582. Using Theorem 5 of this paragraph, find the limits:
1) $\lim _{x \rightarrow 0} \frac{\sin 3 x}{5 x}$;
2) $\lim _{x \rightarrow 0} \frac{\sin 9 x}{\sin 5 x}$
3) $\lim _{x \rightarrow 0} \frac{\cos x-\cos 2 x}{x^{2}}$;
4) $\lim _{x \rightarrow 0} \frac{\tan p x}{\tan q x}$; 5) $\lim _{x \rightarrow 0} \frac{\ln (1+3 x)}{7 x}$; 6) $\lim _{x \rightarrow 0} \frac{\arcsin 5 x}{\arcsin 2 x}$.
Solution. 1) $\sin 3 x \sim 3 x$, therefore, $\lim _{x \rightarrow 0} \frac{\sin 3 x}{5 x}=$ $=\lim _{x \rightarrow 0} \frac{3 x}{5 x}=\frac{3}{5}$. The infinitesimal quantities $\beta=\sin 3 x$ and $\alpha=5 x$ are of the same order.

2) $\lim _{x \rightarrow 0} \frac{\sin 9 x}{\sin 5 x}=\lim _{x \rightarrow 0} \frac{9 x}{5 x}=\frac{9}{5}$.
3) $\lim _{x \rightarrow 0} \frac{\cos x-\cos 2 x}{x^{2}}=\lim _{x \rightarrow 0} \frac{2 \sin \frac{3 x}{2} \cdot \sin \frac{x}{2}}{x^{2}}=2 \lim _{x \rightarrow 0} \frac{\frac{3 x}{2} \cdot \frac{x}{2}}{x^{2}}=\frac{3}{2}$.

Thus, the infinitesimal $\beta=\cos x-\cos 2 x$ is a quantity of the second order compared to the infinitesimal $\alpha=x$.
4) $\operatorname{tg} p x \sim p x, \operatorname{tg} q x \sim q x$, therefore,

$$
\lim _{x \rightarrow 0} \frac{\operatorname{tg} p x}{\operatorname{tg} q x}=\lim _{x \rightarrow 0} \frac{p x}{q x}=\frac{p}{q}
$$

5) $\ln (1+3 x) \sim 3 x$, therefore,

$$
\lim _{x \rightarrow 0} \frac{\ln (1+3 x)}{7 x}=\lim _{x \rightarrow 0} \frac{3 x}{7 x}=\frac{3}{7}
$$

$6) \arcsin 5 x \sim 5 x, \arcsin 2 x \sim 2 x$, therefore,

$$
\lim _{x \rightarrow 0} \frac{\arcsin 5 x}{\arcsin 2 x}=\lim _{x \rightarrow 0} \frac{5 x}{2 x}=\frac{5}{2}
$$