Fonction inverse

Calculer les limites en 00 - Exercice 1

8 min
20
Question 1
Calculer les limites suivantes :

limx0x>04x\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x>0} \end{array}} \frac{-4}{x}

Correction
  • limx0x>01x=+\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x>0} \end{array}} \frac{1}{x} =+\infty
  • limx0x<01x=\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x<0} \end{array}} \frac{1}{x} =-\infty
  • limx0x>04x=limx0x>04×1x\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x>0} \end{array}} \frac{-4}{x}=\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x>0} \end{array}} -4\times\frac{1}{x}
    Or limx0x>01x=+\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x>0} \end{array}} \frac{1}{x}=+\infty nous allons maintenant multiplier\red{\text{multiplier}} la limite par 4-4 qui est une valeur neˊgative\red{\text{négative}}.
    Nous allons donc changer le signe du reˊsultat\blue{\text{Nous allons donc changer le signe du résultat}}.
    Cela nous donne donc :limx0x>04×1x=\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x>0} \end{array}} -4\times\frac{1}{x}=-\infty
    Finalement :
    limx0x>04x=\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x>0} \end{array}} \frac{-4}{x}=-\infty
    Question 2

    limx0x>02x\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x>0} \end{array}} \frac{2}{x}

    Correction
  • limx0x>01x=+\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x>0} \end{array}} \frac{1}{x} =+\infty
  • limx0x<01x=\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x<0} \end{array}} \frac{1}{x} =-\infty
  • limx0x>02x=limx0x>02×1x\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x>0} \end{array}} \frac{2}{x}=\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x>0} \end{array}} 2\times\frac{1}{x}
    Or limx0x>01x=+\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x>0} \end{array}} \frac{1}{x}=+\infty nous allons maintenant multiplier\red{\text{multiplier}} la limite par 22 qui est une valeur positive\red{\text{positive}}.
    Cela nous donne donc :limx0x>02×1x=+\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x>0} \end{array}} 2\times\frac{1}{x}=+\infty
    Finalement :
    limx0x>02x=+\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x>0} \end{array}} \frac{2}{x}=+\infty
    Question 3

    limx0x<06x\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x<0} \end{array}} \frac{6}{x}

    Correction
  • limx0x>01x=+\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x>0} \end{array}} \frac{1}{x} =+\infty
  • limx0x<01x=\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x<0} \end{array}} \frac{1}{x} =-\infty
  • limx0x<06x=limx0x<06×1x\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x<0} \end{array}} \frac{6}{x}=\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x<0} \end{array}} 6\times\frac{1}{x}
    Or limx0x<01x=\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x<0} \end{array}} \frac{1}{x}=-\infty nous allons maintenant multiplier\red{\text{multiplier}} la limite par 66 qui est une valeur positive\red{\text{positive}}.
    Cela nous donne donc :limx0x<06×1x=\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x<0} \end{array}} 6\times\frac{1}{x}=-\infty
    Finalement :
    limx0x<06x=\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x<0} \end{array}} \frac{6}{x}=-\infty
    Question 4

    limx0x<07x\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x<0} \end{array}} \frac{-7}{x}

    Correction
  • limx0x>01x=+\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x>0} \end{array}} \frac{1}{x} =+\infty
  • limx0x<01x=\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x<0} \end{array}} \frac{1}{x} =-\infty
  • limx0x<07x=limx0x<07×1x\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x<0} \end{array}} \frac{-7}{x}=\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x<0} \end{array}} -7\times\frac{1}{x}
    Or limx0x<01x=\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x<0} \end{array}} \frac{1}{x}=-\infty nous allons maintenant multiplier\red{\text{multiplier}} la limite par 7-7 qui est une valeur neˊgative\red{\text{négative}}.
    Nous allons donc changer le signe du reˊsultat\blue{\text{Nous allons donc changer le signe du résultat}}.
    Cela nous donne donc :limx0x<07×1x=+\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x<0} \end{array}} -7\times\frac{1}{x}=+\infty
    Finalement :
    limx0x<07x=+\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x<0} \end{array}} \frac{-7}{x}=+\infty
    Question 5

    limx0x>09x\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x>0} \end{array}} \frac{-9}{x}

    Correction
  • limx0x>01x=+\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x>0} \end{array}} \frac{1}{x} =+\infty
  • limx0x<01x=\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x<0} \end{array}} \frac{1}{x} =-\infty
  • limx0x>09x=limx0x>09×1x\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x>0} \end{array}} \frac{-9}{x}=\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x>0} \end{array}} -9\times\frac{1}{x}
    Or limx0x>01x=+\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x>0} \end{array}} \frac{1}{x}=+\infty nous allons maintenant multiplier\red{\text{multiplier}} la limite par 9-9 qui est une valeur neˊgative\red{\text{négative}}.
    Nous allons donc changer le signe du reˊsultat\blue{\text{Nous allons donc changer le signe du résultat}}.
    Cela nous donne donc :limx0x>09×1x=\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x>0} \end{array}} -9\times\frac{1}{x}=-\infty
    Finalement :
    limx0x>09x=\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x>0} \end{array}} \frac{-9}{x}=-\infty
    Question 6

    limx0x>02x\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x>0} \end{array}} \frac{2}{x}

    Correction
  • limx0x>01x=+\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x>0} \end{array}} \frac{1}{x} =+\infty
  • limx0x<01x=\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x<0} \end{array}} \frac{1}{x} =-\infty
  • limx0x>02x=limx0x>02×1x\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x>0} \end{array}} \frac{2}{x}=\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x>0} \end{array}} 2\times\frac{1}{x}
    Or limx0x>01x=+\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x>0} \end{array}} \frac{1}{x}=+\infty nous allons maintenant multiplier\red{\text{multiplier}} la limite par 22 qui est une valeur positive\red{\text{positive}}.
    Cela nous donne donc :limx0x>02×1x=+\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x>0} \end{array}} 2\times\frac{1}{x}=+\infty
    Finalement :
    limx0x>02x=+\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x>0} \end{array}} \frac{2}{x}=+\infty
    Question 7

    limx0x<011x\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x<0} \end{array}} \frac{-11}{x}

    Correction
  • limx0x>01x=+\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x>0} \end{array}} \frac{1}{x} =+\infty
  • limx0x<01x=\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x<0} \end{array}} \frac{1}{x} =-\infty
  • limx0x<011x=limx0x<011×1x\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x<0} \end{array}} \frac{-11}{x}=\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x<0} \end{array}} -11\times\frac{1}{x}
    Or limx0x<01x=\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x<0} \end{array}} \frac{1}{x}=-\infty nous allons maintenant multiplier\red{\text{multiplier}} la limite par 11-11 qui est une valeur neˊgative\red{\text{négative}}.
    Nous allons donc changer le signe du reˊsultat\blue{\text{Nous allons donc changer le signe du résultat}}.
    Cela nous donne donc :limx0x<011×1x=+\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x<0} \end{array}} -11\times\frac{1}{x}=+\infty
    Finalement :
    limx0x<011x=+\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x<0} \end{array}} \frac{-11}{x}=+\infty
    Question 8

    limx0x<09x\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x<0} \end{array}} \frac{9}{x}

    Correction
  • limx0x>01x=+\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x>0} \end{array}} \frac{1}{x} =+\infty
  • limx0x<01x=\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x<0} \end{array}} \frac{1}{x} =-\infty
  • limx0x<09x=limx0x<09×1x\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x<0} \end{array}} \frac{9}{x}=\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x<0} \end{array}} 9\times\frac{1}{x}
    Or limx0x<01x=\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x<0} \end{array}} \frac{1}{x}=-\infty nous allons maintenant multiplier\red{\text{multiplier}} la limite par 99 qui est une valeur positive\red{\text{positive}}.
    Cela nous donne donc :limx0x<09×1x=\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x<0} \end{array}} 9\times\frac{1}{x}=-\infty
    Finalement :
    limx0x<09x=\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x<0} \end{array}} \frac{9}{x}=-\infty