How do i find the limit to this fraction (n=>infinity)

  • Thread starter devanlevin
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  • #1
devanlevin
[(n+5)/(n-1)]^n

i get [infinity/nifinity]^infinity and i dont see anything to change
 

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  • #2
HallsofIvy
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A little bit more accurately, as n goes to infinity (n+5)/(n+1) goes to 1 so this is of the form [itex]1^{\infty}= 1[/itex].

If this weren't in the "precalculus" section, I would recommend using L'Hopital's rule!
 
  • #3
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A little bit more accurately, as n goes to infinity (n+5)/(n+1) goes to 1 so this is of the form [itex]1^{\infty}= 1[/itex].

If this weren't in the "precalculus" section, I would recommend using L'Hopital's rule!
True enough that it's of the indeterminate form [tex]1^\infty[/tex], but it's not necessarily equal to 1. Another limit with this form is lim (1 + 1/n)^n, for n approaching [tex]\infty[/tex]. The limit here is the natural number, e.
 
  • #4
devanlevin
all true, but the limits anwer is e^6
 
  • #5
HallsofIvy
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!

True enough that it's of the indeterminate form [tex]1^\infty[/tex], but it's not necessarily equal to 1. Another limit with this form is lim (1 + 1/n)^n, for n approaching [tex]\infty[/tex]. The limit here is the natural number, e.
all true, but the limits anwer is e^6
Yes! Mark44 essentially gives that! Dividing, (n+5)/(n-1)= 1+ 6/(n-1) so ((n+5)/(n-1))n= (1+ 6/(n-1))n. Since, for n going to infinity, the difference between n and n-1 is negligible, the limit is the same as the limit of (1+ 6/n)n= (1+ (6/n))6(n/6)= (1+ 1/m)6m where m= 6/n. That is [(1+ 1/m)m]6. As Mark44 said, the limit of (1+ 1/m)m is e so the limit of [(1+ 1/m)m]6 is e6

His response was far more helpful than mine was!
 

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