Determine the infinitesimal order of the function in $x0= 0$ Problem: infinitesimal order. If you can explain me how you solve this
function: 
$$(1+3x)^x -1 -x\ln(1+3x)$$
Mine resolution: $$(1+3x)^x$$ = 1 (taylor series)
                 $$xln(2+3x)$$ = $3x^2$
so $$lim \frac {1-1-3x^2}{x^n}$$= -3  (this is not equal to 0 so the infinitesimal order is 2? Is this the correct solution?)

Note that
$$x\ln(1+3x)=x\left(3x-\frac{9}{2}x^2+9x^3+o(x^3)\right)=3x^2-\frac{9}{2}x^3+9x^4+o(x^4)$$
$$(1+3x)^x=e^{x\ln(1+3x)}=e^{3x^2-\frac{9}{2}x^3+9x^4+o(x^3)}=1+3x^2-\frac{9}{2}x^3+9x^4+\frac{(3x^2-\frac{9}{2}x^3+9x^4+o(x^4))^2}{2}+o(x^4)=1+3x^2-\frac{9}{2}x^3+9x^4+\frac{9}{2}x^4+o(x^4)$$
thus
$$(1+3x)^x -1 -x\ln(1+3x)=1+3x^2-\frac{9}{2}x^3+9x^4+\frac{9}{2}x^4-1-3x^2+\frac{9}{2}x^3-9x^4+o(x^4)=\frac{9}{2}x^4+o(x^4)$$
