I'm trying to compute the following limit by Taylor expansion: $$\lim_{x\to 0}\frac{x^2-\sin^2x}{x^2\sin^2x}$$ However, how to compute the Taylor expansion of $\sin^2(x)$? I don't need the full series, just finite expansion with big-$O$ notation.

I know that $\displaystyle\sin x=x-\frac{1}{6}x^3+O(x^4)$, and then how can I compute


Should I use the lengthy formula $(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$? But how to deal with the $O(x^4)$ term?

It is suggested by the people below to memorize some of the big-$O$ properties. But to keep the minimal amount to memorize, what are the properties we are at least and at best choose to remember?

  • $\begingroup$ What happens if you use that formula? $\endgroup$ Jan 21, 2018 at 8:53
  • $\begingroup$ $x^2+\frac{1}{36}x^6+O(x^4)^2-\frac{1}{3}x^4-\frac{1}{3}x^3O(x^4)+2xO(x^4)$. But what are those terms involving $O(x^4)$? Should I memorize the bunches of big $O$ properties? $\endgroup$
    – Eric
    Jan 21, 2018 at 8:55
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    $\begingroup$ Transform $\sin^2(x)$ in terms of $\cos(2x)$ $\endgroup$ Jan 21, 2018 at 8:55
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    $\begingroup$ As @ClaudeLeibovici said, try considering $\sin^2(x)=\frac{1-\cos(2x)}{2}$. In addition, you do not need to consider the taylor expansion of $\sin^2$ in this computation: rewrite your limit as $\frac{(x+\sin(x))(x-\sin(x))}{(x\sin(x))^2}$ $\endgroup$
    – user515010
    Jan 21, 2018 at 9:00
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    $\begingroup$ Well, if you are going to use O you probably should know its properties? Otherwise you are going to mess up, inevitably. What option do you imagine there is to knowing it properties? $\endgroup$ Jan 21, 2018 at 9:05

2 Answers 2


Since $h(x)=O(x^n)\quad (x\to 0)$ means (by definition) that $$\frac{h(x)}{x^n}$$ is bounded in a neighborhood of $x=0$, and $g(x)=o(x^n) \quad (x\to 0)$ means that $$\lim_{x\to 0} \frac{g(x)}{x^n}=0,$$ then $h(x^n)\cdot x^m=O(x^{n+m})$ and $g(x^n)\cdot x^m=o(x^{n+m})$ since $$\frac{h(x)\cdot x^m}{x^{n+m}}$$ is bounded in a neighborhood of $x=0$ and $$\lim_{x\to 0} \frac{g(x)\cdot x^m}{x^{n+m}}=0.$$

So, long story short: $$O(x^n)\cdot x^m=O(x^{n+m})$$ and $$o(x^n)\cdot x^m=o(x^{n+m}).$$

In the same fashion you can prove a more general property:

$$g(x)=O(x^m) \wedge h(x)=O(x^n) \quad \implies \quad g(x)\cdot h(x)=O(x^{m+n})$$ and the same is true for little-o notation.

Put simpler although maybe a little imprecise: $O(x^m)\cdot O(x^n)=O(x^{m+n})$.

Finally, you could find also useful the property $$m\le n \quad \implies \quad O(x^m)+O(x^n)=O(x^m).$$

All these are proven in the same way, and are also valid for little-o.

Finally, remember that $g(x)=O(x^n)$ and $h(x)=O(x^n)$ does not imply that $g(x)=h(x)$, since "$...=O(x^n)$" is not to be read as an actual equality, but just as what is defined to be (see at the beginning).

Having said that, Taylor expansion of $\sin^2(x)$ is not you're only option here. Maybe you can think of some other path, although this one is fine and you should try it to.

Can you go on with this?

  • $\begingroup$ Please fix your latex code where you are writing g(x) ^x^m $\endgroup$
    – Paramanand Singh
    Jan 21, 2018 at 9:08
  • $\begingroup$ Thanks. So the properties that I must remember are: $O(f(x))\cdot O(g(x))=O(f(x)\cdot g(x))$, $f(x)O(g(x))=O(f(x)\cdot g(x))$ and when $x\to 0$, $m\le n \implies O(x^m)+O(x^n)=O(x^m)$? $\endgroup$
    – Eric
    Jan 21, 2018 at 9:15
  • $\begingroup$ Well, they're valid when $x\to a$ ($a=0$ or not) and when $x \to \infty$ as well. $\endgroup$ Jan 21, 2018 at 9:17
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    $\begingroup$ No, the latter is valid as $x\to0$, as @Eric correctly wrote. If $x\to a$ it's trivially valid, but also useless. And if $x\to+\infty$, it's wrong, since for $n>m$, $O(x^m)+O(x^n)=O(x^n)$. $\endgroup$ Jan 21, 2018 at 9:25
  • $\begingroup$ @Jean-Claude Arbaut: Thanks. So to save the precious time to do the logical inference whenever dealing with big-$O$ notation, it is better to just memorize some of the properties of it. But to keep the minimal amount to memorize, what are the properties we are at least and at best choose to remember? For example, in the "$O(x^m)+O(x^n)=O(x^?)$" case, it has many different situation with different result, if we're going to remember all the variation, it may be tedious. But if we don't memorize any of it, it takes us many effort and time whenver dealing with it. $\endgroup$
    – Eric
    Jan 21, 2018 at 9:31

You have $$\sin x=x(1-{x^2\over6}+?x^4)$$ and therefore, squaring and collecting terms in your head, $$\sin^2 x=x^2(1-{x^2\over3}+?x^4)\ .$$ (Remark: Note that the question mark is not an obscure $O$-term, but each time a stand-in for a complete convergent power series whose coefficients we do not want to bother with. See it this way: For any power series and any given order $r$ we can write $$\sum_{k=0}^\infty a_k x^k=\sum_{k=0}^{r-1} a_k x^k+ x^r\sum_{k=0}^\infty a_{r+k'} x^{k'}=:\ \sum_{k=0}^{r-1} a_k x^k+?x^r\ .\quad)$$

It follows that $x^2-\sin^2 x=x^4({1\over3}+?x^2)$ and therefore $${x^2-\sin^2 x\over x^2\sin^2 x}={x^4({1\over3}+?x^2)\over x^4(1-{x^2\over3}+?x^4)}\to{1\over3}\qquad(x\to0)\ .$$

  • $\begingroup$ Does $\sin^2 x=x^2(1-{x^2\over3}+?x^4)$ means $\sin^2 x=x^2(1-{x^2\over3}+?x^4+?x^5+?x^6+\cdots\cdots)$? $\endgroup$
    – Eric
    Jan 21, 2018 at 10:40
  • $\begingroup$ @Eric: See my edit. $\endgroup$ Jan 21, 2018 at 10:53

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