Question

RIP OpenStudy ;(

Answer 1

Hi steve ;) 
 
you just have to apply simple exponent rule:
 $\frac{x^n}{x^y} =x^{n-m}$

& RIP OS ;-; :( 
#os<3

Answer 2

First note that $\frac{2^n+1}{2^{n+1}} = \frac{2^n}{2^{n+1}} + \frac{1}{2^{n+1}} = \frac{1}{2} + \frac{1}{2^{n+1}}$

If you take limit, then you have $\lim_{n \to \infty}( \frac{1}{2} + \frac{1}{2^{n+1}})= \lim_{n \to \infty}( \frac{1}{2}) +\lim_{n \to \infty}(\frac{1}{2^{n+1}})=\frac{1}{2} +0= \frac{1}{2}$