Mathematics

Question

RIP OpenStudy ;(

Help me with limits.

[tex]\lim_{n \to \infty} \frac{2^n+1}{2^{n+1}} [/tex]

2 Answer

  • 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]



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

    & RIP OS ;-; :( 
    #os<3
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