Suppose we have the following integral: $$\int_{0}^{x_{\text{max}}}\frac{x}{1+e^x}dx,$$ where $x\equiv \frac{\Delta E}{k_BT}$. We are supposed to argue that it is justified to consider $x_{\text{max}}\rightarrow \infty$ for $T\approx 1 \ \text{K}$.
Please note that we are supposed to do this in Physics, so I don't think we need a strict proof. My idea was so far was to substitute x and put in $T \approx 1\ \text{K}$, but I obtain this integral: $$\int_{0}^{E_{max}}\frac{\Delta E}{k_B^2}\frac{1}{1+\exp\{\frac{\Delta E}{k_B}\}}d\left(\Delta E\right),$$ but then the question remains why I can take $E_{max}\to \infty$?
We write $g(x)=\tfrac{x}{e^x+1}$. The approximation you are talking about is valid as long as $x_{max}\gg1$. Since the integral converges in the limit of $x_{max}\rightarrow \infty$ we can then write
$$ I=\int_0^{\infty}dxg(x)-\int_{x_{max}}^{\infty}dxg(x)\sim C-\int_{x_{max}}^{\infty}dx(xe^{-x}+O(xe^{-2x})) $$
where $C$ is some constant. So $$ I\sim C-(x_{max}+1)e^{-x_{max}}+O(x^2_{max}e^{-2x_{max}}) $$ which means that all terms expect of the first are asymtotically negligble
and we have
$$ I\sim C+O(x_{max}e^{-x_{max}}) $$
note again that the above asymptotic simplifications are only valid if our above asssumption holds true (so that we can Taylor expand the second integral).