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

[devanlevin]

[(n+5)/(n-1)]^n

i get [infinity/nifinity]^infinity and i don't see anything to change

 

[HallsofIvy]

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

If this weren't in the "precalculus" section, I would recommend using L'Hopital's rule!

 

[Mark44]

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

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 1^\infty, but it's not necessarily equal to 1. Another limit with this form is lim (1 + 1/n)^n, for n approaching \infty. The limit here is the natural number, e.

 

[devanlevin]

all true, but the limits anwer is e^6

 

[HallsofIvy]

!

True enough that it's of the indeterminate form 1^\infty, but it's not necessarily equal to 1. Another limit with this form is lim (1 + 1/n)^n, for n approaching \infty. 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!