تمارين - 1BACSEF

الأولى باكالوريا العلوم التجريبية – خيار فرنسي

درس : Limite d'une fonction numérique

Exercice 11

Calculer les limites suivantes :

  • lim⁡x→1x2−1x−1\lim\limits_{x\to 1} \dfrac{x^2-1}{x-1}

  • lim⁡x→2x−2x2−3x+2\lim\limits_{x\to 2} \dfrac{x-2}{x^2-3x+2}

  • lim⁡x→1x−1x−1\lim\limits_{x\to 1} \dfrac{\sqrt{x}-1}{x-1}

  • lim⁡x→0x+1−1x2−x\lim\limits_{x\to 0} \dfrac{\sqrt{x+1}-1}{x^2-x}

  • lim⁡x→1x2−1x−1\lim\limits_{x \to 1} \dfrac{x^2 - 1}{x - 1}
lim⁡x→1x2−1x−1=lim⁡x→1(x−1)(x+1)x−1=lim⁡x→1(x+1)=1+1=2\begin{align*} \lim\limits_{x \to 1} \dfrac{x^2 - 1}{x - 1} &=\lim\limits_{x \to 1} \dfrac{(x - 1)(x + 1)}{x - 1}\\ &=\lim\limits_{x \to 1} (x + 1) \\&= 1 + 1 = 2 \end{align*}

Alors

lim⁡x→1x2−1x−1=2\boxed{ \lim\limits_{x \to 1} \dfrac{x^2 - 1}{x - 1}=2 }
  • lim⁡x→2x−2x2−3x+2\lim\limits_{x \to 2} \dfrac{x - 2}{x^2 - 3x + 2}

on a

x2−3x+2=a(x−x1)(x−x2)\begin{align*} x^2 - 3x + 2 &=a(x-x_1)(x-x_2)\\ \end{align*}

avec : x2−3x+2=ax2+bx+cx^2 - 3x + 2=ax^2+bx+c

et donc a=1a=1 , x1=2x_1=2

et x1.x2=ca=21=2x_1.x_2=\frac ca=\frac21=2

Donc x2=1x1=1x_2=\frac1x_1=1

x2−3x+2=(x−1)(x−2)x^2 - 3x + 2 = (x - 1)(x - 2)

Ainsi,

lim⁡x→2x−2x2−3x+2=lim⁡x→2x−2(x−1)(x−2)=lim⁡x→21x−1=12−1=1\begin{align*} \lim\limits_{x \to 2}\dfrac{x - 2}{x^2 - 3x + 2} &=\lim\limits_{x \to 2} \dfrac{x - 2}{(x - 1)(x - 2)} \\&=\lim\limits_{x \to 2}\dfrac{1}{x - 1} \\&=\frac{1}{2-1}=1 \end{align*}

alors :

lim⁡x→2x−2x2−3x+2=1\boxed{ \lim\limits_{x \to 2} \dfrac{x - 2}{x^2 - 3x + 2}=1 }
  • lim⁡x→1x−1x−1\lim\limits_{x \to 1} \dfrac{\sqrt{x} - 1}{x - 1}
x−1x−1=(x−1)(x+1)(x−1)(x+1)=x−1(x−1)(x+1)=1x+1 pour x≠1\begin{align*} \dfrac{\sqrt{x} - 1}{x - 1} &= \dfrac{(\sqrt{x} - 1)(\sqrt{x} + 1)}{(x - 1)(\sqrt{x} + 1)}\\ &= \dfrac{x - 1}{(x - 1)(\sqrt{x} + 1)} \\ &= \dfrac{1}{\sqrt{x} + 1} \text{ pour } x \ne 1 \end{align*}

On a

lim⁡x→11x+1=11+1=12\lim\limits_{x \to 1} \dfrac{1}{\sqrt{x} + 1} = \dfrac{1}{\sqrt{1} + 1} = \dfrac{1}{2}

Alors :

lim⁡x→1x−1x−1=12\boxed{ \lim\limits_{x \to 1} \dfrac{\sqrt{x} - 1}{x - 1}=\dfrac12 }
  • lim⁡x→0x+1−1x2−x\lim\limits_{x \to 0} \dfrac{\sqrt{x + 1} - 1}{x^2 - x}
x+1−1x2−x=(x+1−1)(x+1+1)(x2−x)(x+1+1) =x+1−1(x2−x)(x+1+1) =x(x2−x)(x+1+1) =xx(x−1)(x+1+1) =1(x−1)(x+1+1)\begin{align*} \dfrac{\sqrt{x + 1} - 1}{x^2 - x} &= \dfrac{(\sqrt{x + 1} - 1)(\sqrt{x + 1} + 1)}{(x^2 - x)(\sqrt{x + 1} + 1)} \\~\\ &= \dfrac{x + 1 - 1}{(x^2 - x)(\sqrt{x + 1} + 1)} \\~\\ &= \dfrac{x}{(x^2 - x)(\sqrt{x + 1} + 1)} \\~\\ &=\dfrac{x}{x(x - 1)(\sqrt{x + 1} + 1)} \\~\\ &=\dfrac{1}{(x - 1)(\sqrt{x + 1} + 1)} \end{align*}

On a :

lim⁡x→01(x−1)(x+1+1)=1(0−1)(0+1+1)=−12\lim\limits_{x \to 0} \dfrac{1}{(x - 1)(\sqrt{x + 1} + 1)} = \dfrac{1}{(0 - 1)(\sqrt{0 + 1} + 1)} = -\dfrac{1}{2}

Alors :

lim⁡x→0x+1−1x2−x=−12\boxed{ \lim\limits_{x \to 0} \dfrac{\sqrt{x + 1} - 1}{x^2 - x} =-\dfrac{1}{2} }