How to compute the Taylor expansion of $\sin^2(x)$?

Question

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

$$\left(x-\frac{1}{6}x^3+O(x^4)\right)^2$$

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?

Answer 1

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?

Answer 2

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