x)\mid \le c\mid x\mid$, eventually, for some $c>0$ (which will depend on $f$). The limit $\lim_{x\rightarrow 0} O(x)$ is then interpreted to mean the limit of all such functions $f(x)$ as $x\rightarrow 0$, if it exists. It does in this case, since every $f\in O(x)$ satisfies $\mid f(x)\mid \le c\mid x\mid$, and so has limiting value $0$ as $x\rightarrow 0$.

share|cite|improve this answer

The $O(x)$ does define a set of function, and so $\lim_{x \to 0}O(x)=0$ means that $\lim_{x\to 0}f(x)=0$ for any $f \in O(x)$. You can easily show it using a single function from that set. $id(x)=x$ is obviously in $O(x)$ and satisfies $\lim_{x \to 0}id(x)=0$. Now, let there be $g$ in $O(x)$. So there exists a positive real number $M$ and a real number $\epsilon >0$ such that $0<|g(x)|< Mx$ for all $x$ s.t. $0-\epsilon< x< 0+\epsilon$ (Since I assumed that the $O(x)$ is "Big O around $0$".) Now $g$ converges to $0$ by the sandwich theorm.

share|cite|improve this answer
Thanks! Do you have an answer to my first question aswell perhaps? Then i can accept you answer. – Nga Jun 9 '12 at 17:37
Yes. Similarly, when x approaches 0, x^n becomes smaller for bigger n's. Thus you can easily show that O(x^3) contains O(x^4) and even O(x^3+O(x^4)). Hence the O(x^4) is redundant since the O(x^3) already contains it. – idan Jun 9 '12 at 17:49

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.