📝 Exercices de la série▼
Exercice 1Exercice 2Exercice 3Exercice 4Exercice 5Exercice 6Exercice 7Exercice 8Exercice 9Exercice 10Exercice 11Exercice 12Exercice 13

Exercice 9

Calculer les limites suivantes

1/ lim⁡x→+∞3ln⁡(x)+1x\quad\lim\limits_{x\to+\infty}\dfrac{3\ln(x)+1}{x}

2/ lim⁡x→2ln⁡(x−1)x−2\quad\lim\limits_{x\to2}\dfrac{\ln(x-1)}{x-2}

3/ lim⁡x→0+ln⁡(x2+x+1)x\quad\lim\limits_{x\to0^+}\dfrac{\ln(x^2+x+1)}{x}

1/ lim⁡x→+∞3ln⁡(x)+1x=lim⁡x→+∞3ln⁡(x)x+1x=0\quad\lim\limits_{x\to+\infty}\frac{3\ln(x)+1}{x}=\lim\limits_{x\to+\infty}3\frac{\ln(x)}{x}+\frac1x=0

car {lim⁡x→+∞ln⁡xx=0 lim⁡x→+∞1x=0\Bigg\{ \begin{array}{l} \lim\limits_{x \to +\infty} \frac{\ln x}{x} = 0 \\~\\ \lim\limits_{x \to +\infty} \frac{1}{x} = 0 \end{array}

2/ lim⁡x→2ln⁡(x−1)x−2="00" F.I \quad\lim\limits_{x\to2}\dfrac{\ln(x-1)}{x-2}\quad="\frac{0}{0}"\text{ F.I }

On pose: X=x−2  ⟺  x=X+2X = x - 2 \iff x = X + 2

et donc x→2+⇒X→0+x \to 2^+ \Rightarrow X \to 0^+

Et la limite devient alors:

lim⁡x→0+ln⁡X(X+2)2−4=lim⁡x→0+ln⁡XX2+4X+4−4=lim⁡x→0+ln⁡XX(X+4)=lim⁡x→0+ln⁡XX⋅1X+4=−∞⋅14=−∞\begin{aligned} \lim\limits_{x \to 0^+} \frac{\ln X}{(X+2)^2 - 4} &= \lim\limits_{x \to 0^+} \frac{\ln X}{X^2 + 4X + 4 - 4} \\ &= \lim\limits_{x \to 0^+} \frac{\ln X}{X(X+4)} \\ &= \lim\limits_{x \to 0^+} \frac{\ln X}{X} \cdot \frac{1}{X+4} \\ &= -\infty \cdot \frac{1}{4} = -\infty \end{aligned}

3/ lim⁡x→0+ln⁡(x2+x+1)x="00" F.I \quad\lim\limits_{x \to 0^+} \dfrac{\ln(x^2 + x + 1)}{x}\quad="\frac{0}{0}"\text{ F.I }

lim⁡x→0+ln⁡(x2+x+1)x=lim⁡x→0+ln⁡[1+(x2+x)]x2+x⋅x2+xx=lim⁡x→0+ln⁡[1+(x2+x)]x2+x⋅(x+1)=1⋅1=1\begin{aligned} &\lim\limits_{x \to 0^+} \frac{\ln(x^2 + x + 1)}{x} \\ &= \lim\limits_{x \to 0^+} \frac{\ln[1 + (x^2 + x)]}{x^2 + x} \cdot \frac{x^2 + x}{x} \\ &= \lim\limits_{x \to 0^+} \frac{\ln[1 + (x^2 + x)]}{x^2 + x} \cdot (x + 1) \\ &= 1 \cdot 1 = 1 \end{aligned}

car {lim⁡t→0ln⁡(1+t)t=1 lim⁡x→0+(x+1)=1\Bigg\{ \begin{array}{l} \lim\limits_{t \to 0} \frac{\ln(1+t)}{t} = 1 \\~\\ \lim\limits_{x \to 0^+} (x+1) = 1 \end{array}