## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 1}\left(\frac{\sin (x-1)}{x-1}\right)^{\frac{\sin (x-1)}{x-1-\sin (x-1)}}$
## Solution

Substitution:

$x=y+1 \Rightarrow y=x-1$

$x \rightarrow 1 \Rightarrow y \rightarrow 0$

We get:

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

Using the substitution of equivalent infinitesimals:

$\ln \left(1-\left(1-\frac{\sin y}{y}\right)\right) \sim-\left(1-\frac{\sin y}{y}\right)_{, \text {as }} y \rightarrow 0\left(-\left(1-\frac{\sin y}{y}\right) \rightarrow 0\right)$

$\sin y \sim y$, as $y \rightarrow 0$

We get:

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