Fonction logarithme népérien

Les limites avec la fonction x↦ln⁡(x)x\mapsto \ln \left(x\right) - Exercice 2

15 min
30
Déterminer les limites suivantes :
Question 1

lim⁡x→0+ln⁡(x)+3\mathop{\lim }\limits_{x\to 0^{+} } \ln \left(x\right)+3 que l'on peut aussi écrire lim⁡x→0x>0ln⁡(x)+3\mathop{\lim }\limits_{\begin{array}{l} {x\to 0} \\ {x>0} \end{array}} \ln \left(x\right)+3

Correction
  • lim⁡x→0+ln⁡(x)=−∞\lim\limits_{x\to 0^{+} } \ln \left(x\right)=-\infty
    • lim⁡x→0+ln⁡(x)=−∞lim⁡x→0+3=3}\left. \begin{array}{ccc} {\lim\limits_{x\to 0^{+} }\ln \left(x\right)} & {=} & {-\infty } \\ {\lim\limits_{x\to 0^{+} } 3} & {=} & {3 } \end{array}\right\} par addition\text{\red{par addition}}
    lim⁡x→0+ln⁡(x)+3=−∞\mathop{\lim }\limits_{x\to 0^{+} } \ln \left(x\right)+3=-\infty
    Question 2

    lim⁡x→0+3ln⁡(x)−2\mathop{\lim }\limits_{x\to 0^{+} } 3\ln \left(x\right)-2

    Correction
  • lim⁡x→0+ln⁡(x)=−∞\lim\limits_{x\to 0^{+} } \ln \left(x\right)=-\infty
    • lim⁡x→0+3ln⁡(x)=−∞lim⁡x→0+−2=−2}\left. \begin{array}{ccc} {\lim\limits_{x\to 0^{+} }3\ln \left(x\right)} & {=} & {-\infty } \\ {\lim\limits_{x\to 0^{+} } -2} & {=} & {-2 } \end{array}\right\} par addition\text{\red{par addition}}
    lim⁡x→0+3ln⁡(x)−2=−∞\mathop{\lim }\limits_{x\to 0^{+} } 3\ln \left(x\right)-2=-\infty
    Question 3

    lim⁡x→0+5ln⁡(x)+3x\mathop{\lim }\limits_{x\to 0^{+} } 5\ln \left(x\right)+3x

    Correction
  • lim⁡x→0+ln⁡(x)=−∞\lim\limits_{x\to 0^{+} } \ln \left(x\right)=-\infty
    • lim⁡x→0+5ln⁡(x)=−∞lim⁡x→0+3x=0}\left. \begin{array}{ccc} {\lim\limits_{x\to 0^{+} }5\ln \left(x\right)} & {=} & {-\infty } \\ {\lim\limits_{x\to 0^{+} } 3x} & {=} & {0 } \end{array}\right\} par addition\text{\red{par addition}}
    lim⁡x→0+5ln⁡(x)+3x=−∞\mathop{\lim }\limits_{x\to 0^{+} } 5\ln \left(x\right)+3x=-\infty
    Question 4

    lim⁡x→0+−2ln⁡(x)−7x+1\mathop{\lim }\limits_{x\to 0^{+} } -2\ln \left(x\right)-7x+1

    Correction
  • lim⁡x→0+ln⁡(x)=−∞\lim\limits_{x\to 0^{+} } \ln \left(x\right)=-\infty
    • 2°)  Calculons  dans  un  second  temps  la  limite  de  −2ln⁡(x)‾\underline{\color{black}2°)\;Calculons\;dans\;un\;second\;temps\;la\;limite\;de\;-2\ln(x)}
    lim⁡x→0+ln⁡(x)=−∞lim⁡x→0+−2=−2}\left. \begin{array}{ccc} {\lim\limits_{x\to 0^{+} }\ln \left(x\right)} & {=} & {-\infty } \\ {\lim\limits_{x\to 0^{+} } -2} & {=} & {-2 } \end{array}\right\} par produit\text{\red{par produit}}
    lim⁡x→0+−2ln⁡(x)=+∞\mathop{\lim }\limits_{x\to 0^{+} } -2\ln \left(x\right)=+\infty

    On  peut  donc  conclure  :‾\underline{\color{black}On\;peut\;donc\;conclure\;:}
    lim⁡x→0+−2ln⁡(x)=+∞lim⁡x→0+−7x+1=1}\left. \begin{array}{ccc} {\lim\limits_{x\to 0^{+} }-2\ln \left(x\right)} & {=} & {+\infty } \\ {\lim\limits_{x\to 0^{+} } -7x+1} & {=} & {1 } \end{array}\right\} par somme\text{\red{par somme}}
    lim⁡x→0+−2ln⁡(x)−7x+1=+∞\mathop{\lim }\limits_{x\to 0^{+} } -2\ln \left(x\right)-7x+1=+\infty
    Question 5

    lim⁡x→0+xln⁡(x)−4\mathop{\lim }\limits_{x\to 0^{+} } x\ln \left(x\right)-4

    Correction
  • lim⁡x→0+xln⁡(x)=0\lim\limits_{x\to 0^{+} } x\ln \left(x\right)=0
    • lim⁡x→0+xln⁡(x)=0lim⁡x→0+−4=−4}\left. \begin{array}{ccc} {\lim\limits_{x\to 0^{+} }x\ln \left(x\right)} & {=} & {0 } \\ {\lim\limits_{x\to 0^{+} } -4} & {=} & {-4 } \end{array}\right\} par addition\text{\red{par addition}}
    lim⁡x→0+xln⁡(x)−4=−4\mathop{\lim }\limits_{x\to 0^{+} } x\ln \left(x\right)-4=-4
    Question 6

    lim⁡x→0+xln⁡(x)+6x+8\mathop{\lim }\limits_{x\to 0^{+} } x\ln \left(x\right)+6x+8

    Correction
  • lim⁡x→0+xln⁡(x)=0\lim\limits_{x\to 0^{+} } x\ln \left(x\right)=0
    • lim⁡x→0+xln⁡(x)=0lim⁡x→0+6x+8=8}\left. \begin{array}{ccc} {\lim\limits_{x\to 0^{+} }x\ln \left(x\right)} & {=} & {0 } \\ {\lim\limits_{x\to 0^{+} } 6x+8} & {=} & {8 } \end{array}\right\} par addition\text{\red{par addition}}
    lim⁡x→0+xln⁡(x)+6x+8=8\mathop{\lim }\limits_{x\to 0^{+} } x\ln \left(x\right)+6x+8=8
    Question 7

    lim⁡x→0+ln⁡(x+1)x+2\mathop{\lim }\limits_{x\to 0^{+} } \frac{\ln \left(x+1\right)}{x} +2

    Correction
  • lim⁡x→0+ln⁡(x+1)x=1\lim\limits_{x\to 0^{+} } \frac{\ln \left(x+1\right)}{x}=1
    • lim⁡x→0+ln⁡(x+1)x=1lim⁡x→0+2=2}\left. \begin{array}{ccc} {\lim\limits_{x\to 0^{+} }\frac{\ln \left(x+1\right)}{x}} & {=} & {1 } \\ {\lim\limits_{x\to 0^{+} } 2} & {=} & {2 } \end{array}\right\} par addition\text{\red{par addition}}
    lim⁡x→0+ln⁡(x+1)x+2=3\mathop{\lim }\limits_{x\to 0^{+} } \frac{\ln \left(x+1\right)}{x} +2=3
    Question 8

    lim⁡x→0+xln⁡(x)−7x\mathop{\lim }\limits_{x\to 0^{+} } x\ln \left(x\right)-7x .

    Correction
  • lim⁡x→0+xln⁡(x)=0\lim\limits_{x\to 0^{+} } x\ln \left(x\right)=0
    • lim⁡x→0+xln⁡(x)=0lim⁡x→0+−7x=0}\left. \begin{array}{ccc} {\lim\limits_{x\to 0^{+} }x\ln \left(x\right)} & {=} & {0 } \\ {\lim\limits_{x\to 0^{+} } -7x} & {=} & {0} \end{array}\right\} par addition\text{\red{par addition}}
    lim⁡x→0+xln⁡(x)−7x=0\mathop{\lim }\limits_{x\to 0^{+} } x\ln \left(x\right)-7x=0