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Chapter 8: Problem 74

Use the formal definition of the limit of a sequence to prove the following limits. $$\lim _{n \rightarrow \infty} \frac{n}{n^{2}+1}=0$$

Short Answer

Expert verified
Question: Use the formal definition of the limit of a sequence to prove that the limit of the sequence \(\frac{n}{n^{2}+1}\) as \(n\) approaches infinity is 0. Answer: Based on the formal definition and the steps provided, we showed that there exists an N such that \(N>\frac{1}{\epsilon}\). With this choice of N, we proved that for all \(n>N\), the difference between the sequence and the limit is less than \(\epsilon\), indicating that the limit of the sequence as \(n\) approaches infinity is 0.

Step by step solution

01

Write down the formal definition of the limit

According to the formal definition of the limit of a sequence, we need to show that for any \(\epsilon>0\), there exists a positive integer N such that for all \(n>N\), we have: $$ \left|\frac{n}{n^{2}+1}-0\right|<\epsilon $$ Simplifying, we get: $$ \frac{n}{n^{2}+1}<\epsilon $$
02

Find a suitable inequality for N

We want to find a suitable inequality that helps us find a connection between N and \(\epsilon\). An inequality that can be helpful in this case is: $$ \frac{n}{n^{2}+1}<\frac{1}{n} $$ This inequality holds because the denominator of the left-hand side, \(n^{2}+1\), is always greater than \(n^{2}\), making the entire fraction smaller than \(\frac{1}{n}\).
03

Determine N in terms of 蔚

Now, we want to find N such that for all \(n>N\), we have: $$ \frac{1}{n}<\epsilon $$ For this inequality to be true, we can choose N based on the following inequality: $$ N>\frac{1}{\epsilon} $$
04

Prove the limit using the chosen N

Now that we have chosen N satisfying that \(N>\frac{1}{\epsilon}\), we want to prove the main inequality to show that the limit of the sequence is indeed 0. For all \(n>N\) (which implies \(n>\frac{1}{\epsilon}\)), we have $$ \left|\frac{n}{n^{2}+1}-0\right|=\frac{n}{n^{2}+1}<\frac{1}{n}<\epsilon $$ This shows that the limit of the sequence as \(n\) approaches infinity is indeed 0, as required.

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