Exercice 12 Calculer les limites suivantes : limx→+∞x+2xx−1\lim\limits_{x\to +\infty} \dfrac{x+2\sqrt{x}}{x-1}x→+∞limx−1x+2x limx→2x2−xx+x\lim\limits_{x\to 2} \dfrac{x^2-x}{x+\sqrt{x}}x→2limx+xx2−x Correction limx→+∞x+2xx−1=limx→+∞x(1+2xx)x(1−1x)=limx→+∞1+2xx21−1x=limx→+∞1+21x1−1x=1\begin{align*} \lim\limits_{x\to +\infty} \dfrac{x+2\sqrt{x}}{x-1} &=\lim\limits_{x\to +\infty} \dfrac{x\left(1+2\dfrac{\sqrt x}x\right)}{x\left(1-\dfrac1x\right)} \\ &=\lim\limits_{x\to +\infty} \dfrac{1+2\sqrt\dfrac{x}{x^2}}{1-\dfrac1x}\\ &=\lim\limits_{x\to +\infty} \dfrac{1+2\sqrt\dfrac{1}x}{1-\dfrac1x}\\ &=1 \end{align*}x→+∞limx−1x+2x=x→+∞limx(1−x1)x(1+2xx)=x→+∞lim1−x11+2x2x=x→+∞lim1−x11+2x1=1 car limx→+∞1x=0\lim\limits_{x\to +\infty} \dfrac1x=0x→+∞limx1=0 limx→2x2−xx+x=22−22+2=22+2=2−2\begin{align*} \lim\limits_{x\to 2} \dfrac{x^2-x}{x+\sqrt{x}} &= \dfrac{2^2-2}{2+\sqrt{2}}=\dfrac{2}{2+\sqrt{2}}\\ &=2-\sqrt2 \end{align*}x→2limx+xx2−x=2+222−2=2+22=2−2