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!
 A: You're trying to go from $\frac{a}{1+b}$ to $a(1-b)$, which is an approximation (but a good one) if $|b|\ll 1$. In this case $b=2^{\sqrt{n}}/x^n$, so the approximation works for large $n$ if $|x|>1$.
A: $$\frac{2^{\sqrt{n}}}{1+\frac{2^{\sqrt{n}}}{x^n}}=\frac{x^n2^{\sqrt{n}}}{x^n+2^{\sqrt{n}}}=2^{\sqrt{n}}\left(\frac{x^n}{x^n+2^{\sqrt{n}}}\right)=2^{\sqrt{n}}\left(\frac{x^n+2^{\sqrt{n}}}{x^n+2^{\sqrt{n}}}-\frac{2^{\sqrt{n}}}{x^n+2^{\sqrt{n}}}\right)=2^{\sqrt{n}}\left(1-\frac{2^{\sqrt{n}}}{x^n+2^{\sqrt{n}}}\right)$$
now you have to show that for your values of $x,n$:
$$x^n+2^{\sqrt{n}}\approx x^n$$
A: $\frac{2^{\sqrt n}}{1+ \frac{2^{\sqrt n}}{x^n}}=2^{\sqrt n } \left(1- \frac{2^{\sqrt n}}{x^n} \right)\iff$
$2^{\sqrt n}= 2^{\sqrt n } (1- \frac{2^{\sqrt n}}{x^n})(1+ \frac{2^{\sqrt n}}{x^n})\iff$
$(1- \frac{2^{\sqrt n}}{x^n})(1+ \frac{2^{\sqrt n}}{x^n})=1\iff$
$1- (\frac {2^{\sqrt n}}{x^n})^2 = 1\iff$
$\frac {4^{\sqrt n}}{x^{2n}} = 0$
Does $\frac {4^{\sqrt n}}{x^{2n}} \approx 0$?  Beat's me.  You seem to think so. If $|x|$ is large it does.  If $|x|>1$ and $n$ is large it does too.
But this actually depends on problem.
