Martinovo isprobavanje latexa

latex_demo_01

\documentclass[oneside,a4paper]{book}
\headsep 0.5cm \pagestyle{myheadings}
\usepackage{amssymb,amsmath,latexsym,graphicx,tikz,indentfirst}
%\usepackage{amsart}

\def\t{\hspace{-5mm}{\bf .}\hspace{2mm}}%tacka kod sekcije
\def\bt{\hspace{-2mm}{\bf .}\hspace{2mm}}
\def\sst{\hspace{-4mm}{\bf .}\hspace{2mm}}%tacka kod podsekcije
\def\dj{d\kern-0.4em\char"16\kern-0.1em}
\def\Dj{\mbox{\raise0.3ex\hbox{-}\kern-0.4em D}}
% POSTAVLJANJE BROJA STRANE


\renewcommand{\chaptername}{Glava}
\renewcommand{\contentsname}{Sadr\v zaj}
\renewcommand{\bibname}{Literatura}
\renewcommand{\figurename}{Slika}


\newtheorem{teo}{Teorema}[chapter]
\newtheorem{lem}[teo]{Lema}
\newtheorem{tvr}[teo]{Tvr\dj enje}
\newtheorem{pos}[teo]{Posledica}
\newtheorem{de}[teo]{Definicija}
\newtheorem{pr}[teo]{Primer}
\newtheorem{nap}[teo]{Napomena}
\def\fakef{{}^{\lceil}f{}^{\rceil}}
\def\trqr{\,\theta(\rho_{qr})\,}
\def\tr{\,\theta(\rho)\,}
\def\aqr{\,\alpha_{qr}^*\,}
%\setcounter{ex}

\makeatletter
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\makeatother

\newcommand{\qed}{\hfill $\Box$\vspace*{4mm}}
\newenvironment{dokaz}
{\noindent
{\it Dokaz.}}
{\hspace{\stretch{1}}%
$\Box$}

\newenvironment{prim}
{\noindent
{\bf Primer.}}
{\hspace{\stretch{1}}%
$\Box$}

\newcounter{primer}[chapter]
\newcommand{\p}{\par\refstepcounter{primer}{\bf Primer \arabic{primer}. }}
\renewcommand{\theprimer}{\arabic{primer}}




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\makeatletter 
\tikzset{my loop/.style =  {to path={
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\makeatletter 





\begin{document}

\section{\emph{prva sekcija}}

Dokaz sledece teoreme se moze naci u \cite{teofanov}

\begin{teo}
Realna funkcija $f:\mathbb R\rightarrow \mathbb R$ ako i samo ako za svaki niz $\{x_n\}_{n\in\mathbb N}$ takav da je limes $\displaystyle\lim_{n\rightarrow\infty}x_n=a$ vazi $\displaystyle\lim_{n\rightarrow\infty}f(x_n)=a$.
\end{teo}

\begin{pos}
Funkcija $f:\mathbb R\rightarrow\mathbb R$ definisana sa $f(x)=x^2$ je neprekidna u $0$.

\end{pos}

\begin{dokaz}
Jasno je da $\displaystyle\lim_{n\rightarrow\infty}x_n=0$ povla\v ci $\displaystyle\lim_{n\rightarrow\infty}x_n^2$.
Dakle, na osnovu teoreme sledi da je $f$ neprekidna u $0$.
\end{dokaz}

$
M=
\begin{matrix} 
1 & 2 \\
3 & 4 
\end{matrix}
$

\begin{thebibliography}{9}

\bibitem{teofanov} 
Nenad Teofanov, Uvod u Analizu.

\end{thebibliography}

\end{document}