diff --git a/chapters/multivar_calc.tex b/chapters/multivar_calc.tex
new file mode 100644
index 0000000..ee2c7a2
--- /dev/null
+++ b/chapters/multivar_calc.tex
@@ -0,0 +1,8 @@
+\documentclass[../script.tex]{subfiles}
+%! TEX root = ../../script.tex
+
+\begin{document}
+    \chapter{Multivariable Calculus}
+
+    \subfile{sections/partial_total_diff.tex}
+\end{document}
\ No newline at end of file
diff --git a/chapters/sections/conv_of_funcseq.tex b/chapters/sections/conv_of_funcseq.tex
new file mode 100644
index 0000000..e49384b
--- /dev/null
+++ b/chapters/sections/conv_of_funcseq.tex
@@ -0,0 +1,323 @@
+\documentclass[../../script.tex]{subfiles}
+% !TEX root = ../../script.tex
+
+\begin{document}
+\section{Convergence of Function sequences}
+
+\begin{defi}[Pointwise convergence]
+    Let $M$ be a set, $f_n: M \rightarrow \field ~~\forall n \in \natn$ and $f: M \rightarrow \field$. 
+    The sequence $\seq{f}$ is said to be pointwise convergent to $f$ if 
+    \[
+        \limn f_n(x) = f(x) ~~\forall x \in M
+    \]
+\end{defi}
+
+\begin{eg}
+    Consider
+    \begin{align*}
+        f_n: [0, 1] &\longrightarrow \realn \\
+        x &\longmapsto \begin{cases}
+            1 - nx, & x \in [0, \frac{1}{n}] \\
+            0, & \text{else}
+        \end{cases}
+    \end{align*}
+
+    \begin{center}
+        \begin{tikzpicture}[domain=0:1]
+            \begin{axis}[xmin=0, ymin=0, xmax=1, ymax=1, samples=50]
+                \addplot[black, thick] {1-x} node[above right, pos=0.6] {$f_1$};
+                \addplot[black, thick] {1-2*x} node[above right, pos=0.4] {$f_2$};
+                \addplot[black, thick] {1-3*x} node[right, pos=0.3] {$f_3$};
+                \addplot[black, thick] {1-4*x} node[left, pos=0.23] {$f_4$};
+            \end{axis}
+        \end{tikzpicture}
+    \end{center}
+    The $f_n$ are continuous for all $n \in \natn$ and converge pointwise to 
+    \begin{align*}
+        f: [0, 1] &\longrightarrow \realn \\
+        x &\longmapsto \begin{cases}
+            1, & x = 0 \\
+            0, & x \ne 0
+        \end{cases}
+    \end{align*}
+    $f$ is not continuous.
+\end{eg}
+
+\begin{rem}
+    Let $M$ be a set. Then 
+    \[
+        B(M) = \set[\exists K \in \realn: ~~|f(x)| < K ~~\forall x \in M]{f_n: M \longrightarrow \field}
+    \]
+    is a linear subspace of the space of all functions $M \rightarrow \field$. We can define the supremum norm 
+    \begin{align*}
+        \supnorm{\cdot}: B(M) &\longrightarrow \realn \\
+        f &\longmapsto \sup_{x \in M}\set{\abs{f(x)}}
+    \end{align*}
+\end{rem}
+\begin{proof}
+    We will now proof that $\supnorm{\cdot}$ is a norm. 
+    It is defined, because 
+    \begin{equation}
+        \supnorm{f} = 0 \implies \abs{f(x)} = 0 ~~\forall x \in M
+    \end{equation}
+    This implies 
+    \begin{equation}
+        f(x) = 0 ~~\forall x \in M \implies f = 0
+    \end{equation}
+    The triangle inequality is proven by first considering
+    \begin{equation}
+        \abs{f(x)} \le \supnorm{f} ~~\forall f \in B(M) ~\forall x \in M
+    \end{equation}
+    Let $f, g \in B(M)$, then 
+    \begin{equation}
+        \abs{f(x) + g(x)} \le \abs{f(x)} + \abs{g(x)} \le \supnorm{f} + \supnorm{g} ~~\forall x \in M
+    \end{equation}
+    Which implies 
+    \begin{equation}
+        \supnorm{f + g} = \sup_{x \in M} \abs{f(x) + g(x)} \le \supnorm{f} + \supnorm{g}
+    \end{equation}
+\end{proof}
+
+\begin{defi}[Uniform convergence]
+    A sequence of bounded functions $\seq{f}$,
+    \[
+        f_n: M \longrightarrow \field
+    \]
+    is said to be uniformly convergent to $f: M \rightarrow \field$ if its norm converges.
+    \[
+        \supnorm{f_n - f} \conv{n \rightarrow \infty} 0
+    \]
+\end{defi}
+
+\begin{rem}
+    Formally, pointwise convergence means 
+    \[
+        \forall \epsilon > 0 ~\forall x \in M ~\exists N \in \natn ~\forall n \ge N: ~~\abs{f_n(x) - f(x)} < \epsilon
+    \]
+    and uniform convergence means 
+    \[
+        \forall \epsilon > 0 ~\exists N \in \natn ~\forall x \in M ~\forall n \ge N: ~~\abs{f_n(x) - f(x)} < \epsilon
+    \]
+\end{rem}
+
+\begin{thm}
+    The function space $B(M)$ is complete.
+\end{thm}
+\begin{proof}
+    Let $\anyseqdef[f]{B(M)}$ be a Cauchy sequence in terms of $\supnorm{\cdot}$. Firstly, we have for some fixed $x \in M$
+    \begin{equation}
+        \abs{f_n(x) - f_m(x)} \le \supnorm{f_n - f_m}
+    \end{equation}
+    Since $\seq{f}$ is a Cauchy sequence, $(f_n(x))$ is also a Cauchy sequence in $\field_0$. Because $\field$ is complete,
+    $(f_n(x))$ converges, and we define 
+    \begin{equation}
+        f(x) = \limn f_n(x)
+    \end{equation}
+    thus $\seq{f}$ converges pointwise to $f$. Let $\epsilon > 0$. Then 
+    \begin{equation}
+        \exists N \in \natn: ~~\supnorm{f_n \cdot f_m} < \epsilon ~~\forall n, m \ge N
+    \end{equation}
+    Then $\forall x \in M, ~\forall n, m \ge N$ we have 
+    \begin{equation}
+        \abs{f_n(x) - f_m(x)} \le \supnorm{f_n - f_m} < \epsilon
+    \end{equation}
+    We can find the limit for $m \rightarrow \infty$
+    \begin{equation}
+        \abs{f(x) - f_n(x)} \le \epsilon
+    \end{equation}
+    and 
+    \begin{equation}
+        \supnorm{f} = \sup_{x \in M}\abs{f} \le \sup_{x \in M} \abs{f(x) - f_n(x)} + \sup_{x \in M} \abs{f_n(x)} = \epsilon + \supnorm{f_n}
+    \end{equation}
+    Thus, $f$ is bounded. Furthermore
+    \begin{equation}
+        \supnorm{f - f_n} = \sup_{x \in M} \abs{f(x) - f_n(x)} \le \epsilon
+    \end{equation}
+    which in turn implies 
+    \begin{equation}
+        \supnorm{f - f_n} \conv{n \rightarrow \infty} 0
+    \end{equation}
+\end{proof}
+
+\begin{defi}
+    Let $\metric$ be a metric space, then $C_b(X)$ is said to be the space of all continuous bounded functions.
+\end{defi}
+
+\begin{rem}
+    If $X$ is compact (e.g. a bounded, closed subset of $\realn^n$) then all continuous functions are bounded. We then write $C(X)$ for $C_b(X)$.
+\end{rem}
+
+\begin{thm}
+    Let $\metric$ be a metric space. $C_b(X)$ is closed in $B(X)$. In other words, every uniformly convergent sequence of continuous functions converges to a continuous function.
+\end{thm}
+\begin{proof}
+    Let $\anyseqdef[f]{C_b(X)}$ be a sequence that uniformly converges to $f \in B(X)$. 
+    Let $x \in X$ and $\epsilon > 0$, then 
+    \begin{equation}
+        \exists N \in \natn: ~~\supnorm{f - f_n} y \frac{\epsilon}{3} ~~\forall n \ge N
+    \end{equation}
+    Choose a fixed $n \ge N$. Since $f_n$ is continuous, this means that 
+    \begin{equation}
+        \exists \delta > 0: ~~\abs{f_n(x) - f_n(y)} < \frac{\epsilon}{3} ~~\forall y \in \oball[\delta](x)
+    \end{equation}
+    Then we have for all such $y$
+    \begin{equation}
+        \begin{split}
+            \abs{f(x) - f(y)} &\le \abs{f(x) - f_n(x)} + \abs{f_n(x) - f_n(y)} + \abs{f_n(y) - f(y)} \\
+            &\le 2 \cdot \supnorm{f - f_n} + f_n(x) - f_n(y) < \epsilon
+        \end{split}
+    \end{equation}
+    This proves the continuity of $f$ in $x$. Since $x \in X$ was chosen arbitrarily, $f$ is continuous everywhere.
+\end{proof}
+
+\begin{defi}
+    Let $x_0 \in \field$ and $\anyseqdef[a]{\field}$. Then 
+    \[
+        \series{n} a_n (x - x_0)^n
+    \]
+    is called a power series around $x_0$. The number 
+    \[
+        \rho := \sup\set[\series{n} a_n (x - x_0)^n \text{ converges}]{\abs{x - x_0}}
+    \]
+    is the convergence radius.
+
+    \begin{center}
+        \begin{tikzpicture}
+            \draw[->, thick] (0, 0) -- (0, 4) node [above] {};
+            \draw[->, thick] (0, 0) -- (4, 0) node [right] {};
+            
+            \draw[fill] (2, 2) circle [radius=1pt] node [below right] {$x_0$};
+            \draw[dashed] (2, 2) circle [radius=1.2cm];
+
+            \draw[<->, rotate around={45:(2, 2)}] (2, 2) -- node[above] {$\rho$} (3.2, 2);
+
+            \draw[fill, rotate around={100:(2, 2)}] (3.2, 2) circle [radius=1pt] node [above] {$x$};
+        \end{tikzpicture}
+    \end{center}
+\end{defi}
+
+\begin{rem}
+    All results so far (including proofs) can be extended to $\realn^n$-valued functions, or functions with values in a Banach space in general.
+\end{rem}
+
+\begin{thm}
+    Let $\series{n} a_n (x - x_0)^n$ be a power series with convergence radius $\rho \in [0, \infty) \cup \set{\infty}$.
+    If $\abs{x - x_0} < \rho$ then the series converges absolutely, for $\abs{x - x_0} > \rho$ it diverges.
+    \[
+        \frac{1}{\rho} = \limsupn \sqrt[n]{\abs{a_n}}
+    \]
+\end{thm}
+\begin{proof}
+    W.l.o.g. choose $x_0 = 0$:
+    For $\abs{x} > \rho$ the series diverges by definition. If $\abs{x} < \rho$ then there exists $y \in \field$ such that
+    $\abs{x} < \abs{y} \le \rho$ and $\series{n} a_n y^n$ convergent. Especially, $(a_n y^n)$ is a null sequence.
+    This means $\exists C > 0$ such that $\abs{a_n y^n} \le C ~~\forall n \in \natn$
+    \begin{equation}
+        \series{n} \abs{a_n x^n} = \series{n} \abs{a_n y^n} \abs{\frac{x}{y}}^n \le C \cdot \series{n} \abs{\frac{x}{y}}^n < \infty
+    \end{equation}
+    This statement only holds for $\rho > 0$.
+\end{proof}
+
+\begin{rem}
+    \begin{enumerate}[(i)]
+        \item We have 
+        \[
+            \rho = \sup \set[\series{n}\abs{a_n}a^n \text{ converges}]{a \in [0, \infty)}
+        \]
+
+        \item If the following limit exists, then 
+        \[
+            \rho = \limn \frac{\abs{a_n}}{\abs{a_{n+1}}}
+        \]
+    \end{enumerate}
+\end{rem}
+
+\begin{eg}
+    The series 
+    \[
+        \series{n} x^n
+    \]
+    is convergent on $(-1, 1)$, so $\rho = 1$. The limit function is 
+    \[
+        x \longmapsto \frac{1}{1 - x}
+    \]
+\end{eg}
+
+\begin{thm}
+    Let $\series{n} a_n (x - x_0)^n$ be a power series with convergence radius $\rho > 0$. 
+    Let $0 < a < \rho$. Then this power series converges uniformly on $\cball[a](x_0)$. Especially
+    \begin{align*}
+        f: \oball[\rho](x_0) &\longrightarrow \realn \\
+        x &\longmapsto \series{n} a_n (x_n - x_0)^n
+    \end{align*}
+\end{thm}
+\begin{proof}
+    W.l.o.g. choose $x_0 = 0$. Let $0 < a < \rho$. We know that $\series{n} a_nx^n$ converges on $\cball[a](0)$. 
+    
+    \begin{center}
+        \begin{tikzpicture}
+            \draw[->, thick] (0, -3) -- (0, 3);
+            \draw[->, thick] (-3, 0) -- (3, 0);
+
+            \draw[fill] (0, 0) circle [radius=1pt];
+
+            \draw[solid] (0, 0) circle [radius=1.8cm];
+            \draw[dashed] (0, 0) circle [radius= 2.4cm] node[above right=2.4cm] {$\rho$};
+
+            \node at (2, 0.2) {$a$};
+        \end{tikzpicture}
+    \end{center}
+
+    Define 
+    \begin{equation}
+    \begin{split}
+        f_n: \cball[a](0) &\longrightarrow \field \\
+        x &\longmapsto x^n ~~\forall n \in \natn
+    \end{split}
+    \end{equation}
+    We can see that 
+    \begin{equation}
+        \supnorm{f} = \sup_{x \in \cball[a](0)} \abs{f_n} = \sup_{x \in \cball[a](0)} = a^n
+    \end{equation}
+    and thus 
+    \begin{equation}
+        \series{n} a_n f_n \implies \series{n} \supnorm{a_n f_n} = \series{n} \abs{a_n}^n < \infty
+    \end{equation}
+    because $a < \rho$. The series $\series{n} a_n f_n$ is absolutely convergent in $C(\cball[a](0))$.
+    Since $C(\cball[a](0))$ is complete, $\series{n} a_n f_n$ is convergent because the partial sums $\series[N]{n} a_n f_n$ are continuous $\forall N \in \natn$.
+    Therefore $f$ is also continuous on $\cball[a](0)$. Let $x \in \oball[\rho](0)$. Then there exists some $a > 0$ such that $\abs{x} < a < \rho$.
+    Thus, $f$ is continuous on $\cball[a](0)$. Since $\cball[a](0)$ contains a neighbourhood of $x$, and continuity is a local property,
+    $f$ is also continuous in $x$. Because $x \in \oball[\rho](0)$ was chosen arbitrarily, $f$ is continuous.
+\end{proof}
+
+\begin{rem}
+    $\exp$, $\sin$, $\cos$ are continuous.
+\end{rem}
+
+\begin{eg}
+    The statements above can be extended to Banach space-valued power series (e.g. matrix-valued functions).
+    The norm on $\realn^{n \times n}$ is 
+    \[
+        \norm{A} = \sup\set[{\forall x \in \oball[1](0)}]{\norm{Ax}}
+    \]
+    Define 
+    \[
+        \exp(A) := \series{0} \frac{A^n}{n!}
+    \]
+    This converges $\forall A \in \realn^{n \times n}$, because
+    \begin{align*}
+        \series{n} \norm{\frac{A^n}{n!}} &= \series{n} \frac{1}{n!} \norm{A^n} \le \series{n} \frac{1}{n!} \norm{A}^n \\
+        &= \exp(\norm{A}) < \infty
+    \end{align*}
+    Thus, $\series{n} \frac{A^n}{n!}$ converges absolutely. Now consider the function 
+    \begin{align*}
+        \realn &\longrightarrow \realn^{n \times n} \\
+        t &\longmapsto \exp(At)
+    \end{align*}
+    This is a matrix-valued power series
+    \[
+        \exp(At) = \series{n} \frac{(At)^n}{n!} = \series{n} \frac{A^n}{n!} t^n
+    \]
+    with a convergence radius of $\rho = \infty$. In this case $\exp(A + B)$ doesn't necessarily have to equal $\exp(A) \cdot \exp(B)$.
+\end{eg}
+\end{document}
\ No newline at end of file
diff --git a/chapters/sections/partial_total_diff.tex b/chapters/sections/partial_total_diff.tex
new file mode 100644
index 0000000..e91ae6d
--- /dev/null
+++ b/chapters/sections/partial_total_diff.tex
@@ -0,0 +1,10 @@
+\documentclass[../../script.tex]{subfiles}
+%! TEX root = ../../script.tex
+
+\begin{document}
+\section{Partial and Total Differentiability}
+
+\begin{defi}
+
+\end{defi}
+\end{document}
\ No newline at end of file
diff --git a/chapters/sections/vector_spaces.tex b/chapters/sections/vector_spaces.tex
index 989d2c6..b868d47 100644
--- a/chapters/sections/vector_spaces.tex
+++ b/chapters/sections/vector_spaces.tex
@@ -304,7 +304,7 @@ Let $V$ be a vector space. A non-empty set $W \subset V$ is called a vector subs
 \begin{eg}
 Consider
 \[
-	W = \set[chi \in \realn]{(\chi, \chi) \in \realn^2}
+	W = \set[\chi \in \realn]{(\chi, \chi) \in \realn^2}
 \]
 This is a subspace, because
 \[
diff --git a/chapters/topo_of_metr_spaces.tex b/chapters/topo_of_metr_spaces.tex
index c92cc63..d0fc8c0 100644
--- a/chapters/topo_of_metr_spaces.tex
+++ b/chapters/topo_of_metr_spaces.tex
@@ -8,4 +8,5 @@
     \subfile{sections/seq_ser_limits.tex}
     \subfile{sections/open_closed_sets.tex}
     \subfile{sections/continuity.tex}
+    \subfile{sections/conv_of_funcseq.tex}
 \end{document}
\ No newline at end of file
diff --git a/script.pdf b/script.pdf
index 4f5d3da..7de1477 100644
Binary files a/script.pdf and b/script.pdf differ
diff --git a/script.tex b/script.tex
index de2bf73..cba5588 100644
--- a/script.tex
+++ b/script.tex
@@ -156,5 +156,6 @@
 \subfile{chapters/linear_algebra.tex}
 \subfile{chapters/real_analysis_2.tex}
 \subfile{chapters/topo_of_metr_spaces.tex}
+\subfile{chapters/multivar_calc.tex}
 
 \end{document}