Example 7. Find $\lim _{x \rightarrow 0} \frac{(1+x)^{a}-1}{x}$.
Solution. When $x \rightarrow \mathbf{0}$, the quantity $(1+x)^{a}-1$ is an infinitesimal. Let

$$
(1+x)^{a}-1=z
$$

Since

$$
z \sim \ln (1+z)
$$

then

$$
(1+x)^{a}-1 \sim \ln \left[1+(1+x)^{a}-1\right]=\ln (1+x)^{a}=a \ln (1+x)
$$

Therefore,

$$
\lim _{x \rightarrow 0} \frac{(1+x)^{a}-1}{x}=\lim _{x \rightarrow 0} \frac{a \ln (1+x)}{x}=a
$$

## Problems

1. Compare the infinitesimal quantities $a x, c x^{3}, b \sqrt[3]{x}$ with the infinitesimal $x$.
2. Prove the equivalence of infinitesimal quantities:
1) $\operatorname{tg} c x$ and $c x$
2) $\arcsin c x$ and $c x$
3) $\frac{1}{3} \operatorname{tg} 3 x$ and $x$
4) $x+2 x^{2}$ and $x$.
3. Find the limits:
1) $\lim _{x \rightarrow 0} \frac{\operatorname{tg} a x}{\operatorname{tg} b x}$
2) $\lim _{x \rightarrow 0} \frac{\operatorname{tg}^{3} a x}{\sin ^{3} b x}$
3) $\lim _{x \rightarrow 0} \frac{\sqrt[5]{1+x}-1}{x}$
4) $\lim _{x \rightarrow 0} \frac{\frac{1}{\sqrt[3]{1+x}}-1}{x}$
5) $\lim _{x \rightarrow 0} \frac{\ln (1+a x)}{x}$
6) $\lim _{x \rightarrow 2} \frac{\ln \left(3 x^{2}+5 x-21\right)}{x^{2}-6 x+8}$.

## Answers

1. $a x$ and $x$ are infinitesimals of the same order; $c x^{3}$ is an infinitesimal of a higher order compared to $x$; $b \sqrt[3]{x}$ is an infinitesimal of a lower order compared to $x$. 2. Hint. Use formula (1.72).
2. 3) $\frac{a}{b}$; 2) $\frac{a^{3}}{b^{3}}$; 3) $\frac{1}{5}$. Hint. See example 7; 4) $-\frac{1}{3}$; 5) $\left.a ; 6\right)-\frac{17}{2}$.

## § 1.8. Continuity of a Function

A function $y=f(x)$ is called continuous at a point $x_{1}$ if, at this point, an infinitely small increment of the argument $\Delta x$ corresponds to an infinitely small increment of the function $\Delta y$, i.e.,

$$
\lim _{\Delta x \rightarrow 0} \Delta y=\lim _{\Delta x \rightarrow 0}\left[f\left(x_{1}+\Delta x\right)-f\left(x_{1}\right)\right]=0
$$

A function $y=f(x)$ is called continuous at a point $x_{1}$ if, as $x \rightarrow x_{1}$, the limit of the function exists and is equal to its value at the point $x_{1}$, i.e.,

$$
\lim _{x \rightarrow x_{1}} f(x)=f\left(x_{1}\right)
$$

These two definitions of the continuity of a function at a point are equivalent to each other.

A function is called continuous in an interval if it is continuous at all points of this interval.

## Properties of Continuous Functions

1. The sum of several continuous functions is a continuous function.
2. The product of several continuous functions is a continuous function.
3. The quotient of two continuous functions is a continuous function at all points where the denominator is not zero.
4. If $y=f(z)$ and $z=\varphi(x)$ are continuous functions of their arguments, then the composite function $y=f[\varphi(x)]$ is also continuous.
5. If the function $y=f(x)$ is continuous and there exists an inverse function $x=\varphi(y)$, then the latter is also continuous.

A function $y=f(x)$ is called discontinuous at a point $x_{1}$ if it is defined in arbitrarily close points, but at the point $x_{1}$ it does not satisfy the condition of continuity.

If for the function $y=f(x)$ there exist finite limits

$$
\lim _{x \rightarrow x_{1}-0} f(x) \text { and } \lim _{x \rightarrow x_{1}+0} f(x)
$$

and not all three numbers $f\left(x_{1}\right), \lim _{x \rightarrow x_{1}-0} f(x), \lim _{x \rightarrow x_{1}+0} f(x)$ are equal to each other, then $x_{1}$ is called a point of discontinuity of the first kind. In particular, if

$$
\lim _{x \rightarrow x_{1}-0} f(x)=\lim _{x \rightarrow x_{1}+0} f(x)
$$

then $x_{1}$ is called a removable point of discontinuity.

Points of discontinuity of a function that are not points of discontinuity of the first kind are called points of discontinuity of the second kind. At such points, at least one of the one-sided limits is infinite or does not exist.