continuité en un point – continuité sur un intervalle

\(y=mx-5\) alors \(y=mx+2\) donc \(ax^2 + bx + c = 0\)

•   \(\cos (2\theta) = \cos^2 \theta ‐ \sin^2 \theta\)
•   \(a \bmod b)\)

\begin{pmatrix}1 & 2 \\3 & 4 \\\end{pmatrix}

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\begin{align}\cos 2x  &= \cos^2x – \sin^2x  \tag{4} \label{cos2x} \\1        &= \cos^2x + \sin^2x  \tag{5} \label{5}     \\\end{align}

We deduce from the formulas \ref{cos2x} and \ref{5} :\begin{align}\cos 2x &= \cos^2x + \sin^2x – \sin^2x – \sin^2x     \\ \\&= 1 – 2\sin^2x\end{align}

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\(\bbox[8px,border:2px solid red]{   e^x=\lim_{n\to\infty} \left( 1+\frac{x}{n} \right)^n}\)

\(\class{cmjx-highlight}{e^x=\lim_{n\to\infty} \left( 1+\frac{x}{n} \right)^n}\)