Example 8. Find the limits

\[
\begin{gathered}
\lim _{x \rightarrow 2-0} \operatorname{arctg} \frac{1}{2-x} \text { and } \\
\lim _{x \rightarrow 2+0} \operatorname{arctg} \frac{1}{2-x}
\end{gathered}
\]
Solution. Let $x \rightarrow 2$ from the left, i.e., remaining less than 2, then the quantity $(2-x)$ will be infinitely small, taking positive values, and the quantity $\frac{1}{2-x}$ will be infinitely large, also taking positive values. Considering the definition of the function $z=\operatorname{arctg} y$ and the fact that $\operatorname{tg} z \rightarrow \infty$ as $z \rightarrow \frac{\pi}{2}$, we conclude that

$$
\lim _{x \rightarrow 2-0} \operatorname{arctg} \frac{1}{2-x}=\frac{\pi}{2}
$$

If, however, $x \rightarrow 2$ from the right, i.e., remaining greater than 2, then the quantity $(2-x)$ will be a negative infinitely small value, and the quantity $\frac{1}{2-x}$ will be a negative infinitely large value.

Since $\operatorname{tg} z \rightarrow -\infty$ as $z \rightarrow -\frac{\pi}{2}$, then

$$
\lim _{x \rightarrow 2+0} \operatorname{arctg} \frac{1}{2-x}=-\frac{\pi}{2}
$$

The limit of the function $y=\operatorname{arctg} \frac{1}{2-x}$ does not exist as $x \rightarrow 2$ in any arbitrary manner, since the one-sided limits at this point are not equal to each other.

The graph of the function is shown in Fig. 1.24.

![](https://cdn.mathpix.com/cropped/2024_05_22_7ef2114a321cf223ba99g-038.jpg?height=403&width=503&top_left_y=53&top_left_x=28)

Fig. 1.24

## Problems

1. Starting from which number do the values of each of the following sequences:
1) $x_{n}=\frac{1}{n}$
2) $y_{n}=\frac{1}{n^{2}}$
3) $z_{n}=\frac{1}{n^{4}}(n=1,2,3, \ldots)$

become and remain less than $\varepsilon=0.0001$? Show that each sequence has a limit of zero.

2. Prove that each of the sequences

$$
x_{n}=\frac{1}{2^{n}}, y_{n}=-\frac{1}{2^{n}}, z_{n}=\frac{(-1)^{n}}{2^{n}}
$$

has a limit of zero. Starting from which number do the values of each of them remain less than $\varepsilon=0.001$ in absolute value?

3. Given three sequences:

$$
x_{n}=2-\frac{1}{4^{n}} ; y_{n}=-3+\frac{1}{5^{n}} ; z_{n}=7+\frac{(-1)^{n}}{6^{n}}
$$

Prove that:

$$
\text { 1) } \lim _{n \rightarrow \infty} x_{n}=2 \text {; 2) } \lim _{n \rightarrow \infty} y_{n}=-3 ; \text { 3) } \lim _{n \rightarrow \infty} z_{n}=7
$$

4. Prove that:
1) $\lim _{x \rightarrow \infty} \frac{4 x+5}{5 x}=\frac{4}{5}$
2) $\lim _{x \rightarrow 3}(4 x-7)=5$;
3) $\lim _{x \rightarrow-1}(5 x+8)=3$;
4) $\lim _{x \rightarrow 2}\left(3 x^{2}-4 x+6\right)=10$.
5. Find the one-sided limits as $x \rightarrow 2$ from the left and right for the following functions:
1) $f(x)=\frac{8}{2-x}$
2) $f(x)=\frac{4}{(x-2)^{2}}$;
3) $f(x)=2^{\frac{1}{x-2}}$;
4) $f(x)=\operatorname{arctg} \frac{1}{2-x}$.

## Answers

1. 2) $N_{1}=10000$;
2) $N_{2}=100$;
3) $N_{3}=10$.
2. $N=10$.
3. 4) $\lim _{x \rightarrow 2-0} \frac{8}{2-x}=\infty, \quad \lim _{x \rightarrow 2+0} \frac{8}{2-x}=-\infty ; \quad$ 2) $\quad \lim _{x \rightarrow 2-0} \frac{4}{(x-2)^{2}}=+\infty$, $\lim _{x \rightarrow 2+0} \frac{4}{(x-2)^{2}}=+\infty$.

## § 1.4. Finding Limits

If the variable quantities $f_{1}(x), f_{2}(x), f_{3}(x)$ have finite limits as $x \rightarrow a$, then

$$
\begin{gathered}
\lim _{x \rightarrow a}\left[f_{1}(x)+f_{2}(x)-f_{3}(x)\right]=\lim _{x \rightarrow a} f_{1}(x)+\lim _{x \rightarrow a} f_{2}(x)-\lim _{x \rightarrow a} f_{3}(x) \\
\lim _{x \rightarrow a}\left[f_{1}(x) \cdot f_{2}(x) \cdot f_{3}(x)\right]=\lim _{x \rightarrow a} f_{1}(x) \cdot \lim _{x \rightarrow a} f_{2}(x) \cdot \lim _{x \rightarrow a} f_{3}(x) \\
\lim _{x \rightarrow a}[c f(x)]=c \lim _{x \rightarrow a} f(x)(c=\text { const }) \\
\lim _{x \rightarrow a}[f(x)]^{n}=\left[\lim _{x \rightarrow a} f(x)\right]^{n}(n-\text { integer, } n>0) \\
\lim _{x \rightarrow a} \frac{f_{1}(x)}{f_{2}(x)}=\frac{\lim _{x \rightarrow a} f_{1}(x)}{\lim _{x \rightarrow a} f_{2}(x)}\left(\lim _{x \rightarrow a} f_{2}(x) \neq 0\right) \\
\lim _{x \rightarrow a} \sqrt[n]{f_{1}(x)}=\sqrt[n]{\lim _{x \rightarrow a} f_{1}(x)} \\
\lim _{x \rightarrow a} c=c \quad(c=\text { const })
\end{gathered}
$$

If $c$ is a constant, and $c>0$, then:

$$
\begin{aligned}
& \lim _{x \rightarrow \infty} c x=\infty \\
& \lim _{x \rightarrow \infty} \frac{x}{c}=\infty \\
& \lim _{x \rightarrow-0} \frac{c}{x}=-\infty
\end{aligned}
$$

$$
\begin{gathered}
\lim _{x \rightarrow+0} \frac{c}{x}=+\infty \\
\lim _{x \rightarrow 0} \frac{c}{x}=\infty \\
\lim _{x \rightarrow \infty} \frac{c}{x}=0 \\
\lim _{x \rightarrow+\infty} c^{x}=\left\{\begin{array}{c}
0, \text { if } 01
\end{array}\right. \\
\lim _{x \rightarrow-\infty} c^{x}=\left\{\begin{array}{r}
+\infty, \text { if } 01
\end{array}\right.
\end{gathered}
$$