\chapter{Example with sin}

\begin{figure}
\begin{center}
\vskip -39pt
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\normalsize
%\put(40,0){\ing{oblakoV}}
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\put(-18,1006){\sx{6}{\(\frac{1}{2}\)}}
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\put(-38, 4){\sx{5}{\(\frac{-\!1}{2}\)}}
\put(44, -32){\sx{5}{\(-0.4\)}}
\put(244,-32){\sx{5}{\(-0.2\)}}
\put(508,-32){\sx{5}{\(0\)}}
\put(688,-32){\sx{5}{\(0.2\)}}
\put(888,-32){\sx{5}{\(0.4\)}}
\put(1088,-32){\sx{5}{\(0.6\)}}
\put(1288,-32){\sx{5}{\(0.8\)}}
\put(1488,-32){\sx{5}{\(1.0\)}}
\put(1688,-32){\sx{5}{\(1.2\)}}
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\put(532,930){\sx{4}{\rot{90}\(u\!=\!0\) \ero} }
\put(474,880){\sx{4}{\rot{0}\(v\!=\!1\) \ero} }
\put(740,1024){\sx{4}{\rot{32}\(v\!=\!0.8\) \ero} }
\put(822,886){\sx{4}{\rot{24}\(v\!=\!0.6\) \ero} }
\put(820,764){\sx{4}{\rot{10}\(v\!=\!0.4\) \ero} }
\put(30,659){\sx{4}{\rot{0}\(v\!=\!0.2\) \ero} }
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\put(1862,470){\sx{4}{\rot{89}\(u\!=\!1.2\) \ero} }
\end{picture}}
\end{center}
\vskip -4pt
\caption{Example \(u+\mathrm i v=f(x\!+\!\mathrm i y)\ \) ; \(\ y<x^2\) is shaded.\label{oblako}}
\vskip -4pt
\end{figure}

In order to show that term "asymptotic" is not so obvious, consider the example with elementary function:
\be
f(x)= \left\{  
\begin{array}{cc}
x^2 \ \sin(1/x)  &,\text{ if } x\ne 0 \\
0                      &, \text{ if } x= 0 
\end{array}
\right.
\ee

Can 0 be interpreted as asympototic of this function at zero?

Can 0 be interpreted as asympototic of derivative of this function at zero?

The answer depends on the set of values that are allowed for \(x\).\\

Zero appears as asymptotic of function \(f(z)\) at \(z\to \)
being considered at the range
\be
\{ z \in \bC : \Im(z)<\Re(z)^2 \}
\ee

The above seems to be simplest example, where, in order to get the asymptotic of the function, we beed to set some restriction on the range of values of its argument.