1/ $\quad \lim\limits_{x\to+\infty}\dfrac{\ln(x+3)}{x^2+x+1}$

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

car :

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

2/ $\quad \lim\limits_{x\to+\infty}(\ln(x))^2-x$

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

car :

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

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

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

car :

$$\lim\limits_{x\to0^+}\sqrt x\ln(\sqrt x)= \lim\limits_{x\to0^+}t\ln(t)=0$$