limit of a sequence (Using definition)

I'm taking real-analysis and wanted to know if the way I've proved the problem is okay.

Using the definition for convergence of a sequence

The sequence $$s_n$$ converges to its limit $$L$$ if $$\forall\;\epsilon>0\;\exists$$ an $$N(\epsilon)\in\Bbb{N}$$ such that $$N $$\implies|s_n-L|.$$

Prove $$\displaystyle\lim_{n\rightarrow\infty}\frac{n+6}{n^2-6}=0$$.

Scratchwork:

$$\left|\frac{n+6}{n^2-6}-0\right|<\epsilon\implies\left|\frac{n+6}{n^2-6}\right|<\epsilon$$

Assume $$n>2$$, then

$$\frac{n+6}{n^2-6}<\frac{n}{n^2}=\frac{1}{n}<\epsilon$$

thus $$\frac{1}{n}<\epsilon\implies n>\frac{1}{\epsilon}$$

$$\therefore \forall\;\epsilon>0,$$ there is an $$N(\epsilon)=\frac{1}{\epsilon}$$ such that $$N. QED

• Your step $\frac{n+6}{n^2-6}<\frac1n$ is false. E.g. $\frac{3+6}{3^2-6}=\frac93=3>\frac13$ Feb 5 '20 at 4:08

$$\frac{n+6}{n^2-6}=\frac{\frac{6}{n^2}+\frac{1}{n}}{1-\frac{6}{n^2}}$$
Then easily when $$n→∞$$, the limit value will approach to $$0/1 = 0$$
Has I said in a comment $$\frac{n+6}{n^2-6}<\frac1n$$ is false. You could try with $$n=3$$ to see it.
The get a bigger fraction we can either increase the numerator or decrease the denominator. Here we will do the later. If $$n>6$$, we have $$0<\frac{n+6}{n^2-6}<\frac{n+6}{n^2-36}=\frac1{n-6}$$ Let $$\epsilon>0$$, if we take $$N(\epsilon)=\frac1\epsilon+6$$, then For $$N(\epsilon), we have $$n>\frac1\epsilon+6\implies n-6>\frac1\epsilon\implies \frac1{n-6}<\epsilon$$ Thus, for $$N(\epsilon), we have $$\left|\frac{n+6}{n^2-6}-0\right|<\epsilon$$
• Okay, could you do $n+6 \leq 7n$ for the numerator and $n^2-6\geq\frac{n^2}{2}$ for the denominator to get $n<\frac{14}{\epsilon}$ ? Feb 5 '20 at 4:34