## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 3}\left(2-\frac{x}{3}\right)^{\sin (\pi x)}$
## Solution

$$
\begin{aligned}
& \lim _{x \rightarrow 3}\left(2-\frac{x}{3}\right)^{\sin (\pi x)}=\lim _{x \rightarrow 3}\left(e^{\ln \left(2-\frac{x}{3}\right)}\right)^{\sin (\pi x)}= \\
& =\lim _{x \rightarrow 3} e^{\sin (\pi x) \cdot \ln \left(2-\frac{x}{3}\right)}=\exp \left\{\lim _{x \rightarrow 3} \sin (\pi x) \cdot \ln \left(2-\frac{x}{3}\right)\right\}= \\
& =\exp \left\{\lim _{x \rightarrow 3} \sin (\pi x) \cdot \ln \left(1+\left(1-\frac{x}{3}\right)\right)\right\}=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

$$
\ln \left(1+\left(1-\frac{x}{3}\right)\right) \sim 1-\frac{x}{3}, \text { as } \quad x \rightarrow 3\left(1-\frac{x}{3} \rightarrow 0\right)
$$

We get:

$$
\begin{aligned}
& =\exp \left\{\lim _{x \rightarrow 3} \sin (\pi x) \cdot\left(1-\frac{x}{3}\right)\right\}=\exp \left\{\sin (\pi \cdot 3) \cdot\left(1-\frac{3}{3}\right)\right\}= \\
& =\exp \{0\}=e^{0}=1
\end{aligned}
$$

## Problem Kuznetsov Limits 20-7