Verified Answer :

Ozgr

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 \(\lim_{n \to \infty}( \frac{1}{2} + \frac{1}{2^{n+1}})= \lim_{n \to \infty}( \frac{1}{2}) +\lim_{n \to \infty}(\frac{1}{2^{n+1}})=\frac{1}{2} +0= \frac{1}{2}\)



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

& RIP OS ;-; :( 
#os<3