Example. Calculate the limit of the function

$$
\lim _{x \rightarrow 1}\left(\frac{2 x-1}{x}\right)^{\ln (3+2 x) / \ln (2-x)}
$$
Solution. As $x \rightarrow 1$, the expression under the limit sign represents a power, the base of which tends to one:

$$
\lim _{x \rightarrow 1} \frac{2 x-1}{x}=1
$$

and the exponent tends to infinity:

$$
\lim _{x \rightarrow 1} \frac{\ln (3+2 x)}{\ln (2-x)}=\infty
$$

1. To use the table of equivalent infinitesimals, we make a variable substitution $t=x-1$ (then $t \rightarrow 0$ as $x \rightarrow 1$) and transform the expression under the limit sign:

$$
\begin{aligned}
\lim _{x \rightarrow 1}\left(\frac{2 x-1}{x}\right)^{\ln (3+2 x) / \ln (2-x)}=\lim _{t \rightarrow 0}\left(\frac{2 t+1}{t+1}\right)^{\ln (5+2 t) / \ln (1-t)}= \\
=\lim _{t \rightarrow 0} \exp \left[\frac{\ln (5+2 t)}{\ln (1-t)} \ln \frac{2 t+1}{t+1}\right]
\end{aligned}
$$

2. Since the exponential function $e^{x}$ is continuous, we can take the limit under the sign of this function. We have

$$
\lim _{t \rightarrow 0} \exp \left[\frac{\ln (5+2 t)}{\ln (1-t)} \ln \frac{2 t+1}{t+1}\right]=\exp \left[\lim _{t \rightarrow 0} \frac{\ln (5+2 t)}{\ln (1-t)} \ln \frac{2 t+1}{t+1}\right]
$$

3. We compute the limit of the exponent by replacing infinitesimal functions with equivalent ones:

$$
\begin{aligned}
\lim _{t \rightarrow 0}\left[\frac{\ln (5+2 t)}{\ln (1-t)} \ln \frac{2 t+1}{t+1}\right]=\lim _{t \rightarrow 0}\left[\frac{\ln (5+2 t)}{\ln (1-t)}\right. & \left.\ln \left(1+\frac{t}{t+1}\right)\right]= \\
& =\lim _{t \rightarrow 0} \frac{t \ln (5+2 t)}{-t(t+1)}=-\ln 5
\end{aligned}
$$

4. Finally, we get

$$
\lim _{x \rightarrow 1}\left(\frac{2 x-1}{x}\right)^{\ln (3+2 x) / \ln (2-x)}=e^{-\ln 5}=\frac{1}{5}
$$

Answer. $\lim _{x \rightarrow 1}\left(\frac{2 x-1}{x}\right)^{\ln (3+2 x) / \ln (2-x)}=\frac{1}{5}$.

Conditions of the problems. Calculate the limits.

1. $\lim _{x \rightarrow 2}\left(\frac{4-x}{2}\right)^{\operatorname{tg}(\pi x / 4)}$.
2. $\lim _{x \rightarrow 1}\left(\frac{\sin x}{\sin 1}\right)^{1 /(x-1)}$.
3. $\lim _{x \rightarrow 3}\left(\frac{\cos x}{\cos 3}\right)^{1 /(x-3)}$.
4. $\lim _{x \rightarrow 1}(2-x)^{\operatorname{tg}(\pi x / 2)}$.
5. $\lim _{x \rightarrow 1}\left(3 e^{x-1}-2\right)^{x /(x-1)}$.
6. $\lim _{x \rightarrow 2 \pi}(\cos x)^{1 / \sin ^{2} x}$.
7. $\lim _{x \rightarrow \pi / 4}(\operatorname{tg} x)^{1 / \cos 2 x}$.
8. $\lim _{x \rightarrow 1}(3-2 x)^{1 / \ln (2-x)}$
9. $\lim _{x \rightarrow 2}\left(\frac{4-x}{x}\right)^{1 / \ln (3-x)}$.
10. $\lim _{x \rightarrow 3}\left(\frac{3}{x}\right)^{1 / \ln (4-x)}$.

Answers. 1. $e^{2 / \pi}$. 2. $e^{\operatorname{ctg} 1} \cdot$ 3. $e^{-\operatorname{tg} 3} \cdot$ 4. $e^{2 / \pi} \cdot$ 5. $e^{3} \cdot 6 \cdot e^{-1 / 2}$. 7. $e^{-1}$. 8. $e^{2}$. 9. e. $10 . e^{1 / 3}$.

### 3.12. Calculation of $\lim _{x \rightarrow a} F(u(x) v(x)+f(x))$

Problem Statement. Calculate the limit

$$
\lim _{x \rightarrow a} F(u(x) v(x)+f(x))
$$

where $F(x)$ is continuous on $\mathbb{R}$, $f(x)$ is continuous at the point $x=a$, $u(x)$ is an infinitesimal function at the point $x=a$, and $v(x)$ is a function bounded in some neighborhood of the point $x=a$.

## PLAN OF SOLUTION.

1. Since $F(x)$ is continuous on $\mathbb{R}$, by the theorem on passing to the limit under the sign of a continuous function, we have

$$
\lim _{x \rightarrow a} F(u(x) v(x)+f(x))=F\left(\lim _{x \rightarrow a}(u(x) v(x)+f(x))\right)
$$

2. Since $u(x)$ is an infinitesimal function at the point $x=a$ and $v(x)$ is a bounded function in some neighborhood of the point $x=a$, then $u(x) v(x)$ is an infinitesimal function at the point $x=a$, i.e.,

$$
\lim _{x \rightarrow a} u(x) v(x)=0
$$

3. Since $f(x)$ is continuous at the point $a$, then

$$
\lim _{x \rightarrow a} f(x)=f(a)
$$

Using the basic properties of the limit of a function at a point, we get

$$
\lim _{x \rightarrow a} F(u(x) v(x)+f(x))=F(f(a))
$$