\stepcounter{subsection}
\subsection{Continuity}
\compactdef{Convergence in $\R^n$} Let $(x_k)_{k \in \N}$ where $x_k \in \R^n$ with $x_k = (x_{k, 1}, \ldots, x_{k, n})$ and let $y = (y_1, \ldots, y_n) \in \R^n$.
$(x_k)$ converges to $y$ as $k \rightarrow +\infty$ if $\forall \varepsilon > 0 \smallhspace \exists N \geq 1$ s.t. $\forall n \geq N$ we have $||x_k - y|| < \varepsilon$
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\shortlemma $(x_k)$ converges to $y$ as $k \rightarrow +\infty$ iff one of following equiv. statements holds:
\bi{(1)} $\forall 1 \leq i \leq n$, the sequence $(x_{k, i})$ with $x_{k, i} \in \R$ converges to $y_i$
\bi{(2)} $(||x_k - y||)$ converges to $0$ as $k \rightarrow +\infty$
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\compactdef{Continuity} Let $X \subseteq \R^n$ and $f: X \rightarrow \R^m$.
\bi{(1)} Let $x_0 \in X$. $f$ continuous in $\R^n$ if $\forall \varepsilon > 0 \smallhspace \exists \delta > 0$ s.t. if $x \in X$ satisfies $||x - x_0|| < \delta$,
then $||f(x) - f(x_0)|| < \varepsilon$
\bi{(2)} $f$ continuous \textit{on} $X$ if continuous at $x_0 \smallhspace \forall x_0 \in X$
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\shortproposition Let $X$ and $f$ as prev. Let $x_0 \in X$. $f$ continuous at $x_0$ iff $\forall (x_k)_{k \geq 1}$ in $X$ s.t.
$x_k \rightarrow x_0$ as $k \rightarrow +\infty$, $(f(x_k))_{k \geq 1}$ in $\R^m$ converges to $f(x)$\\
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\compactdef{Limit} Let $X$, $f$ and $x_0$ as prev. and $y \in \R^m$. $f$ \textit{has limit} $y$ as $x \rightarrow x_0$ with $x \neq x_0$ if
$\forall \varepsilon > 0 \smallhspace \exists \delta > 0$ s.t. $\forall x \neq x_0 \in X, ||x - x_0|| < \delta$ we have $||f(x) - y|| < \varepsilon$.
We write $\lim_{\elementstack{x \rightarrow x_0}{x \neq x_0}} f(x) = y$
\shortremark Also possible without ass. that $x_0 \in X$
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\shortproposition Let $X$, $f$, $x_0$ and $y$ as prev. We have $\lim_{\elementstack{x \rightarrow x_0}{x \neq x_0}} f(x) = y$ 
iff $\forall (x_k)$ in $X$ s.t. $x_k \rightarrow x$ as $k \rightarrow +\infty$ and $x_k \neq x_0$ $(f(x_k))$ in $\R^m$ converges to $y$
\stepLabelNumber{all}
\shortproposition Let $X \subseteq \R^n$, $y \subseteq \R^m$, $p \in \N$ and let $f: X \rightarrow Y$ and $g: Y \rightarrow \R^p$ be cont. Then $g \circ f$ is continuous
