# What's wrong with this use of Taylor's expansions?

I'm trying to find the value of the following limit: $$\lim_{x \to 0} \frac{x^2\cos x - \sin(x\sin x)}{x^4}$$ Which I know equals to $-\dfrac13$.

I tried to do the following: $$\lim_{x \to 0} \frac{x^2(1 - \frac{x^2}{2} + o(x^4)) - \sin(x(x + o(x^3)))}{x^4}\\ = \lim_{x \to 0} \frac{x^2 - \frac{x^4}{2} + o(x^4) - \sin(x^2 + o(x^4)))}{x^4}\\ = \lim_{x \to 0} \frac{x^2 - \frac{x^4}{2} + o(x^4) - x^2 + o(x^4)}{x^4} = -\frac12$$ The result is clearly wrong. I suspect the mistake to be in the expansion of $\sin (x \sin x)$ but I don't get it.

What's wrong?

• Two things: 1) $x^p o(x^n)=o(x^{p+n})$. 2) You cannot neglect the $o(x^3)$, which should be an $o(x^4)$, term since the denominator is $x^4$. – GFR Jan 7 '16 at 10:59
• @GFR you're talking about the first expansion of $\sin x$? So I should go past $x^n$ with the Taylor expansion if $n$ is the power of the denominator? – mattecapu Jan 7 '16 at 11:03
• In the end you're computing $\lim (-x^4+O(x^4)/(2x^4))$ which gives $-1/2+O(1)$. So basically anything. To get a meaningful result you obviously have to expand further. – Bort Jan 7 '16 at 11:06
• @Bort that's a really nice explanation – mattecapu Jan 7 '16 at 11:08
• You want to be sure to get terms up to $o(x^4)$. Actually I think the right notation is $O(x^4)$ as $O(x^n)/x^n\rightarrow$ constant, while you use the little $o$ if the ratio goes to zero. Your $x^2 O(x^4)=O(x^6)$ vanishes divided by $x^4$ in the limit. The term $x O(x^3)=O(x^4)$ does not so you need to explicitly evaluate it. – GFR Jan 7 '16 at 11:09

Since you have $x^4$ in the denominator, you need to expand the numerator up to $x^4$. You did that for $x^2\cos x$ but not for $\sin (x \sin x)$.

So, you need to include terms up to $x^4$ in the expansion of $\sin (x \sin x)$: $$\sin (x \sin x) = x^2-x^4/6+o(x^6)$$ This comes for $$\sin (x) = x-x^3/6+o(x^5)$$ $$x\sin (x) = x^2-x^4/6+o(x^6)$$ which you then feed to $$\sin (y) = y+o(y^3)$$

• Yes, probably. But what's wrong in my attempt? Where's the error? If I didn't check on WA I wouldn't suspect it to be wrong, is such a nice result. – mattecapu Jan 7 '16 at 10:59
• @mattecapu, you have live terms in $x^4$ for $x^2\cos x$ but not for $\sin (x \sin x)$. – lhf Jan 7 '16 at 11:03
• now I understand the problem. If you add your last comment to your answer I'll be happy to accept it – mattecapu Jan 7 '16 at 11:06

Hint $$\sin(x\sin(x))=x^2-\frac{x^4}{6}+O\left(x^5\right)$$ would help.

In fact, and this is a general rule, if the denominator is $x^n$, you must develop the numerator at least to order $n$.

I suspect the mistake to be in the expansion of $~\sin(x\sin x),$ but I don't get it. What's wrong ?

Your error consists in using the double-approximation $\sin(x\sin x)\simeq x\sin x\simeq x^2,$ by applying

$\sin t\simeq t$ twice instead of just once, yielding the more accurate $\sin(x\sin x)\simeq x\sin x.$ The latter

leads to $~\lim\limits_{x\to0}~\dfrac{\cos x-\dfrac{\sin x}x}{x^2}=-\dfrac13,~$ which is different from $~\lim\limits_{x\to0}~\dfrac{\cos x-\color{red}1}{x^2}=-\dfrac12.$