TITLE: Limitations of approximating $\sin(x) = x$
QUESTION [16 upvotes]: $$\lim _{x\rightarrow 0}{\frac {\cos \left( x \right) \sin \left( x
 \right) -x}{ \left( \sin \left( x \right)  \right) ^{3}}}$$

I know that the real limit is $-2/3$
However, I've noticed that by approximating $\sin(x)$ as $x$ and $\cos(x)$ as $1-(x^2/2)$ I get the following:
$(1-(x^2/2))x - x)/ x^3 = (1- (x^2/2) -1 )(1/x^2) = -x^2/(2x^2) = -1/2 $
also if I only partially approximate x like: $(x\cos(x) - x)/(x^3)$  = $(\cos(x)-1)/x^2$ and then use L'hospital's rule to chisel this down I get:
(L'hospital) = $-\sin x / 2x $ = (L'hospital) = $-\cos(x) / 2 = -1/2 $
Why does this conflict with doing L'hospital the whole way through without approximating $\sin(x)$ as $x$ ?  Why is approximating $\sin(x)^3$ as $x^3$ wrong? Isn't this always approaching zero?

REPLY [2 votes]: You need to relate this issue to the "order of approximation" of a Taylor series. You can show that if $P_n^a(x)$ is the Taylor Polynomial of order $n$ around $x=a$ of $F(x)$ then
$$\lim_{x\to a}\frac{F(x)-P_{n}^a}{(x-a)^n}=0$$
Just to make things clear $$P_n^a(x) = \sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(x-a)^k$$
In $\mathcal{o}$ notation
$$f(x) = P_n^a(x)+o((x-a)^n)$$
Let's put this into practice:
We can readily check that
$$\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - 1 - x}}{x} = 0$$
But what if we take
$$\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - 1 - x}}{{{x^2}}}$$
Now the limit is not zero ($=1/2$). So what do we need to make things "right"? Adding the next term in the Taylor series does the work. So we have:
$$\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - 1 - x - \frac{{{x^2}}}{2}}}{{{x^2}}} = 0$$
Similarily for the sine, when dealing with higher order powers of $x$, it doesn't suffice to put only $\sin x \sim x$. Although since for the sine the even powers are zero, some formulas do work, like
$$\mathop {\lim }\limits_{x \to 0} \frac{{\sin x - x}}{{{x^2}}} = 0$$
But for $x^3$ we need the next approximation:
$$\mathop {\lim }\limits_{x \to 0} \frac{{\sin x - x - \frac{{{x^3}}}{3!}}}{{{x^3}}} = 0$$
Thus you see $\sin(x) = x + o(x^3) $ is not correct, although $\sin(x) = x + o(x^2) $ and $\sin(x) = x + o(x)$ are.
IMPORTANT: Do not confuse Big-O with Little-O notation.