## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 1} \frac{3-\sqrt{10-x}}{\sin 3 \pi x}$
## Solution

Substitution:

$$
\begin{aligned}
& x=y+1 \Rightarrow y=x-1 \\
& x \rightarrow 1 \Rightarrow y \rightarrow 0
\end{aligned}
$$

We get:

$$
\begin{aligned}
& \lim _{x \rightarrow 1} \frac{3-\sqrt{10-x}}{\sin 3 \pi x}=\lim _{y \rightarrow 0} \frac{3-\sqrt{10-(y+1)}}{\sin 3 \pi(y+1)}= \\
& =\lim _{y \rightarrow 0} \frac{3-\sqrt{9-y}}{\sin (3 \pi y+3 \pi)}=\lim _{y \rightarrow 0} \frac{3-\sqrt{9-y}}{\sin (3 \pi y+\pi)}= \\
& =\lim _{y \rightarrow 0} \frac{3-\sqrt{9-y}}{-\sin 3 \pi y}=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

$\sin 3 \pi y \sim 3 \pi y, \text{ as } y \rightarrow 0 (3 \pi y \rightarrow 0)$

We get:

$$
\begin{aligned}
& =\lim _{y \rightarrow 0} \frac{3-\sqrt{9-y}}{-3 \pi y}=\lim _{y \rightarrow 0} \frac{(3-\sqrt{9-y})(3+\sqrt{9-y})}{-3 \pi y(3+\sqrt{9-y})}= \\
& =\lim _{y \rightarrow 0} \frac{9-(9-y)}{-3 \pi y(3+\sqrt{9-y})}=\lim _{y \rightarrow 0} \frac{y}{-3 \pi y(3+\sqrt{9-y})}= \\
& =\lim _{y \rightarrow 0} \frac{1}{-3 \pi(3+\sqrt{9-y})}=\frac{1}{-3 \pi(3+\sqrt{9-0})}=-\frac{1}{18 \pi}
\end{aligned}
$$

## Problem Kuznetsov Limits 13-29