تمارين - 2BACSEF

التانية باكالوريا العلوم التجريبية – خيار فرنسي

درس : Fonction logarithme népérien

Exercice 11

Calculer les limites suivantes

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

2/ lim⁡x→+∞(ln⁡(x))2−x\quad\lim\limits_{x\to+\infty}(\ln(x))^2-x

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

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

="+∞+∞"    F.I=\quad "\dfrac{+\infty}{+\infty}" ~~~~\text{F.I}
lim⁡x→+∞ln⁡(x+3)x2+x+1=lim⁡x→+∞ln⁡(x+3)x+3×x+3x2+x+1=0\begin{align*} \lim\limits_{x\to+\infty}\dfrac{\ln(x+3)}{x^2+x+1}&= \lim\limits_{x\to+\infty}\dfrac{\ln(x+3)}{x+3}\times \dfrac{x+3}{x^2+x+1} \\ &=0 \end{align*}

car :

{lim⁡x→+∞ln⁡(x+3)x+3=lim⁡t→+∞ln⁡(t)t=0 lim⁡x→+∞x+3x2+x+1=lim⁡x→+∞xx2=lim⁡x→+∞1x=0\left\{ \begin{align*} &\lim\limits_{x\to+\infty}\dfrac{\ln(x+3)}{x+3}=\lim\limits_{t\to+\infty}\dfrac{\ln(t)}{t}=0 \\~\\ &\lim\limits_{x\to+\infty} \dfrac{x+3}{x^2+x+1} =\lim\limits_{x\to+\infty}\dfrac{x}{x^2} =\lim\limits_{x\to+\infty}\dfrac{1}x=0 \end{align*} \right.

2/ lim⁡x→+∞(ln⁡(x))2−x\quad \lim\limits_{x\to+\infty}(\ln(x))^2-x

="+∞−∞"    F.I=\quad "+\infty-\infty" ~~~~\text{F.I}
lim⁡x→+∞(ln⁡(x))2−x=lim⁡x→+∞x[(ln⁡(x)x)2−1]=lim⁡x→+∞x[(ln⁡(x2)x)2−1]=lim⁡x→+∞x[4(ln⁡(x)x)2−1]=−∞\begin{align*} \lim\limits_{x\to+\infty}(\ln(x))^2-x&=\lim\limits_{x\to+\infty}x\left[\left(\dfrac{\ln(x)}{\sqrt x}\right)^2-1\right] \\ &=\lim\limits_{x\to+\infty}x\left[\left(\dfrac{\ln(\sqrt x^2)}{\sqrt x}\right)^2-1\right]\\ &=\lim\limits_{x\to+\infty}x\left[4\left(\dfrac{\ln(\sqrt x)}{\sqrt x}\right)^2-1\right] \\&=-\infty \end{align*}

car :

lim⁡x→+∞ln⁡(x)x=lim⁡t→+∞ln⁡(t)t=0\begin{align*} &\lim\limits_{x\to+\infty}\dfrac{\ln(\sqrt x)}{\sqrt x}=\lim\limits_{t\to+\infty}\dfrac{\ln(t)}{t}=0 \end{align*}

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

="0×+∞"     F.I\quad =\quad"0\times+\infty"~~~~~\text{F.I}
lim⁡x→0+x(ln⁡(x))2=lim⁡x→0+(xln⁡(x2))2=lim⁡x→0+4[xln⁡(x)]2=0\begin{align*} \lim\limits_{x\to0^+}x(\ln(x))^2 &=\lim\limits_{x\to0^+}(\sqrt x\ln(\sqrt x^2))^2\\ &=\lim\limits_{x\to0^+}4\left[\sqrt x\ln(\sqrt x)\right]^2 \\&=0 \end{align*}

car :

lim⁡x→0+xln⁡(x)=lim⁡x→0+tln⁡(t)=0\lim\limits_{x\to0^+}\sqrt x\ln(\sqrt x)= \lim\limits_{x\to0^+}t\ln(t)=0