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

1. Nov 12, 2008

### devanlevin

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

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

2. Nov 12, 2008

### HallsofIvy

Staff Emeritus
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!

3. Nov 12, 2008

### Staff: Mentor

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.

4. Nov 12, 2008

### devanlevin

all true, but the limits anwer is e^6

5. Nov 12, 2008

### HallsofIvy

Staff Emeritus
!

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!