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. 

{0,1/2,2/3,3/4,…}, which is defined by f(n) = (n−1) / n, for n=1,2,3,…

Video transcript

Welcome back, everyone. Consider the sequence 49 divided by 2, 14 divided by 3, 19 divided by 4, 24 divided by 5. Defined by the function F of N is equal to 5 N minus 1 divided by N. 4 N equals. 123 and so on. What is the limit of this sequence? We're given for answer choices A 0, B4, C5, and D 6. If we want to identify the limit of the sequence, all that we have to do is simply consider the general form. What we have to do is evaluate the limit as x approaches positive infinity of F of X. In this case, our F of X is simply F of N, so we have to evaluate the limit as N approaches infinity of F of N, and we're going to substitute F of N into this expression. Limits n approaches infinity of 5 m minus 1 divided by n. First of all, we have to apply the right substitution. We're going to get 5 multiplied by infinity minus 1 divided by infinity, which is simply infinity divided by infinity, and this is an indeterminate form. Whenever we get this Indeterminate form and we have the rational function. Well, one of the ways to get the actual finite limit is to divide both sides of the fraction by the variable with the highest exponent, and that's N to the power of 1. So let's go ahead and do that. Limit as n approaches infinity. And now what we're going to do is simply split our fraction into two terms. Instead of dividing both sides by M, we're going to simply split this fraction into 5 M divided by M. Minus 1 divided by n and we can do that based on the properties of fractions, right, because we can always split the addition or the subtraction of two terms into two separate fractions. And now let's apply the properties of limits. We can get two limits. The first one is going to be limit as N approaches infinity of 5 because we can cancel out the n terms. So that'd be limit of 5 minus limit. As n approaches infinity of 1 divided by n, and since n approaches infinity, then 1 divided by n approaches 0. So that specific limit is going to be equal to 0 because we have a number divided by an infinitely large number and we end up with only one limit, limit of 5 as n approaches infinity, which is 5. Because our function is a constant function which is independent of m and therefore the correct answer to this problem corresponds to the answer choice C. Thank you for watching.