Question

Use the formal definition of the limit of a sequence to prove the following limits. $$\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}=0$$

Short answer

**Question:** Prove using the formal definition of the limit of a sequence that: $$\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}=0$$ **Answer:** We can prove that $$\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}=0$$ by the following steps: 1. Write the definition of the limit of a sequence: $$\underset{n\to \mathrm{\infty }}{lim}{a}_n=L$$ if for any real number $ϵ>0$, there exists a positive integer $N$ such that for all $n>N$, we have $$|a_n-L|<ϵ$$ 2. Apply the definition to the given limit: In our case, $a_n=\frac{1}{n}$ and $L=0$. We need to find $N$ such that for all $n>N$, $$|\frac{1}{n}-0|<ϵ$$ 3. Simplify the inequality: We can rewrite the inequality as $$\frac{1}{n}<ϵ$$ 4. Find the value of $N$: By solving the inequality for $n$, we get $$n>\frac{1}{ϵ}$$. We choose the smallest integer $N$ greater than $\frac{1}{ϵ}$, that is, $$N=\lceil \frac{1}{ϵ}\rceil$$ 5. Conclude the proof: Since we've found an integer $N$ such that for all $n>N$, $\frac{1}{n}<ϵ$, we have proven that $$\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}=0$$ by the formal definition of the limit of a sequence.

Step-by-step solution

Write down the definition of the limit of a sequence

According to the formal definition, the limit of a sequence is: $$\underset{n\to \mathrm{\infty }}{lim}{a}_n=L$$ if for any real number $ϵ>0$, there exists a positive integer $N$ such that for all $n>N$, we have: $$|a_n-L|<ϵ$$

Apply the definition to the given limit

We are asked to prove that $$\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}=0$$, so our $a_n=\frac{1}{n}$ and $L=0$. We need to find $N$ such that for all $n>N$: $$|\frac{1}{n}-0|<ϵ$$

Simplify the inequality

The inequality can be simplified to: $$\left|\frac{1}{n}\right|<ϵ$$ Since $n$ is a positive integer, and $\frac{1}{n}$ is also positive, we can drop the absolute value without any issue: $$\frac{1}{n}<ϵ$$

Find the value of N

To find $N$, let's solve the inequality for $n$: $$n>\frac{1}{ϵ}$$ Since $n$ must be an integer, we can choose $N$ as the smallest integer greater than $\frac{1}{ϵ}$. That is: $$N=\lceil \frac{1}{ϵ}\rceil$$

Conclude the proof

We've found an integer $N$ such that for all $n>N$, $\frac{1}{n}<ϵ$. This means we have shown that: $$\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}=0$$ by the formal definition of the limit of a sequence.