Question

Show that $\frac{x}{1+x}<\ln (1+x)<x\forall x>0$

Answer 1

Let $f\left(x\right)=\frac{x}{1+x}-ln(1+x)$

$f^{\prime }\left(x\right)=\frac{1}{1+x}+\frac{-x}{(1+x{)}^2}-\frac{1}{1+x}$

$=\frac{-x}{(1+x{)}^2}$

Thus $f^{\prime }\left(x\right)<0\forall x\in R$

Thus $f\left(x\right)$ is decreasing $f{x}^n$

$f\left(0\right)=0$

Thus $f\left(x\right)<f\left(0\right)\forall x>0$

$\Rightarrow \frac{x}{1+x}-ln(1+x)<0$

$\Rightarrow \frac{x}{1+x}<ln(1+x)\forall x>0$ ……(i)

Let $g\left(x\right)=ln(1+x)-x$

$g^{\prime }\left(x\right)=\frac{1}{1+x}-1=\frac{1-(1+x)}{1+x}=\frac{-x}{1+x}<0\forall x>0$

$g\left(x\right)=$ is decreasing $f{x}^h\forall x>0$

$\Rightarrow g\left(x\right)<g\left(0\right)\forall x>0$

$\Rightarrow ln(1+x)-x<0\forall x>0$

$\Rightarrow ln(1+x)<x$ ……..(ii)

Proved.