calculus - Limitations of approximating $sin(x) = x$

$$\lim _{x\rightarrow 0}{\frac {\cos \left( x \right) \sin \left( x
\right) -x}{ \left( \sin \left( x \right) \right) ^{3}}}$$

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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?

Answer

As you know, $\sin x$ is only approximately equal to $x$, and $\cos x$ is only approximately $1 - \frac12 x^2$. How do you know when the approximation is good enough? One way is to keep track of how big the terms we've thrown away are. If you look at the Taylor series of $\sin$ and $\cos$, you find that $\sin x = x + O(x^3)$, and $\cos x = 1 - \frac12 x^2 + O(x^4)$, where $O(x^n)$ means something on the order of $x^n$ whose exact value we don't care about. So your limit is

$$\begin{align}
\lim_{x\to 0} \frac {\cos x \sin x - x}{\sin^3 x} &= \lim_{x\to 0} \frac{\big(1 - \frac12 x^2 + O(x^4)\big)\big(x + O(x^3)\big) - x}{\big(x + O(x^3)\big)^3} \\
&= \lim_{x\to 0} \frac{\big(x - \frac12 x^3 + O(x^3)\big) - x}{x^3 + O(x^5)} \\
&= \lim_{x\to 0} \frac{-\frac12 x^3 + O(x^3)}{x^3 + O(x^5)} \\
&= \lim_{x\to 0} -\frac12 + O(1)
\end{align}$$
As you can see, one of the terms we ignored produces an error that doesn't go away as we approach $0$, so our solution is no good. If you trace backward to where the error came from, you'll find that it's the $O(x^3)$ term in the approximation of $\sin$. Then you would be wise to replace it with its true value, giving $\sin x = x - \frac16 x^3 + O(x^5)$, and evaluate the limit again. This time you should get the right answer, with an extra term that goes to zero as $x \to 0$.