[Math] Verify the limit using the definition of convergence of a sequence

Question

I know that the definition is:

"A sequence $(s_n)$ is said to converge to the real number $s$ provided that for every $\epsilon > 0$ there exists a natural number $N$ such that for all $n \in \mathbb{N}$, $N \leq n$ implies that $|s_n – s | < \epsilon$".

So here is an exercise in the problem section of the chapter:

"Verify the limit using the definition of convergence of a sequence that:

$\lim_{n \to \infty} \frac{\sqrt[]{n}}{n+2} = 0$"

I don't know how to really start. I guess, given any $\epsilon$ we want to make $|\frac{\sqrt[]{n}}{n+2} – 0| < \epsilon$. Looking at:The definition of a sequence converging.

So, lets pick a value for epsilon, $\epsilon = 0.5$. Now we have to find a natural number $N$ s.t. for $n \in \mathbb{N}$, $|s_n – s| < \epsilon$. So, we have to solve for $n$ in $\frac{\sqrt[]{n}}{n+2} < \frac{1}{2}$.

Couldn't that be quite messy with other limit problems? I suppose we could use the fact if $a < b$ and $b < c$, then $a < c$, but i'm not sure how exactly to find something larger than $\frac{\sqrt[]{n}}{n+2}$ .

Answer 1

HINT:

$$\frac{\sqrt{n}}{n+2}<\frac{\sqrt{n}}{n}=\frac{1}{\sqrt{n}}.$$

Now compare $$\frac{1}{\sqrt{n}}\sim \varepsilon.$$