A sequence is an infinite, ordered list of numbers that is oft...

Question

A sequence is an infinite, ordered list of numbers that is often defined by a function. For example, the sequence {2,4,6,8,…} is specified by the function f(n) = 2n, where n=1,2,3,….The limit of such a sequence is lim n→∞ f(n), provided the limit exists. All the limit laws for limits at infinity may be applied to limits of sequences. Find the limit of the following sequences or state that the limit does not exist.

{2,3/4,4/9,5/16,…}, which is defined by f(n) = (n+1) / n^2, for n=1,2,3,…

{2,3/4,4/9,5/16,…}, which is defined by f(n) = (n+1) / n^2, for n=1,2,3,…

Accepted Answer

Welcome back, everyone. Consider the sequence 25 divided by 6, 14 divided by 27, 3 divided by 8, and so on. Defined by the function f of n is equal to 4N plus 2 divided by 3n 2, or n values +123, and so on. What is the limit of this sequence? We're given for answer choices A says 2. B says 4 divided by 3, C says 2 divided by 3, and D says 0. So, now, let's consider the given expression. It says that F of M. is equal to 4 M + 2 divided by. Wem squared and this is the expression of the general term. If we want to find the limit of a sequence, we simply want to evaluate the limit of F of N as N approaches infinity, right? We want to identify a number towards which our terms converge. Let's go ahead and substitute. F of N so limit as n approaches infinity, and now we want to substitute 4 n + 2 divided by 3 m2. This is the limit that we want to evaluate and the result will essentially correspond to the final answer of this problem. If we apply the direct substitution, we're going to get 4 multiplied by infinity plus 2, which is infinity, and in the denominator 3 multiplied by infinity squared is also infinity. This is an indeterminate form, right? So what we're going to do is basically split our fraction into two fractions because we have addition in the numerator and in the denominator we only have one term. So based on the properties of fractions, we're going to get two limits. The first one is limits and approaches infinity of 4 N divided by 3N2. And the second one is going to be limit. As n approaches infinity of 2 divided by 3 n squad. Now, let's consider each limit. For the first one, we can cancel out the N in the numerator and in the denominator we can cancel out the square, right? So now let's rewrite the limits. We get limit as N approaches infinity of 4 divided by 3 N. For the second limit, we're going to get limit as N approaches infinity of 2 divided by 3N2. Analyzing the first limit, we have 4 divided by 3 N as N approaches infinity. So that would be 4 divided by infinity, because 3 multiplied by infinity. Simply approaches infinity, and a number divided by infinity approaches 0, so we can immediately say that the first limit is going to be equal to 0. And now for the second limit, once again, we have 2 divided by 3 and 2, that's an approach is infinity. So that's 2 divided by infinity again, which also approaches 0 because we have a number divided by an infinitely large number. And now 0 + 0 gives us the overall result of 0, meaning the limit of the sequence is equal to 0, and the final answer to this problem is option D 0. Thank you for watching.

Key Concepts

Sequences

Limits

Limit Laws

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