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

In summary, the limit of [(n+5)/(n-1)]^n, as n goes to infinity, is equal to e^6. This is based on the fact that the limit of (1+ 1/m)^m, as m goes to infinity, is equal to e. This can be solved using L'Hopital's rule or by recognizing the limit as a form of the natural number, e.
  • #1
devanlevin
[(n+5)/(n-1)]^n

i get [infinity/nifinity]^infinity and i don't see anything to change
 
Physics news on Phys.org
  • #2
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
HallsofIvy said:
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
all true, but the limits anwer is e^6
 
  • #5
!

Mark44 said:
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.

devanlevin said:
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!
 

1. What is the definition of a limit in mathematics?

A limit is a fundamental concept in calculus that describes the behavior of a function as its input value approaches a certain point. It is denoted by the symbol "lim" and is used to determine the value that a function approaches as its input value gets closer and closer to a specified point.

2. How do I know if a limit exists for a given function?

A limit exists for a given function if the two-sided limit and the one-sided limits from both directions approach the same value. In other words, the function must have a consistent behavior as the input value gets closer to the specified point.

3. How do I find the limit of a fraction with an infinity in the denominator?

If the denominator of a fraction approaches infinity, the limit is equal to zero. This is because as the denominator gets larger and larger, the fraction becomes smaller and smaller, approaching zero as its limit.

4. Can I use algebraic manipulation to find the limit of a fraction?

Yes, algebraic manipulation can be used to find the limit of a fraction. This involves simplifying the fraction by factoring and canceling out common terms in the numerator and denominator. However, it is important to note that this method only works if the limit exists for the given function.

5. Are there any other methods for finding the limit of a fraction?

Yes, there are other methods for finding the limit of a fraction, such as using L'Hôpital's rule or the squeeze theorem. These methods are particularly useful for more complex fractions or functions where algebraic manipulation may not be possible.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
10
Views
604
  • Precalculus Mathematics Homework Help
Replies
13
Views
576
  • Precalculus Mathematics Homework Help
Replies
21
Views
840
  • Precalculus Mathematics Homework Help
Replies
3
Views
1K
  • Precalculus Mathematics Homework Help
Replies
9
Views
621
Replies
6
Views
675
  • Precalculus Mathematics Homework Help
Replies
5
Views
1K
  • Precalculus Mathematics Homework Help
Replies
14
Views
2K
  • Precalculus Mathematics Homework Help
Replies
3
Views
240
  • Precalculus Mathematics Homework Help
Replies
6
Views
775
Back
Top