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

Thread starter GreenPrint
Start date Sep 19, 2011
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Limit Sequence

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

Sep 19, 2011

GreenPrint

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.

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Sep 19, 2011

Dick

Science Advisor

Homework Helper

Divide numerator and denominator by 3^n to begin with. Then tell me what you think.

Sep 19, 2011

GreenPrint

ah thanks it's been a while <_<

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

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.

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.

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.

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.

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.