## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 0} \frac{\tan \left(\pi\left(1+\frac{x}{2}\right)\right)}{\ln (x+1)}$
## Solution

We will use the substitution of equivalent infinitesimals:

\(\ln (1+x) \sim x\), as \(x \rightarrow 0\)

\(\operatorname{tg} \frac{\pi x}{2} \sim \frac{\pi x}{2}\), as \(\frac{\pi x}{2} \rightarrow 0\)

We get:

\[
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\operatorname{tg}\left(\pi\left(1+\frac{x}{2}\right)\right)}{\ln (x+1)}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 0} \frac{\operatorname{tg}\left(\pi+\frac{\pi x}{2}\right)}{x}= \\
& =\lim _{x \rightarrow 0} \frac{\operatorname{tg}\left(\frac{\pi x}{2}\right)}{x}=\lim _{x \rightarrow 0} \frac{\frac{\pi x}{2}}{x}=\lim _{x \rightarrow 0} \frac{\pi}{2}=\frac{\pi}{2}
\end{aligned}
\]

## Problem Kuznetsov Limits 12-29