Behavior of $\sin x/x$ as $x$ approaches 0? Which is the limiting behavior, $x \to 0\ \frac{\sin x}{x}$, in terms of x:
1) $\lim\limits_{x \to 0} \frac{\sin x}{x}$ = $\lim\limits_{x \to 0}\frac{\sin'x}{x'} = \frac{\lim\limits_{x \to 0}\cos x}{1} \rightarrow 1 - x^2/2 + O(x^4)$ by L'Hospitals' rule, or,
2) As $x \to 0,\  \frac{\sin x}{x}\sim\frac{x - x^3/3! + O(x^5)}{ x } = 1 - x^2/6 + O(x^4)$ expanding $\sin x$ by Taylor series.
I know each has a limit of 1, but what is the behavior in terms of the small $x$, and why don't both approaches give the same answer?
 A: Because L'Hopital's theorem says:

Let $f,g$ be two differentiable functions such that $g'(x)\ne 0$ in a neighbourhood of $x_0$, $\lim_{x\to x_0}\frac{f'(x)}{g'(x)}=L\in\Bbb R\cup\{-\infty,\infty\}$, and either $\lim_{x\to x_0} g(x)=\infty$ or $\lim_{x\to x_0}f(x)=\lim_{x\to x_0} g(x)=0$. Then, $$\lim_{x\to x_0}\frac{f(x)}{g(x)}=L$$

As you can see, the information it provides for the first-and-subsequent-order expansion of $\frac{f(x)}{g(x)}$ amounts to $0$.
Added: If by chance you wished to apply it to go further in the expansion of $\frac fg$ (let's say, first order and $f(x)\to 0,\ g(x)\to 0$), you'd end up with something like $$\lim_{x\to x_0} \frac{f(x)-Lg(x)}{xg(x)}\stackrel?=\lim_{x\to x_0} \frac{f'(x)+Lg'(x)}{g(x)+xg'(x)}$$ and there is no evident manipulation of the RHS that yields an explicit limit, nor it is obvious that $g+xg'$ is not frequently $0$ in the neighbourhoods of $x_0$.
A: !) is incorrect as stated by another person .
2)is correct provided you remove all the limit signs in the equations . The limit sign when added makes the expression denote just the  number which is the expressed limit. 
