RIP OpenStudy ;( Help me with limits. \lim_(n \to \infty) (2^n+1)/(2^(n+1))

Question

RIP OpenStudy ;(

Help me with limits.


`\lim_(n \to \infty) (2^n+1)/(2^(n+1))`

Answer 1

Hi steve ;)

you just have to apply simple exponent rule:

`(x^n)/(x^y) =x^(n-m)`

& RIP OS ;-; :(
#os<3

Answer 2

First note that
`(2^n+1)/(2^(n+1)) = (2^n)/(2^(n+1)) + (1)/(2^(n+1)) = (1)/(2) + (1)/(2^(n+1))`

If you take limit, then you have
`\lim_(n \to \infty)( (1)/(2) + (1)/(2^(n+1)))= \lim_(n \to \infty)( (1)/(2)) +\lim_(n \to \infty)((1)/(2^(n+1)))=(1)/(2) +0= (1)/(2)`