Showing that $\frac{2^{\sqrt n}}{1+2^{\sqrt n}/x^n}$ is approximated by $2^{\sqrt n}\left(1- \frac{2^{\sqrt n}}{x^n} \right)$ Sorry for this inconvenient question but really I can't see how to show that $$\frac{2^{\sqrt n}}{1+ \frac{2^{\sqrt n}}{x^n}}\quad\text{is approximated by}\quad 2^{\sqrt n } \left(1- \frac{2^{\sqrt n}}{x^n} \right)$$
I know how to move from right to left but I can't achieve the inverse!

$$\frac{2^{\sqrt{n}}}{1+\frac{2^{\sqrt{n}}}{x^n}}=\frac{x^n2^{\sqrt{n}}}{x^n+2^{\sqrt{n}}}=2^{\sqrt{n}}\left(\frac{x^n}{x^n+2^{\sqrt{n}}}\right)=2^{\sqrt{n}}\left(\frac{x^n+2^{\sqrt{n}}}{x^n+2^{\sqrt{n}}}-\frac{2^{\sqrt{n}}}{x^n+2^{\sqrt{n}}}\right)=2^{\sqrt{n}}\left(1-\frac{2^{\sqrt{n}}}{x^n+2^{\sqrt{n}}}\right)$$
now you have to show that for your values of $x,n$:
$$x^n+2^{\sqrt{n}}\approx x^n$$
