calculer les limites suivantes :

• \(\lim _{x \rightarrow+\infty} \sqrt{x^2-1}-2 x\)
• \(\lim _{x \rightarrow+\infty} \sqrt{x^2-1}-x\)
• \(\lim _{x \rightarrow+\infty} \sqrt{4 x^2-x-1}-2 x+1\)
• \(\lim _{x \rightarrow+\infty} \sqrt[3]{x^3-1}-2 x\)

CORRECTION

Essayer de faire l'exercice avant de voir la correction

\(\begin{aligned} \lim _{x \rightarrow+\infty} \sqrt{x^2-1}-2 x &=\lim _{x \rightarrow+\infty} \sqrt{x^2\left(1-\frac{1}{x^2}\right)}-2 x \\ &=\lim _{x \rightarrow+\infty}|x| \sqrt{1-\frac{1}{x^2}}-2 x \\ &=\lim _{x \rightarrow+\infty} x \sqrt{1-\frac{1}{x^2}}-2 x \\ &=\lim _{x \rightarrow+\infty} x\left(\sqrt{1-\frac{1}{x^2}}-2\right) \\ &=-\infty \end{aligned}\)

\(\begin{aligned} \lim _{x \rightarrow+\infty} \sqrt[3]{x^3-1}-2 x &=\lim _{x \rightarrow+\infty} \sqrt{x^3\left(1-\frac{1}{x^3}\right)}-2 x \\ &=\lim _{x \rightarrow+\infty} x \sqrt{1-\frac{1}{x^3}}-2 x \\ &=\lim _{x \rightarrow+\infty} x\left(\sqrt{1-\frac{1}{x^3}}-2\right) \\ &=-\infty \end{aligned}\)

\(\begin{aligned} \lim _{x \rightarrow+\infty} \sqrt{x^2-1}-x &=\lim _{x \rightarrow+\infty} \frac{\left(\sqrt{x^2-1}-x\right)\left(\sqrt{x^2-1}+x\right)}{\sqrt{x^2-1}+x} \\ &=\lim _{x \rightarrow+\infty} \frac{\sqrt{x^2-1}-x^2}{\sqrt{x^2-1}+x} \\ &=\lim _{x \rightarrow+\infty} \frac{x^2-1-x^2}{\sqrt{x^2-1}+x} \\ &=\lim _{x \rightarrow+\infty} \frac{-1}{\sqrt{x^2-1}+x} \\ &=0 \end{aligned}\)

\(\lim _{x \rightarrow+\infty} \sqrt{4 x^2-x-1}-2 x+1\)

\(=\lim _{x \rightarrow+\infty} \frac{\left(\sqrt{4 x^2-x-1}-2 x\right)\left(\sqrt{4 x^2-x-1}+2 x\right)}{\sqrt{4 x^2-x-1}+2 x}+1\)

\(=\lim _{x \rightarrow+\infty} \frac{4 x^2-x-1-4 x^2}{\sqrt{4 x^2-x-1}+2 x}+1\)
\(=\lim _{x \rightarrow+\infty} \frac{-x-1}{\sqrt{x^2\left(4-\frac{1}{x}-\frac{1}{x^2}\right)}+2 x}+1\)
\(=\lim _{x \rightarrow+\infty} \frac{-x-1}{x \sqrt{4-\frac{1}{x}-\frac{1}{x^2}}+2 x}+1\)
\(=\lim _{x \rightarrow+\infty} \frac{-x-1}{x \sqrt{4-\frac{1}{x}-\frac{1}{x^2}}+2 x}+1\)
\(=\lim _{x \rightarrow+\infty} \frac{x\left(-1-\frac{1}{x}\right)}{x\left(\sqrt{4-\frac{1}{x}-\frac{1}{x^2}}+2\right)}+1\)
\(=\lim _{x \rightarrow+\infty} \frac{-1-\frac{1}{x}}{\sqrt{4-\frac{1}{x}-\frac{1}{x^2}}+2}+1\)

\(=-\frac{1}{4}+1\)
\(=\frac{3}{4}\)