📝 Exercices de la série▼
Exercice 1Exercice 2Exercice 3Exercice 4Exercice 5Exercice 6Exercice 7Exercice 8Exercice 9Exercice 10Exercice 11Exercice 12Exercice 13Exercice 14

Exercice 10

Calculer les limites :

  • lim⁡x→+∞x2+3x2−35\lim\limits_{x\to+\infty} \sqrt[5]{\dfrac{x^2+3}{x^2-3}}

  • lim⁡x→+∞4x3+x2−13x\lim\limits_{x\to+\infty} \dfrac{\sqrt[3]{4x^3+x^2-1}}{x}

  • lim⁡x→0x+13−1x\lim\limits_{x\to0} \dfrac{\sqrt[3]{x+1}-1}{x}

Correction

  • On a
lim⁡x→+∞x2+3x2−3=lim⁡x→+∞x2x2=1\lim\limits_{x\to+\infty} \frac{x^2+3}{x^2-3}=\lim\limits_{x\to+\infty} \frac{x^2}{x^2}=1

Donc lim⁡x→+∞x2+3x2−35=15=1\lim\limits_{x\to+\infty} \sqrt[5]{\frac{x^2+3}{x^2-3}} =\sqrt[5]{1}=1

lim⁡x→+∞4x3+x2−13x=lim⁡x→+∞4x3+x2−1x33\lim\limits_{x\to+\infty} \frac{\sqrt[3]{4x^3+x^2-1}}{x}=\lim\limits_{x\to+\infty} \sqrt[3]{\frac{4x^3+x^2-1}{x^3}}

Comme lim⁡x→+∞4x3+x2−1x3=lim⁡x→+∞4x3x3=4\lim\limits_{x\to+\infty}\frac{4x^3+x^2-1}{x^3}=\lim\limits_{x\to+\infty}\frac{4x^3}{x^3}=4

Donc lim⁡x→+∞4x3+x2−13x=43\lim\limits_{x\to+\infty} \frac{\sqrt[3]{4x^3+x^2-1}}{x}=\sqrt[3]{4}

lim⁡x→0x+13−1x=lim⁡x→0(x+13−1)(x+132+x+13+1)x(x+132+x+13+1)=lim⁡x→0x+133−13x(x+132+x+13+1)=lim⁡x→01x+132+x+13+1=13\begin{align*} &\lim\limits_{x\to0} \frac{\sqrt[3]{x+1}-1}{x} \\ &=\lim\limits_{x\to0} \frac{(\sqrt[3]{x+1}-1)(\sqrt[3]{x+1}^2+\sqrt[3]{x+1}+1)}{x(\sqrt[3]{x+1}^2+\sqrt[3]{x+1}+1)} \\ &=\lim\limits_{x\to0} \frac{\sqrt[3]{x+1}^3-1^3}{x(\sqrt[3]{x+1}^2+\sqrt[3]{x+1}+1)} \\ &=\lim\limits_{x\to0} \frac{1}{\sqrt[3]{x+1}^2+\sqrt[3]{x+1}+1} \\ &=\frac{1}{3} \end{align*}