تمارين - 2BACSEF

التانية باكالوريا العلوم التجريبية – خيار فرنسي

درس : Fonction logarithme népérien

Exercice 3

Calculer les limites suivantes :

  • lim⁡x→+∞ln⁡(x)+1x\lim\limits_{x\to+\infty} \dfrac{\ln(x)+1}{x}
  • lim⁡x→0+2x+ln⁡(x)\lim\limits_{x\to0^+} 2x+\ln(x)
  • lim⁡x→+∞ln⁡(x)−x\lim\limits_{x\to+\infty} \ln(x)-x
  • lim⁡x→+∞ln⁡(2x2+x+3)x\lim\limits_{x\to+\infty} \dfrac{\ln(2x^2+x+3)}{x}
lim⁡x→+∞ln⁡(x)+1x=lim⁡x→+∞ln⁡(x)x+1x=0\displaystyle \lim\limits_{x\to+\infty} \dfrac{\ln(x)+1}{x} = \lim\limits_{x\to+\infty} \dfrac{\ln(x)}{x} + \dfrac{1}{x} = 0

car lim⁡x→+∞ln⁡(x)x=0\displaystyle \lim\limits_{x\to+\infty} \dfrac{\ln(x)}{x} = 0 et lim⁡x→+∞1x=0\displaystyle \lim\limits_{x\to+\infty} \dfrac{1}{x} = 0

lim⁡x→0+2x+ln⁡(x)=−∞\displaystyle \lim\limits_{x\to0^+} 2x + \ln(x) = -\infty

car lim⁡x→0+ln⁡(x)=−∞\displaystyle \lim\limits_{x\to0^+} \ln(x) = -\infty

lim⁡x→+∞ln⁡(x)−x=lim⁡x→+∞x(ln⁡(x)x−1)=−∞\begin{aligned} \displaystyle \lim\limits_{x\to+\infty} \ln(x) - x &= \lim\limits_{x\to+\infty} x\left( \dfrac{\ln(x)}{x} - 1 \right) \\&= -\infty \end{aligned}

car lim⁡x→+∞ln⁡(x)x=0\displaystyle \lim\limits_{x\to+\infty} \dfrac{\ln(x)}{x} = 0

lim⁡x→+∞ln⁡(2x2+x+3)x=lim⁡x→+∞ln⁡(x2(2+1x+3x2))x=lim⁡x→+∞ln⁡(x2)+ln⁡(2+1x+3x2)x=lim⁡x→+∞2ln⁡(x)x+ln⁡(2+1x+3x2)x=0\begin{aligned} &\lim\limits_{x\to+\infty} \dfrac{\ln(2x^2+x+3)}{x}\\ &= \lim\limits_{x\to+\infty} \dfrac{\ln\left(x^2\left(2+\dfrac{1}{x}+\dfrac{3}{x^2}\right)\right)}{x} \\ &= \lim\limits_{x\to+\infty} \dfrac{\ln(x^2) + \ln\left(2+\dfrac{1}{x}+\dfrac{3}{x^2}\right)}{x} \\ &= \lim\limits_{x\to+\infty} \dfrac{2\ln(x)}{x} + \dfrac{\ln\left(2+\dfrac{1}{x}+\dfrac{3}{x^2}\right)}{x} \\ &= 0 \end{aligned}

car lim⁡x→+∞ln⁡(x)x=0\displaystyle \lim\limits_{x\to+\infty} \dfrac{\ln(x)}{x} = 0
et lim⁡x→+∞1x=0\displaystyle \lim\limits_{x\to+\infty} \dfrac{1}{x} = 0