55–70. More sequences
Find the limit of the following se...

Question

55–70. More sequences

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

{(75n⁻¹ / 99ⁿ) + (5ⁿsinn / 8ⁿ)}

Accepted Answer

Hi everyone. Let's take a look at this practice problem. This problem says to evaluate the limit of the sequence CN, which is equal to 40, rates to the quantity of n minus 1 n quantity, divided by 64 rates to the power n, plus 2 ton multiplied by sin of n, divided by 9 rates to the power n as N approaches infinity, and we're given 4 possible choices as our answers. For choice A we have minus 1. For choice B, we have 0. for choice C we have 1, and for choice D we have infinity. So we need to determine the limit as n approaches infinity of C sub N. So this is going to be equal to the limit as n approaches infinity of the quantity of 40 rates of the quantity of n minus 1 n quantity divided by 64 rates to the power n plus 2n multiplied by sine of n divided by 9 rates to the power nn quantity. So what we're going to do is valuate this limit for each of our terms separately. So, the first term, we're going to have the limit as N approaches infinity of 40, raised to the quantity of N minus 1 in quantity, divided by 64 raised to the power N. And we're going to rewrite this as being equal to the limit, as N approaches infinity of 1 divided by 40. Multiplied by the quantity, and here we're gonna have 40 divided by 64, which is 5 divided by 8, in quantity raised to the power N. Now here we have a fraction raised to the power n, and 5 divided by 8 is actually less than 1. And recall that if you have a fraction that is less than 1, raise to the power n as in approaches infinity, that that quantity approaches 0. So that means that the limit asn approaches infinity of the quantity of 5 divided by 8 and quantity rates of the power n is going to be equal to 0. So applying this limit, We see that this is going to be equal to 1 divided by 40, multiplied by 0, which is going to be equal to 0. Now, for the second term, we'll have the limit. As N approaches infinity. Of 2 ton, multiplied by sin of N divided by 9 raised to the power N. And again, we can rewrite this as being equal to the limit as N approaches infinity. Of sign of N. Multiplied by the quantity of 2 divided by 9, and quantity raised to the power N. Now, as N approaches infinity, the function sine of n is actually bounded between -1 and 1. So -1 is less than or equal to sin of N, which is less than or equal to 1. So that means that this here is a finite number. So the behavior of this limit that we're trying to calculate is going to be controlled by the quantity of 2 divided by 9 in quantity rates to the power N. And again, we have a fraction here raised to the power N, and this fraction, which is 2 divided by 9, is less than 1. So that means that this limit is going to approach 0. So, that means that overall, that this limit here is just going to be equal to 0. So, if we look at the limit, As n approaches infinity of CN, that this is just going to be equal to 0 + 0, which is just equal to 0. So that is the answer that we were looking for for this problem. And if we compare that to our answer choices, we see that the correct answer choice corresponds to answer choice B. So I hope that this explanation has been useful, and I'll see you in the next video.

55–70. More sequences
Find the limit of the following sequences or determine that the sequence diverges.

{(75n⁻¹ / 99ⁿ) + (5ⁿsinn / 8ⁿ)}

Key Concepts

Limits of Sequences

Exponential Growth and Decay

Behavior of Oscillatory Functions in Sequences

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