1/ $\quad\lim\limits_{x \to 0^+} \left(\dfrac{1}x- \ln x \right)=+\infty$

car $\Bigg\{ \begin{array}{l} \lim\limits_{x \to 0^+} \dfrac{1}{x} = +\infty \\~\\ \lim\limits_{x \to 0^+} \ln x = -\infty \end{array}$

2/ $\quad \lim\limits_{x \to 0} \frac{\ln(1+x) - \ln(1-x)}{x}=\quad "\dfrac{0}{0}" ~~~~\text{F.I}$

$$\begin{aligned} \bullet\quad&\lim\limits_{x \to 0} \frac{\ln(1+x) - \ln(1-x)}{x} \\ &= \lim\limits_{x \to 0} \frac{\ln(1+x)}{x} + \frac{\ln(1-x)}{-x} \\ &= 1 + 1 = 2 \end{aligned}$$

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

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

$$\begin{aligned} &=\lim\limits_{x\to+\infty} \dfrac{\ln\left[x^2\left(1+\dfrac2{x^2}\right)\right]}{x+2} \\~\\ &=\lim\limits_{x\to+\infty} \dfrac{\ln(x^2)+\ln\left(1+\dfrac{2}{x^2}\right)}{x+2} \\~\\ &=\lim\limits_{x\to+\infty} \dfrac{2\ln(x)}{x+2}+\dfrac{\ln\left(1+\dfrac{2}{x^2}\right)}{x+2} \\~\\ &=\lim\limits_{x\to+\infty} \dfrac{2\ln(x)}{x}.\dfrac{x}{x+2}+\dfrac{\ln\left(1+\dfrac{2}{x^2}\right)}{x+2} \\~\\ &=0\times2+0=0 \end{aligned}$$

car :

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