Example 6. Find $\lim _{x \rightarrow+\infty} x\left(\operatorname{arctg} x-\frac{\pi}{2}\right)$.
Solution. Let $\operatorname{arctg} x=\alpha$, then $x=\operatorname{tg} \alpha$, if $x \rightarrow+\infty$, then $\alpha \rightarrow \frac{\pi}{2}$.

Consequently,

$$
\begin{gathered}
\lim _{x \rightarrow+\infty} x\left(\operatorname{arctg} x-\frac{\pi}{2}\right)=\lim _{\alpha \rightarrow \frac{\pi}{2}} \operatorname{tg} \alpha\left(\alpha-\frac{\pi}{2}\right)= \\
=-\lim _{\alpha \rightarrow \frac{\pi}{2}} \frac{\sin \alpha}{\cos \alpha}\left(\frac{\pi}{2}-\alpha\right)=-\lim _{\alpha \rightarrow \frac{\pi}{2}} \sin \alpha \frac{\frac{\pi}{2}-\alpha}{\sin \left(\frac{\pi}{2}-\alpha\right)}= \\
=-\lim _{\alpha \rightarrow \frac{\pi}{2}} \sin \alpha \cdot \lim _{\alpha \rightarrow \frac{\pi}{2}} \frac{\frac{\pi}{2}-\alpha}{\sin \left(\frac{\pi}{2}-\alpha\right)}=-1 \cdot 1=-1
\end{gathered}
$$

## Problems

Find the limits:

1. $\lim _{x \rightarrow+\infty}\left(\sqrt{x^{2}+5 x+4}-\sqrt{x^{2}+x}\right)$.
2. $\lim _{x \rightarrow-4}\left(\frac{1}{x+4}-\frac{8}{16-x^{2}}\right)$.
3. $\lim _{x \rightarrow 0}\left(\frac{\cos x}{\sin ^{2} x}-\operatorname{ctg}^{2} x\right)$.
4. $\lim _{x \rightarrow 0}(\operatorname{ctg} x-\operatorname{cosec} x)$.
5. $\lim _{x \rightarrow 0} x \operatorname{ctg} \frac{\pi}{2} x$.
6. $\lim _{x \rightarrow \frac{\pi}{2}}\left(x-\frac{\pi}{2}\right) \operatorname{tg} x$.
7. $\lim _{x \rightarrow 0} \frac{e^{-x}-1}{3 x}$.
8. $\lim _{m \rightarrow \infty} m[\ln (m+4)-\ln m]$.
9. $\lim _{x \rightarrow 0} \frac{\sqrt{x+9}-3}{\sin 6 x}$.
10. $\lim _{n \rightarrow \infty} \frac{5-2^{n}}{5+2^{n+1}}$.
11. $\lim _{x \rightarrow 0}(\cos 2 x)^{\operatorname{ctg}^{2} 2 x}$.
12. $\lim _{x \rightarrow 0} \frac{1+x^{2}-\cos x}{\sin ^{2} x}$.

## Answers

1.2.2. $-\frac{1}{8}$. 3. $\frac{1}{2}$. 4. 0. 5. $\frac{2}{\pi}$. 6. -1 . Hint. Let $x=\frac{\pi}{2}+\alpha$. 7. $-\frac{1}{3}$. 8.4.9. $\frac{1}{36} \cdot 10 .-\frac{1}{2} \cdot 11 \cdot \frac{1}{\sqrt{e}} \cdot$ 12. $\frac{3}{2}$.

## § 1.7. Comparison of Infinitesimals

1. Two infinitesimals $\alpha$ and $\beta$ are called infinitesimals of the same order if the limit of their ratio is not zero, i.e.

$$
\lim \frac{\alpha}{\beta}=a(a \neq 0)
$$

2. The quantity $\alpha$ is called an infinitesimal of higher order compared to $\beta$ if the limit of the ratio of $\alpha$ to $\beta$ is zero, i.e.

$$
\lim \frac{\alpha}{\beta}=0
$$

3. The quantity $\alpha$ is called an infinitesimal of lower order compared to $\beta$ if the ratio of $\alpha$ to $\beta$ is an infinitely large quantity, i.e.

$$
\lim \frac{\alpha}{\beta}=\infty
$$

4. Infinitesimals $\alpha$ and $\beta$ are called equivalent if the limit of their ratio is one, i.e.

$$
\lim \frac{\alpha}{\beta}=1
$$

Notation for equivalent infinitesimals $\alpha$ and $\beta$:

$$
\alpha \sim \beta
$$

Equivalent infinitesimals have the following properties:

1) the difference of two equivalent infinitesimals is an infinitesimal of higher order compared to each of them;
2) when finding the limit of the ratio of two infinitesimals, each of them (or only one) can be replaced by another infinitesimal equivalent to it, i.e., if $\alpha \sim \alpha_{1}$ and $\beta \sim \beta_{1}$, then

$$
\lim \frac{\alpha}{\beta}=\lim \frac{\alpha_{1}}{\beta}=\lim \frac{\alpha}{\beta_{1}}=\lim \frac{\alpha_{1}}{\beta_{1}}
$$

Remark 1. The product of two infinitesimals is an infinitesimal of higher order compared to each of them.

Remark 2. To compare infinitesimals, find the limit of their ratio. If this ratio does not have a limit, then the quantities are incomparable.