Asymptotic equivalence and $\lim_{x\to 0} \frac{\sin x}{x}=1$

Question

I know that for $x\sim0$ $\sin x$ can be approximated by $x$, hence they are 'asymptotic equivalent in the neighborhood of $x=0$'.

According to the definition of asymptotic equivalence, two functions $f(x)$ and $g(x)$ who are asymptotic equivalent in the neighbourhood of $x=0$ must satisfy the relationship

$$\lim_{x\to 0}\frac{f(x)}{g(x)}=1$$

which indeed $f(x)=\sin x$ and $g(x)=x$ satisfy.

Now, my question is

Can the limit $\displaystyle \lim_{x\to 0}\frac{\sin x}{x}=1$ also be justified using the definition of asymptotic equivalence?

Answer 1

No, asymptotic equivalence doesn't justify the limit $\lim_{x \to 0}\frac{\sin x}{x} = 1$. The limit justifies the equivalence.