Evaluate Limit of Sequence: 3^n/(3^n + 2^n)

In summary, the general formula for evaluating the limit of a sequence is lim(n→∞) a_n = L, where n represents the term number and L is the limit value. A sequence is convergent if its limit exists and is a finite value, and divergent if its limit does not exist or approaches infinity. The limit of the sequence 3^n/(3^n + 2^n) is 1, determined by dividing both the numerator and denominator by the highest power of n. L'Hopital's rule cannot be used to evaluate the limit of a sequence. To prove that the limit of a sequence is equal to a specific value, it must be shown that the terms get arbitrarily close to that value as n approaches
  • #1
GreenPrint
1,196
0
How do I evaluate

lim n->inf 3^n/(3^n + 2^n)

l hospitals rule (or however you spell his last name lol) doesn't work and so I have a hard time proving that it equals one. Thanks for any help anyone can provide.
 
Physics news on Phys.org
  • #2
Divide numerator and denominator by 3^n to begin with. Then tell me what you think.
 
  • #3
ah thanks it's been a while <_<
 

Related to Evaluate Limit of Sequence: 3^n/(3^n + 2^n)

1. What is the general formula for evaluating the limit of a sequence?

The general formula for evaluating the limit of a sequence is:
lim(n→∞) a_n = L, where n represents the term number and L is the limit value.

2. How do you determine if a sequence is convergent or divergent?

A sequence is said to be convergent if its limit exists and is a finite value. This means that as the term number approaches infinity, the terms of the sequence get closer and closer to a specific value. A sequence is divergent if its limit does not exist or if it approaches infinity or negative infinity.

3. What is the limit of the sequence 3^n/(3^n + 2^n)?

The limit of the sequence 3^n/(3^n + 2^n) is 1. This can be determined by dividing both the numerator and denominator by the highest power of n, in this case 3^n. This results in 3^n/3^n + 2^n/3^n. As n approaches infinity, 3^n/3^n becomes equal to 1 and 2^n/3^n approaches 0. Therefore, the limit is 1.

4. Can you use L'Hopital's rule to evaluate the limit of this sequence?

No, L'Hopital's rule is only applicable for evaluating the limit of a function, not a sequence. The limit of a sequence is evaluated by finding the behavior of the terms as n approaches infinity, not by taking the derivative of the function.

5. How can you prove that the limit of a sequence is equal to a specific value?

To prove that the limit of a sequence is equal to a specific value L, you must show that for any value ε > 0, there exists an N such that for all n > N, |a_n - L| < ε. This means that the terms of the sequence get arbitrarily close to L as n approaches infinity.

Similar threads

  • Calculus and Beyond Homework Help
Replies
1
Views
438
  • Calculus and Beyond Homework Help
Replies
8
Views
912
  • Calculus and Beyond Homework Help
Replies
17
Views
767
  • Calculus and Beyond Homework Help
Replies
9
Views
983
  • Calculus and Beyond Homework Help
Replies
2
Views
272
  • Calculus and Beyond Homework Help
Replies
13
Views
2K
  • Calculus and Beyond Homework Help
Replies
4
Views
952
  • Calculus and Beyond Homework Help
Replies
17
Views
1K
  • Calculus and Beyond Homework Help
Replies
11
Views
2K
  • Calculus and Beyond Homework Help
Replies
9
Views
2K
Back
Top