13–52. Limits of sequences
Find the limit of the followi...

Question

13–52. Limits of sequences

Find the limit of the following sequences or determine that the sequence diverges.

{(1 / n)¹⁄ⁿ}

Accepted Answer

Welcome back, everyone. Evaluate the limit of the sequence given by AN equals 2 divided by N. Erase the power of 1 divided by N. SN approaches infinity or determine that the sequence diverges A 0 B 1 C E, and D diverges. For this problem, let's go ahead and write L, which is going to be a limit, as N approaches infinity of 2 divided by N raises the power of 1 divided by N. We can notice that 2 divided by n is going to approach 0 as N approaches infinity, right, because we're dividing a constant by an infinitely large number and our exponent is also approaching 01 divided by an infinitely large number. So we get 0, raise the power of 0, which is an indeterminate form. So, what we can do is simply take the natural log of L. We're going to get L and L is equal to limit as N approaches infinity. Of LN of 2 divided by N raises the power of 1 divided by N. And now we can apply the properties of logarithms. We're going to get the limit. As n approaches infinity of 1 divided by N multiplied by Ln of 2 divided by n. Or basically, we can rewrite it as limit as an approach is infinity of LN of 2 divided by N. Divided by N, right? And now what we can do from here is simply analyze the behavior of each function. In addition to that, we can expand it as limit as N approaches infinity of. In the numerator we have LN 2 minus LN of N using the properties of logarithms. And we're dividing by m. So now what can we tell? Well, basically we can split it into two limits to clearly see the outcome. We get limit as n approaches infinity of LN of T divided by N minus limit as n approaches infinity of L and of N divided by n. Now, the first limit is going to be 0. Alan of 2 is a constant and n approaches infinity where dividing a constant by an infinitely large number. Considering the second limit as n approaches infinity, we have to understand that. Polynomial functions grow faster than logarithms, right? So N grows faster. Because it is a polynomial function. And Alanovn grows slower because it is a logarithmic function. So as N approaches infinity, polynomial functions grow quicker than logarithmic functions, meaning if we have a faster growth in the denominator, then the overall limit is going to be 0. So we're going to get 0 for the first limit minus 0 for the second limit, and 0 minus 0 is 0. Notice that we have calculated LN of L, right? So if we want to solve for L, which is the actual limit, Then L is going to be raised to the I'm sorry, our limit is going to be. Ease the power of L, right? So E erase the power of L is going to be limit, as N approaches infinity. Of 2 divided by n erase the power of 1 divided by n. So, what we're going to do is simply take the value of L, which is 0. And we get er erase the power of 0, which is 1. Meaning the correct answer for this problem is option B. Thank you for watching.

13–52. Limits of sequences
Find the limit of the following sequences or determine that the sequence diverges.

{(1 / n)¹⁄ⁿ}

Key Concepts

Limits of Sequences

Properties of Exponents and Roots

Use of Logarithms in Limit Evaluation

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