Exercice 8 Calculer les limites suivantes : limx→−∞(x2−x5+2x−1)\lim\limits_{x\to -\infty} (x^2-x^5+2x-1)x→−∞lim(x2−x5+2x−1) limx→+∞(−x4−x3+2x+3)\lim\limits_{x\to +\infty} (-x^4-x^3+2x+3)x→+∞lim(−x4−x3+2x+3) limx→0(x2−3x−1)\lim\limits_{x\to 0} (x^2-3x-1)x→0lim(x2−3x−1) limx→+∞x5−3x−13x4−3x2+1\lim\limits_{x\to +\infty} \dfrac{x^5-3x-1}{3x^4-3x^2+1}x→+∞lim3x4−3x2+1x5−3x−1 Correction limx→−∞(x2−x5+2x−1)\lim\limits_{x \to -\infty} (x^2 - x^5 + 2x - 1)x→−∞lim(x2−x5+2x−1) limx→−∞(x2−x5+2x−1)=limx→−∞(−x5)=−∞\begin{align*} \lim\limits_{x \to -\infty} (x^2 - x^5 + 2x - 1) &= \lim\limits_{x \to -\infty} (-x^5) \\&= -\infty \end{align*}x→−∞lim(x2−x5+2x−1)=x→−∞lim(−x5)=−∞ limx→+∞(−x4−x3+2x+3)\lim\limits_{x \to +\infty} (-x^4 - x^3 + 2x + 3)x→+∞lim(−x4−x3+2x+3) limx→+∞(−x4−x3+2x+3)=limx→+∞(−x4)=−∞\begin{align*} \lim\limits_{x \to +\infty} (-x^4 - x^3 + 2x + 3) &= \lim\limits_{x \to +\infty} (-x^4) \\&= -\infty \end{align*}x→+∞lim(−x4−x3+2x+3)=x→+∞lim(−x4)=−∞ limx→0(x2−3x−1)\lim\limits_{x \to 0} (x^2 - 3x - 1)x→0lim(x2−3x−1) limx→0(x2−3x−1)=02−3(0)−1=−1\begin{align*} \lim\limits_{x \to 0} (x^2 - 3x - 1) &= 0^2 - 3(0) - 1 \\&= -1 \end{align*}x→0lim(x2−3x−1)=02−3(0)−1=−1 limx→+∞x5−3x−13x4−3x2+1\lim\limits_{x \to +\infty} \dfrac{x^5 - 3x - 1}{3x^4 - 3x^2 + 1}x→+∞lim3x4−3x2+1x5−3x−1 limx→+∞x5−3x−13x4−3x2+1=limx→+∞x53.x4=limx→+∞x3=+∞\begin{align*} \lim\limits_{x \to +\infty} \dfrac{x^5 - 3x - 1}{3x^4 - 3x^2 + 1} &= \lim\limits_{x \to +\infty} \dfrac{x^5}{3.x^4} \\ &=\lim\limits_{x \to +\infty} \dfrac{x}{3} \\ &= +\infty \end{align*}x→+∞lim3x4−3x2+1x5−3x−1=x→+∞lim3.x4x5=x→+∞lim3x=+∞