Fonction inverse

Calculer les limites en −∞-\infty ou en +∞+\infty - Exercice 1

4 min
10
Question 1
Calculer les limites suivantes :

lim⁡x→+∞2x\mathop{\lim }\limits_{x\to +\infty } \frac{2}{x}

Correction
  • lim⁡x→−∞1x=0\mathop{\lim }\limits_{x\to -\infty } \frac{1}{x} =0
  • lim⁡x→+∞1x=0\mathop{\lim }\limits_{x\to +\infty } \frac{1}{x} =0
    • lim⁡x→+∞2x=lim⁡x→+∞2×1x\mathop{\lim }\limits_{x\to +\infty } \frac{2}{x} =\mathop{\lim }\limits_{x\to +\infty } 2\times\frac{1}{x}
    Nous savons que lim⁡x→+∞1x=0\mathop{\lim }\limits_{x\to +\infty } \frac{1}{x}=0 ainsi lim⁡x→+∞2×1x=2×0\mathop{\lim }\limits_{x\to +\infty } 2\times\frac{1}{x}=2\times0
    Finalement :
    lim⁡x→+∞2×1x=0\mathop{\lim }\limits_{x\to +\infty } 2\times\frac{1}{x}=0
    ou encore
    lim⁡x→+∞2x=0\mathop{\lim }\limits_{x\to +\infty } \frac{2}{x}=0
    Question 2

    lim⁡x→−∞−5x\mathop{\lim }\limits_{x\to -\infty } \frac{-5}{x}

    Correction
  • lim⁡x→−∞1x=0\mathop{\lim }\limits_{x\to -\infty } \frac{1}{x} =0
  • lim⁡x→+∞1x=0\mathop{\lim }\limits_{x\to +\infty } \frac{1}{x} =0
    • lim⁡x→−∞−5x=lim⁡x→−∞−5×1x\mathop{\lim }\limits_{x\to -\infty } \frac{-5}{x} =\mathop{\lim }\limits_{x\to -\infty } -5\times\frac{1}{x}
    Nous savons que lim⁡x→−∞1x=0\mathop{\lim }\limits_{x\to -\infty } \frac{1}{x}=0 ainsi lim⁡x→−∞−5×1x=−5×0\mathop{\lim }\limits_{x\to -\infty } -5\times\frac{1}{x}=-5\times0
    Finalement :
    lim⁡x→−∞−5×1x=0\mathop{\lim }\limits_{x\to -\infty } -5\times\frac{1}{x}=0
    ou encore
    lim⁡x→−∞−5x=0\mathop{\lim }\limits_{x\to -\infty } \frac{-5}{x}=0
    Question 3

    lim⁡x→−∞7x\mathop{\lim }\limits_{x\to -\infty } \frac{7}{x}

    Correction
  • lim⁡x→−∞1x=0\mathop{\lim }\limits_{x\to -\infty } \frac{1}{x} =0
  • lim⁡x→+∞1x=0\mathop{\lim }\limits_{x\to +\infty } \frac{1}{x} =0
    • lim⁡x→−∞7x=lim⁡x→−∞7×1x\mathop{\lim }\limits_{x\to -\infty } \frac{7}{x} =\mathop{\lim }\limits_{x\to -\infty } 7\times\frac{1}{x}
    Nous savons que lim⁡x→−∞1x=0\mathop{\lim }\limits_{x\to -\infty } \frac{1}{x}=0 ainsi lim⁡x→−∞7×1x=7×0\mathop{\lim }\limits_{x\to -\infty } 7\times\frac{1}{x}=7\times0
    Finalement :
    lim⁡x→−∞7×1x=0\mathop{\lim }\limits_{x\to -\infty } 7\times\frac{1}{x}=0
    ou encore
    lim⁡x→−∞7x=0\mathop{\lim }\limits_{x\to -\infty } \frac{7}{x}=0
    Question 4

    lim⁡x→+∞−9x\mathop{\lim }\limits_{x\to +\infty } -\frac{9}{x}

    Correction
  • lim⁡x→−∞1x=0\mathop{\lim }\limits_{x\to -\infty } \frac{1}{x} =0
  • lim⁡x→+∞1x=0\mathop{\lim }\limits_{x\to +\infty } \frac{1}{x} =0
    • lim⁡x→+∞−9x=lim⁡x→+∞−9×1x\mathop{\lim }\limits_{x\to +\infty } -\frac{9}{x} =\mathop{\lim }\limits_{x\to +\infty } -9\times\frac{1}{x}
    Nous savons que lim⁡x→+∞1x=0\mathop{\lim }\limits_{x\to +\infty } \frac{1}{x}=0 ainsi lim⁡x→+∞−9×1x=−9×0\mathop{\lim }\limits_{x\to +\infty } -9\times\frac{1}{x}=-9\times0
    Finalement :
    lim⁡x→+∞−9×1x=0\mathop{\lim }\limits_{x\to +\infty } -9\times\frac{1}{x}=0
    ou encore
    lim⁡x→+∞−9x=0\mathop{\lim }\limits_{x\to +\infty } -\frac{9}{x}=0