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

Answer: The limit of the sequence is 0.

Step-by-step solution

Recall the formal definition of the limit of a sequence

The formal definition of the limit of a sequence says that a sequence \((a_n)\) has a limit L as n approaches infinity if, for every positive real number ε, there exists a positive integer N such that for all n greater than N, the absolute value of the difference between \(a_n\) and L is less than ε: $$\forall ε > 0, \exists N > 0 \text{ such that } \forall n > N, |a_n - L| < ε$$

Apply the formal definition to the given exercise

In this problem, we are given the sequence \(a_n = \frac{n}{n^2 + 1}\) and the limit L = 0. We need to show that for every ε > 0, there is a positive integer N with the following property: $$\forall n > N, \left|\frac{n}{n^2 + 1} - 0\right| < ε$$

Simplify the inequality

We can simplify the inequality by removing the subtraction by zero and taking the absolute value of the function: $$\frac{n}{n^2 + 1} < ε$$

Solve for N in terms of ε

In order to find the N that satisfies the inequality for a given ε, we can rewrite the inequality as follows: $$n < ε(n^2 + 1)$$ Now, we can try to bound the inequality from above: $$n < ε(n^2 + 1) < εn^2$$ Divide both sides of the inequality by ε: $$\frac{n}{ε} < n^2$$ Now, to find the N in terms of ε, we can take the square root of both sides: $$\sqrt{\frac{n}{ε}} < n$$ Let N be the smallest integer greater than \(\sqrt{\frac{n}{ε}}\). This means that for \(\forall n > N\), the inequality \(\frac{n}{n^2 + 1} < ε\) holds. Therefore, the proof is complete and the limit of the sequence is 0: $$\lim _{n \rightarrow \infty} \frac{n}{n^{2}+1}=0$$