Example 4. Let $f(x)=\left\{\begin{array}{c}-x+1 \text { for } x \leqslant 2, \\ 2 x+1 \text { for } x>2 .\end{array}\right.$ Find $f(2+0)$ and $f(2-0)$ :
Solution.

$$
\begin{gathered}
f(2+0)=\lim _{x \rightarrow 2+0} f(x)=\lim _{x \rightarrow 2}(2 x+1)=5 \\
f(2-0)=\lim _{x \rightarrow 2-0} f(x)=\lim _{x \rightarrow 2}(-x+1)=-1
\end{gathered}
$$

## Exercises

Find the right and left limits of the functions.

1. $f(x)=\frac{1}{5+3^{1 / x}}$ as $x \rightarrow 0$.
2. $f(x)=\left\{\begin{array}{l}x^{2} \text { if } x<3, \\ x+6 \text { if } x \geqslant 3\end{array}\right.$ as $x \rightarrow 3$.
3. $f(x)=\sqrt{\sin x}$ as $x \rightarrow 0$. 4. $f(x)=\frac{x}{|x|}$ as $x \rightarrow 0$
Find one-sided limits.
4. $\lim _{x \rightarrow 4 \pm 0} \frac{|x-4|}{x-4}$.
5. $\lim _{x \rightarrow 3 \pm 0} \frac{x+3}{9-x^{2}}$
6. $\lim _{x \rightarrow \pi / 4 \pm 0} \frac{|\operatorname{tg}(\pi-4 x)|}{x-\pi / 4}$.
7. $\lim _{x \rightarrow 1 \pm 0} \arcsin \frac{x}{2}$.
8. $\lim _{x \rightarrow 1 \pm 0} \arccos x$.
9. $\lim _{x \rightarrow 1 \pm 0} \operatorname{arctg} \frac{1}{x-1}$.
10. $\lim _{x \rightarrow 1 \pm 0} 2^{1 /(1-x)}$.
11. $\lim _{x \rightarrow 2 \pm 0} \frac{x^{2}-4}{|x-2|}$
12. $\lim _{x \rightarrow 0 \pm 0} \frac{|\sin x|}{x}$.
13. $\lim _{x \rightarrow \pi / 4 \pm 0} \frac{|\cos (x-\pi / 4)|}{|x-\pi / 4|}$
14. $\lim _{x \rightarrow 0 \pm 0} \frac{\arcsin 3 x}{x}$.

## Answers

1. $f^{+}(0)=0, f^{-}(0)=\frac{1}{5}$. 2. $f(3+0)=f(3-0)=9$. 3. $f^{-}(0)$ does not exist; $f^{+}(0)=0.4$. $f^{-}(0)=-1 ; f^{+}(0)=1$. 5. $\pm 1$. 6. $\mp \infty$. 7. $\pm 4$. 8. $\frac{\pi}{6}$.9. $\frac{\pi}{2} \cdot 10 . \pm \frac{\pi}{2} \cdot 11.0 ;+\infty .12 . \pm 4$. 13. $\pm 1.14 .+\infty$. 15. 3 .

## § 6. Continuous Functions

$1^{\circ}$. Let us recall (see $\S 1$), that a function $f(x)$ is called continuous at a point $x_{0}$ if the following three conditions are met:

1) the function $f(x)$ is defined in every point of some neighborhood of the point $x_{0}$
2) the function has a limit at this point: $\lim _{x \rightarrow x_{0}} f(x)=B$;
3) this limit is equal to $f\left(x_{0}\right)$, i.e., $\lim _{x \rightarrow x_{0}} f(x)=f\left(x_{0}\right)$.

The violation of any of the conditions listed here means that $f(x)$ is discontinuous at the point $x_{0}$. The classification of points of discontinuity is given in $\S 1$.

The definition of continuity applies to functions defined by different formulas in different intervals. A function that is continuous at every point of an interval is called continuous in this interval. Continuity of a function at the end of a segment $[a, b]$ is accepted as one-sided:

$$
\lim _{x \rightarrow a+0} f(x)=f(a), \quad \lim _{x \rightarrow b-0} f(x)=f(b)
$$

Below we will use the statement that every elementary function is continuous at all points of its domain of definition.

## Examples with Solutions