Respuesta :
First note that [tex] \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}}[/tex]
If you take limit, then you have [tex] \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} [/tex]
If you take limit, then you have [tex] \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} [/tex]
Hi steve ;)
you just have to apply simple exponent rule:
[tex] \frac{x^n}{x^y} =x^{n-m} [/tex]
& RIP OS ;-; :(
#os<3
you just have to apply simple exponent rule:
[tex] \frac{x^n}{x^y} =x^{n-m} [/tex]
& RIP OS ;-; :(
#os<3