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

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

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

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HallsofIvy
Homework Helper
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
Mentor
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.