115. Using the fact that when finding the limit of the ratio of two infinitesimals, they can be replaced by equivalent infinitesimals (property II), find the following limits:
1) $\lim _{x \rightarrow 0} \frac{\sin 4 x}{\sin 3 x}$
2) $\lim _{x \rightarrow 0} \frac{\tan^{2} 2 x}{\sin ^{2} \frac{x}{3}}$;
3) $\lim _{x \rightarrow 0} \frac{x \sin 2 x}{(\arctan 5 x)^{2}}$;

4)* $\lim _{n \rightarrow+\infty} \frac{\tan^{3} \frac{1}{n} \cdot \arctan \frac{3}{n \sqrt{n}}}{\frac{2}{n^{3}} \cdot \tan \frac{1}{\sqrt{n}} \cdot \arcsin \frac{5}{n}}$.
Solution. Using the fact that $\sin \alpha \approx \tan \alpha \approx \arcsin \alpha \approx \arctan x \approx \alpha$ as $\alpha \rightarrow 0$, which follows from the solution of problem 114, and applying the property of equivalent infinitesimals, we get:

1) $\lim _{x \rightarrow 0} \frac{\sin 4 x}{\sin 3 x}=\lim \frac{4 x}{3 x}=\frac{4}{3}$;
2) $\lim _{x \rightarrow 0} \frac{\tan^{2} 2 x}{\sin ^{2} \frac{x}{3}}=\lim \frac{(2 x)^{2}}{\left(\frac{x}{3}\right)^{2}}=36$;
3) $\lim _{x \rightarrow 0} \frac{x \sin 2 x}{(\arctan 5 x)^{2}}=\lim \frac{x}{5 x} \cdot \frac{2 x}{5 x}=\frac{2}{25}$;

4)* $\lim _{n \rightarrow+\infty} \frac{\tan^{3} \frac{1}{n} \cdot \arctan \frac{3}{n \sqrt{n}}}{\sin \frac{2}{n^{3}} \cdot \tan \frac{1}{\sqrt{n}} \cdot \arcsin \frac{5}{n}}=\lim \frac{\left(\frac{1}{n}\right)^{3} \cdot \frac{3}{n \sqrt{n}}}{\frac{2}{n^{3}} \cdot \frac{1}{\sqrt{n}} \cdot \frac{5}{n}}=$

$$
=\lim \frac{3 n^{4} \sqrt{n}}{10 n^{4} \sqrt{n}}=0.3
$$

1i6. Prove that as $x \rightarrow 0$ :

1) $\sqrt{6 x+1}-1 \approx 3 x$
2) $\sin x+\tan x \approx 2 x$
3) $\sqrt[3]{x+8}-2 \approx \frac{x}{12}$
4) $1-\cos \frac{x}{m} \approx \frac{x^{2}}{2 m^{2}}$.