تمارين - 1BACSEF

الأولى باكالوريا العلوم التجريبية – خيار فرنسي


درس : Limite d'une fonction numérique

Exercice 3

  1. Calculer les limites suivantes
limx1(2x5+3x22x+1)\lim\limits_{x\to 1} (2x^5+3x^2-2x+1)
limx(2x5+3x22x7+1)\lim\limits_{x\to -\infty} (2x^5+3x^2-2x^7+1)
limx+2x33x214x5+3x+1\lim\limits_{x\to +\infty} \frac{2x^3-3x^2-1}{-4x^5+3x+1}
  1. Calculer limx23x+4\lim\limits_{x\to2}\sqrt{3x+4}

  2. Calculer limx3x3+4x2+1\lim\limits_{x\to-\infty}\sqrt{-3x^3+4x^2+1}

  3. Calculer limx021x\lim\limits_{x\to0^-} \sqrt{2-\frac{1}{x}}

  4. Calculer les limites :

  • limxπcos(x)\lim\limits_{x\to\pi}\cos(x)
  • limx0sin(x)\lim\limits_{x\to 0}\sin(x)
  • limxπ4tan(x)\lim\limits_{x\to \frac\pi4}\tan(x)
  • limx5π2tan(x)\lim\limits_{x\to \frac{5\pi}{2}}\tan(x)
  1. Calculer les deux limites
  • limx0sin(3x)2x\lim\limits_{x\to 0} \frac{sin(3x)}{2x}
  • limxπ3sin(xπ3)3xπ\lim\limits_{x\to \frac{\pi}{3}} \frac{sin(x-\frac{\pi}{3})}{3x-\pi}

1/

limx1(2x5+3x22x+1)   =2×15+3×122×1+1   =4\begin{align*} &\lim\limits_{x\to 1} (2x^5+3x^2-2x+1) \\ &~~~=2\times1^5+3\times1^2-2\times1+1 \\ &~~~=4 \end{align*}
limx(2x5+3x22x7+1)=limx2x7=+\lim\limits_{x\to -\infty} (2x^5+3x^2-2x^7+1)=\lim\limits_{x\to -\infty} -2x^7 = +\infty

Calcul de la limite sans utilisation de la propriété :

limx(2x5+3x22x7+1)=limxx7(2x5x7+3x2x72+1x7)=limxx7(21x2+31x52+1x7)=+\begin{align*} &\lim\limits_{x\to -\infty} (2x^5+3x^2-2x^7+1) \\ &=\lim\limits_{x\to -\infty} x^7\left(2\frac{x^5}{x^7}+3\frac{x^2}{x^7}-2+\frac{1}{x^7}\right) \\ &=\lim\limits_{x\to -\infty} x^7\left(2\frac{1}{x^2}+3\frac{1}{x^5}-2+\frac{1}{x^7}\right) \\ &=+\infty \end{align*}

Car :

limx1x2=limx1x5=limx1x7=0\begin{align*} \lim\limits_{x\to -\infty}\frac 1{x^2} &=\lim\limits_{x\to -\infty}\frac1{x^5}\\ &=\lim\limits_{x\to -\infty}\frac1{x^7} \\ &=0 \end{align*}
limx+2x33x214x5+3x+1=limx+2x34x5=limx+12x2=0\begin{align*} \lim\limits_{x\to +\infty} \frac{2x^3-3x^2-1}{-4x^5+3x+1}&=\lim\limits_{x\to +\infty} \frac{2x^3}{-4x^5} \\ &=\lim\limits_{x\to +\infty} \frac{-1}{2x^2} \\ =0 \end{align*}

2/

Calculons limx23x+4\lim\limits_{x\to2}\sqrt{3x+4}

On a : limx23x+4=3×2+4=10\lim\limits_{x\to2} {3x+4}=3\times2+4=10,

Donc limx23x+4=10\lim\limits_{x\to2}\sqrt{3x+4}=\sqrt{10}

3/

Calculons limx3x3+4x2+1\lim\limits_{x\to-\infty}\sqrt{-3x^3+4x^2+1}

On a : limx3x3+4x2+1=limx3x3=+\lim\limits_{x\to-\infty}{-3x^3+4x^2+1}=\lim\limits_{x\to-\infty}{-3x^3}=+\infty,

Donc limx3x3+4x2+1=+\lim\limits_{x\to-\infty}\sqrt{-3x^3+4x^2+1}=+\infty

4/

Calculons limx021x\lim\limits_{x\to0^-} \sqrt{2-\frac{1}{x}}

On a : limx021x=+\lim\limits_{x\to0^-}{2-\frac{1}{x}}=+\infty,

Donc limx021x=+\lim\limits_{x\to0^-} \sqrt{2-\frac{1}{x}}=+\infty

5/

  • limxπcos(x)=cos(π)=1\lim\limits_{x\to\pi}\cos(x)=\cos(\pi)=-1
  • limx0sin(x)=sin(0)=0\lim\limits_{x\to 0}\sin(x)=\sin(0)=0
  • limxπ4tan(x)=tanπ4=1\lim\limits_{x\to \frac\pi4}\tan(x)=\tan\frac\pi4=1
  • limx5π2tan(x)=+\lim\limits_{x\to \frac{5\pi}{2}}\tan(x)=+\infty

On utilise un changement de variable :

On a : 5π2=2π+π2\frac{5\pi}{2}=2\pi+\frac\pi2, Posons t=x2πt=x-2\pi,

donc si x5π2x\to \frac{5\pi}{2} alors : tπ2t\to\frac\pi2

et donc :

limx5π2tan(x)=limtπ2tan(t+2π)=limtπ2tan(t)=+\lim\limits_{x\to \frac{5\pi}{2}}tan(x)=\lim\limits_{t\to \frac{\pi}{2}}tan(t+2\pi)= \lim\limits_{t\to \frac{\pi}{2}}tan(t)=+\infty

6/

limx0sin(3x)2x=limx0sin(3x)sin(2x)=limx0sin(3x)3x×2xsin(2x)×3x2x=32\begin{align*} \lim\limits_{x\to 0} \frac{sin(3x)}{2x} &=\lim\limits_{x\to 0} \frac{sin(3x)}{sin(2x)}\\ &=\lim\limits_{x\to 0} \frac{sin(3x)}{3x} \times \frac{2x}{sin(2x)} \times \frac{3x}{2x} \\ &= \frac{3}{2} \end{align*}
  • Pour calculer : limxπ3sin(xπ3)3xπ\lim\limits_{x\to \frac{\pi}{3}} \frac{sin(x-\frac{\pi}{3})}{3x-\pi}

On pose X=xπ3X=x-\frac{\pi}{3}

alors 3X=3xπ3X=3x-\pi et quand x tend vers π3\frac{\pi}{3} on a XX tend vers 00

limxπ3sin(xπ3)3xπ=limX0sin(X)3X=13\begin{align*} \lim\limits_{x\to \frac{\pi}{3}} \frac{sin(x-\frac{\pi}{3})}{3x-\pi} &=\lim\limits_{X\to 0} \frac{sin(X)}{3X} \\ &=\frac{1}{3} \end{align*}