Taylor series of $\ln(1+x)$ bounded for 2nd order 
Prove that for all $x > 0 \Rightarrow $ $x - \frac{x^2}{2} < \ln(1+x) < x- \frac{x^2}{2} + \frac{x^3}{3}$  

My Attempt -
I want to show the boundaries of the Maclaurin Series of $\ln(1+x)$ up to $x^2$. As such - 
$P_n(x) = f(x) + \dfrac{f'(x)}{1!}(x)+\dfrac{f''(x)}{2!}(x)^2 + R_n(x)$
$P_n(x) = x-\dfrac{x^2}{2} + R_n(x)$
I know that $ 0<R_n(x) \le \dfrac{f^{(3)}(c)(x^3)}{3!} \,\, = \, \dfrac{x^3}{3}$ (That is i am not sure how to justify instead of $f$ being differentiable and continuous, and is it greater or greater than or equal ?).
Therefore we can conclude that due to $x>0 $ and $R_n(x) $ boundaries$ \,\,\Rightarrow $ $$x - \frac{x^2}{2} <\ln(1+x) < x- \frac{x^2}{2} + \frac{x^3}{3}.$$
Thank you very much in advance, would love to hear you thoughts about this.

Let $f (x)=\ln (1+x) $.
$$f'(x)=\frac {1}{1+x} $$
$$f''(x)=\frac {-1}{(1+x)^2} $$
$$f^{(3)}(c)=\frac {2}{(1+c)^3} $$
we know that $0 <c <x $, so
$$0<f^{(3)}(c)<2$$
and $$0<R_n (x)<\frac {2x^3}{6} $$
Done.
