# Compute $\lim_{x\rightarrow 0} \frac{\sin{x}-\arctan{x}}{x(\cos{2x}-1)}$

Using McLaurin expansion for the respective functions I get

$$\sin{x}=x-\frac{x^3}{6}+O(x^5), \\\arctan{x}=x-\frac{x^3}{3}+O(x^5) \\ \cos{t}=1-\frac{t^2}{2}+O(t^4)$$

The expansion for $\cos{t}$ gives

$$x(\cos{2x}-1)= x(1-2x^2+O(16x^4)-1)=-2x^3+xO(16x^4).$$

Combining everything gives the limit

$$\lim_{x\rightarrow 0} \frac{x-\frac{x^3}{6}+O(x^5)-x+\frac{x^3}{3}-O(x^5)}{-2x^3+xO(16x^4)}=\lim_{x\rightarrow 0}\frac{\frac{x^3}{2}+O(x^5)}{x^3(-2+O(16x^4))}=\lim_{x\rightarrow0}\frac{\frac{1}{2}+O(x^5)}{-2+O(16x^4)} \\ =\frac{1}{2}\cdot\frac{1}{-2}=-\frac{1}{4}.$$

Correct answer is $-\frac{1}{12}$.

Questions:

1) Where is my error?

2) Is it correct to multiply an ordo term with $x$ as I've done?

3) How do I know how many terms is enough to expand to when I want to compute a limit?

• can you use the rules of L'Hospital? Dec 27 '17 at 14:24
• Write numerator as $$\sin x-x+x-\arctan x$$ and use math.stackexchange.com/questions/387333/… Dec 27 '17 at 14:24

The mistake is here: $$-\frac{1}{6}+\frac{1}{3}=\frac{1}{6},$$ but you got $\frac{1}{2}.$

• Ahh basic arithmetic. But Otherwise, are my calculations correct or have I done something mathematically incorrect but accidently got right answer? Dec 27 '17 at 14:43
• I think, without this mistake your calculations are right. Dec 27 '17 at 14:44
• Thanks a lot mate! Dec 27 '17 at 14:45
• You are welcome! Dec 27 '17 at 14:45

using one times the rules of L'Hospital we get $${\frac {\cos \left( x \right) {x}^{2}+\cos \left( x \right) -1}{ \left( {x}^{2}+1 \right) \left( \cos \left( 2\,x \right) -1-2\,x\sin \left( 2\,x \right) \right) }}$$

For your 2nd question, your calculation with the $O$ term in the denominator is incorrect, but it does not affect the result. Also, you can remove constant factors (by definition). Here's a correct calculation: $$x(\cos 2x-1)=x\bigl(1-2x^2+O(x^4)-1\bigr)=-2x^3+xO(x^4)=-2x^3+O(x^5).$$

As to your 3rd question, there's no general rule: you only have to obtain ax expansion which is not reduced to its $O$ term. In ither words, you have to obtain an equivalent of the numerator/denominator.

• But If I have $O(t^4)$ and $t=2x$, won't I get $O((2x)^4)=O(16t^4)?$. One more question, say I have $(x+1)O(x^4)$, how does this simplify? Dec 27 '17 at 15:31
• $O(16x^4)=O(x^4)$ anyway, so why keep the coefficient? Inside $O$ it is meaningless. As to $(x+1)O(x^4)$, just expand it: $xO(x^4)+O(x^4)=O(x^5)+O(x^4)=O(x^4)$ (near $0$, of course, not near $\infty$). Dec 27 '17 at 15:43
• Thanks a lot Bernard! Cleard some things up here. Dec 27 '17 at 15:44