distances before and after tikzpicture

Question

I have the problem that before and after a tikzpicture plot the distance is much higher than it should be

\documentclass[BCOR=3mm,11pt,headsepline,footsepline]{scrbook}
\usepackage[ngerman]{babel}
\usepackage[utf8]{inputenc}
\usepackage{multirow}
\usepackage{pgf,tikz}
\usepackage[framemethod=tikz]{mdframed}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage[dvipdfm]{geometry}
\usepackage{tabulary}
\usepackage{subfigure}
\usetikzlibrary{arrows}
\usetikzlibrary{shapes,snakes}
\usepackage{ntheorem}
\geometry{
   inner=2cm,
   outer=2cm,
   top=2cm,
   bottom=2.5cm,
   head=2cm,
   footnotesep=1cm,
%   bindingoffset=1cm,
   } 
\mdtheorem[%
apptotikzsetting={\tikzset{mdfbackground/.append style=%
{top color=white,
bottom color=white},
mdfframetitlebackground/.append style =%
{top color=gray!10!white,
bottom color=gray!10!white}
}},
,roundcorner=10pt,middlelinewidth=1.3pt,
frametitlerule=true,frametitlerulewidth=1.3pt,
innertopmargin=10pt,%
]{wissen}{Fehlerschwerpunkt}
\begin{document}
\subsection*{Aufgabe 5} 
\begin{itemize}
\item[•] $\displaystyle\lim\limits_{x\to 1} \ \frac{4x^2-3x-1}{x^2-1}=\lim\limits_{x\to 1} \ \frac{8x-3}{2x}=2,5$\\[3mm]
Die Regel von L'Hospital darf angewendet werden, weil der Ausdruck $\frac{0}{0}$ entsteht
\clearpage
\begin{center}
\begin{tikzpicture}[scale=0.6, >=latex,y=.5cm,x=1.7cm]
\draw[-angle 45,line width = 0.8pt] (-4,0) -- (4,0) node[right] {\scriptsize $x$};
\draw[-angle 45,line width = 0.8pt] (0,-5) -- (0,14) node[above] {\scriptsize $y$};
\foreach \x in {-3,...,-1,1,2,3} \draw (\x, 3pt) -- (\x,-3pt) node[anchor=north] {\tiny \x};
\foreach \y in {-4,-2,2,4,6,8,10,12} \draw (3pt,\y) -- (-3pt,\y) node[anchor=east] {\tiny \y};
\draw[color=red, line width = 1pt, domain=-0.6:4, samples=100]     plot (\x,{(4*(\x)^2-3*(\x)-1)/((\x)^2-1)});
\draw[color=red, line width = 1pt, domain=-4:-1.3, samples=100]   plot (\x,{(4*(\x)^2-3*(\x)-1)/((\x)^2-1)});
\draw [red] (1,2.5) circle (4pt);
\end{tikzpicture}
\end{center}
\item[•] $\displaystyle\lim\limits_{x\to 0} \ \frac{e^x-1}{4x}=\lim\limits_{x\to 0} \ \frac{e^x}{4}=0,25$\\[3mm]
Die Regel von L'Hospital darf angewendet werden, weil der Ausdruck $\frac{0}{0}$ entsteht
\begin{center}
\begin{tikzpicture}[scale=0.6, >=latex,y=1.5cm,x=1.7cm]
\draw[-angle 45,line width = 0.8pt] (-4,0) -- (4,0) node[right] {\scriptsize $x$};
\draw[-angle 45,line width = 0.8pt] (0,-1) -- (0,4) node[above] {\scriptsize $y$};
\foreach \x in {-3,...,-1,1,2,3} \draw (\x, 3pt) -- (\x,-3pt) node[anchor=north] {\tiny \x};
\foreach \y in {1,2,3} \draw (3pt,\y) -- (-3pt,\y) node[anchor=east] {\tiny \y};
\draw[color=red, line width = 1pt, domain=-4:-0.001, samples=100]     plot (\x,{(exp(\x)-1)/(4*(\x))});
\draw[color=red, line width = 1pt, domain=0.001:4, samples=100]     plot (\x,{(exp(\x)-1)/(4*(\x))});
\draw [red] (0,0.25) circle (4pt);
\end{tikzpicture}
\end{center}
\end{itemize} 
\end{document}