Question

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

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

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: $$ \frac{n}{n^{2}+1}<\epsilon $$

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 \(\frac{1}{n}\).

Determine N in terms of ε

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: $$ N>\frac{1}{\epsilon} $$

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.