How can I prove that if $\lim_{n \to \infty}s_n=s$ then $|s_n-s|< \epsilon$ is equivalent to $s-\epsilon <s_n <s+ \epsilon$

Question

My professor casually mentioned this in class and told us to prove it if we weren't convinced, however, I cannot find how to prove it.

Answer 1

From the definition of absolute values we have:

$$|s_n-s| < \varepsilon \Longleftrightarrow -\varepsilon < s_n - s < \varepsilon.$$

Adding $s$ to each piece of the inequality, we get

$$|s_n-s| < \varepsilon \Longleftrightarrow s - \varepsilon < s_n < s + \varepsilon.$$

Answer 2

If $|s_n-s|<\epsilon$, then for $s>s_n$ we see that

$$s-s_n<\epsilon\implies s_n>s-\epsilon$$

For $s<s_n$ we see that

$$s_n-s<\epsilon \implies s_n<s+\epsilon$$

Putting $(1)$ and $(2)$ together gives

$$s-\epsilon<s_n<s+\epsilon$$

Answer 3

I'm not sure what you mean, but it is true that $|s_n-s| < \varepsilon$ is equivalent to $s-\varepsilon < s_n < s + \varepsilon$.

Simply subtracting $s$ from the latter gives you $-\varepsilon < s_n-s < \varepsilon$, which is the same as saying $|s_n-s| < \varepsilon$.

Is this what you wanted? :)