Exercice 1
Calculer :
- \(\lim_{x\to-\infty}(-3x^3 + x^2 - 7)\)
- \(\lim_{x\to+\infty}\frac{4x^3 - 1}{2x^3 + x}\)
- \(\lim_{x\to-\infty}\frac{x^2 + 1}{x - 3}\)
Correction
- \(-3x^3\to+\infty\) : limite \(+\infty\).
- \(\frac{4x^3}{2x^3} = 2\).
- \(\frac{x^2}{x} = x\to-\infty\).
Exercice 2
Calculer \(\lim_{x\to-2}\frac{x^2 - 4}{x^2 + 3x + 2}\).
Correction
\(\frac{(x - 2)(x + 2)}{(x + 1)(x + 2)} = \frac{x - 2}{x + 1}\to\frac{-4}{-1} = 4\).
Exercice 3
Calculer \(\lim_{x\to0}\frac{\sqrt{x + 4} - 2}{x}\).
Correction
\(\frac{x}{x(\sqrt{x + 4} + 2)} = \frac{1}{\sqrt{x + 4} + 2}\to\frac14\).
Exercice 4
Calculer \(\lim_{x\to3^-}\frac{x}{3 - x}\) et \(\lim_{x\to3^+}\frac{x}{3 - x}\).
Correction
Numérateur \(\to3\). À gauche, \(3 - x\to0^+\) : \(+\infty\). À droite : \(-\infty\).
Exercice 5
Calculer \(\lim_{x\to0}\frac{\sin 4x}{\sin 2x}\) et \(\lim_{x\to0}\frac{1 - \cos 2x}{x^2}\).
Correction
\(\frac{\sin 4x}{4x}\cdot\frac{2x}{\sin 2x}\cdot2\to2\). \(\frac{1 - \cos 2x}{(2x)^2}\cdot4\to\frac12\times4 = 2\).