Prove $\lim_{n\to \infty}\frac{n}{n+1} = 1$ using epsilon delta

Question

$\lim_{n\to \infty}\frac{n}{n+1} = 1$

Prove using epsilon delta.

Answer 1

By your notation I believe you're talking about the sequence $(a_n)$ of elements of $\mathbb{R}$ defined by:

$$a_n =\frac{n}{n+1}$$

Now, limit for sequences has the following definition: "given $\varepsilon >0$ there's some $n_0 \in \mathbb{N}$ such that if $n > n_0$ then $|a_n - L|<\varepsilon$". So we want some $n_0$ such that:

$$\left|\frac{n}{n+1} - 1\right|<\varepsilon$$

Rewrite this as:

$$\left|\frac{n}{n+1} - \frac{n+1}{n+1}\right| = \left|\frac{1}{n+1}\right|$$

But $n$ is natural so that the thing inside of the module sign is already positive and so we want in truth that:

$$\frac{1}{n+1}<\varepsilon \Longrightarrow n>\frac{1-\varepsilon}{\varepsilon}$$

This part is the deduction part. Now we prove, we say: given $\varepsilon > 0$ take $n_0 =(1-\varepsilon)/\varepsilon$, then we have that for $n > n_0$:

$$\left|\frac{n}{n+1}-1\right| = \left|\frac{1}{n+1}\right| = \frac{1}{n+1}$$

But $n>n_0$ so that $1/(n+1) < 1/(n_0 + 1)$ and hence:

$$\left|\frac{n}{n+1} - 1\right| < \frac{1}{n_0 + 1} = \frac{1}{\frac{1-\varepsilon}{\varepsilon} + 1} = \varepsilon$$

The important part for you to note is that first we deduce which $n_0$ works, after that we throw this part away usually and just say: "take this $n_0$" so that we show that it really works as predicted.