1- a) Montrer que : $\forall t \in\left[0,+\infty\left[; \frac{1}{(2+t)^{2}} \leq \frac{1}{1+t} \leq \frac{1}{2}\left(1+\overline{(1+t)^{2}}\right)\right.\right.$ b) En déduire que : $\forall x \in\left[0,+\infty\left[; \frac{2 x}{2+x} \leq \ln (1+x) \leq \frac{1}{2}\left(\frac{x^{2}+2 x}{1+x}\right)\right.\right.$ 2- Soit $g$ la fonction numérique de la variable réelle $x$ définie sur $] 0,+\infty[$ par \[ g(x)=\frac{\ln (1+x)}{x} \] Montrer que: $\lim _{x \rightarrow 0} \frac{g(x)-1}{x}=\frac{-1}{2}$

Step 1 ：\(\forall t \in\left[0, +\infty\right[\), \text{we have to show that} \frac{1}{(2+t)^{2}} \leq \frac{1}{1+t} \leq \frac{1}{2}\left(1+\overline{(1+t)^{2}}\right)\)

Step 2 ：\(\text{Since} \frac{1}{(2+t)^{2}} \leq \frac{1}{1+t} \iff (1+t)(2+t)^2 \geq 1\)

Step 3 ：\(\text{Expanding the inequality, we get} \ t^3 + 4t^2 + 4t \geq 0\)

Step 4 ：\(\text{Since} \ t \geq 0, \text{the inequality holds true}\)

Step 5 ：\(\text{Now, we have to show that} \frac{1}{1+t} \leq \frac{1}{2}\left(1+\overline{(1+t)^{2}}\right) \iff 2(1+t) \leq 1 + (1+t)^2\)

Step 6 ：\(\text{Expanding the inequality, we get} \ t^2 - t \geq 0\)

Step 7 ：\(\text{Since} \ t(t-1) \geq 0, \text{the inequality holds true}\)

Step 8 ：\(\text{Now, we have to show that} \frac{2 x}{2+x} \leq \ln (1+x) \leq \frac{1}{2}\left(\frac{x^{2}+2 x}{1+x}\right)\)

Step 9 ：\(\text{Using the inequalities we proved above, we get} \int_{0}^{x} \frac{2}{2+t} dt \leq \int_{0}^{x} \frac{1}{1+t} dt \leq \int_{0}^{x} \frac{1}{2}(1+\overline{(1+t)^{2}}) dt\)

Step 10 ：\(\text{Integrating, we get} \frac{2 x}{2+x} \leq \ln (1+x) \leq \frac{1}{2}\left(\frac{x^{2}+2 x}{1+x}\right)\)

Step 11 ：\(\text{Now, let's find the limit} \lim _{x \rightarrow 0} \frac{g(x)-1}{x} = \lim _{x \rightarrow 0} \frac{\frac{\ln (1+x)}{x} - 1}{x}\)

Step 12 ：\(\text{Using L'Hopital's rule, we get} \lim _{x \rightarrow 0} \frac{\frac{1}{1+x} - \frac{\ln (1+x)}{x^2}}{1}\)

Step 13 ：\(\text{Taking the limit as} \ x \rightarrow 0, \text{we get} \frac{-1}{2}\)

Step 14 ：\(\boxed{\lim _{x \rightarrow 0} \frac{g(x)-1}{x} = \frac{-1}{2}}\)