Prove an inequality (Using Taylor expansion) 
Prove: $\frac{x}{1+x} < \ln(1+x) < x$.

I thought a good practice would be to prove it using Taylor Expansion.  
Here's my try:
$$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3}...$$
The n=1 Taylor polynomial is:
$$T_1(x) = x$$
and 
$$ ln(1+x) = T_1(x) +  R_1(x)$$
Lets evaluate $R_1(x)$ by Cauchy's remainder formula:  
$$R_1(x) = \frac{f^{(2)}(\xi)}{2!}\cdot x^2 = \frac{\frac{-1}{(\xi+1)^2}}{2!}\cdot x^2 = \frac{-x^2}{2(\xi+1)^2} < 0$$
Now, it does prove the right-hand side because $x + R_1(x) < x$ ($R_1(x)$ is negative).
I'm not so sure what should I do for the left-hand side. I'd also like to get general critique for my current work.  
Thanks!
 A: We apply the mean value theorem on the function $t\mapsto \ln t$ on the interval $[1,1+x]$: there's $\zeta\in(1,1+x)$ such that
$$\ln(1+x)=\frac x\zeta$$
and notice that
$$\frac1{1+x}<\frac1\zeta<1$$
A: Using the elementary inequality 
$$
1+x\le e^x
$$
one directly obtains one side of the inequality chain. Replace $x$ by $-y$ and invert to obtain
$$
\frac1{1-y}\ge e^y
$$
and then set $1+x=\frac1{1-y}=1+\frac{y}{1-y}$ or $y=\frac{x}{1+x}$ to obtain
$$
1+x\ge e^{\frac{x}{1+x}}
$$
for the other part of the inequality chain.
A: Proof:  

Assuming you argued and justified that $ f (x) = Ln (1 + x) $ is
  continuous and derivable $ \forall x> 0 $.

This is equivalent to writing the interval $ (0, x) $.
By the Mean Value theorem $\exists c \in (0,x)$ such that $f'(c)=\frac{f(b)-f(a)}{b-a}$, i.e.
 $\frac{1}{1+c}=\frac{Ln(1+x)-Ln(1)}{x-0}=\frac{Ln(1+x)}{x}$.
We can delimit  $\frac{1}{1+c}=f'(c)$ (*), then:
$\frac{1}{1+x}<f'(c)<1$
But we know that  $f'(c)=\frac{1}{1+c}=\frac{Ln(1+x)}{x}$ (**)
Replacing in (*) we get:
$\frac{1}{1+x}<\frac{Ln(1+x)}{x}<1$
Multiplying the inequality by $x$, we obtain the desired result:
$\frac{x}{1+x}<Ln(1+x)<x$.
