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Does #a_n=(5^n)/(1+(6^n) #converge? If so what is the limit?

Answer 1

Ezra Adams

Yes, it converges to zero.

First of all, I want to prove that #1+6^n# and #6^n# are asymptotically equivalent. To do so, we need to show that

#lim_{n\to\infty} (1+6^n)/6^n=1#

And a way to show it is factoring #6^n#:

#lim_{n\to\infty} (1+6^n)/6^n = lim_{n\to\infty} (cancel(6^n)(1+1/6^n))/cancel(6^n)= lim_{n\to\infty} 1+1/6^n = 1#

Since the two expressions are equivalent when #n->infty#, we can claim that

#lim_{n\to\infty} 5^n/(1+6^n) =lim_{n\to\infty} 5^n / 6^n = lim_{n\to\infty}(5/6)^n#

And now use the fact that #a_n = k^n# converges to zero if and only if #|k|<1#, which is our case.

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Answer 2

Elijah Abram

To determine if the sequence ( a_n = \frac{5^n}{1 + 6^n} ) converges, we can analyze its behavior as ( n ) approaches infinity.

As ( n ) tends towards infinity, the term ( 6^n ) in the denominator becomes much larger than the term ( 5^n ) in the numerator. Therefore, the fraction ( \frac{5^n}{1 + 6^n} ) tends towards zero.

Thus, the sequence converges, and its limit as ( n ) approaches infinity is ( 0 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.