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 $N<n$ $\implies|s_n-L|.$

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



Assume $n>2$, then


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

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

I don't know much about convergence of sequence, but, this limit can be solved in the usual process.


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 $N(\epsilon)<n$, we have $$n>\frac1\epsilon+6\implies n-6>\frac1\epsilon\implies \frac1{n-6}<\epsilon$$ Thus, for $N(\epsilon)<n$, we have $$\left|\frac{n+6}{n^2-6}-0\right|<\epsilon$$

  • $\begingroup$ 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 $n<\frac{14}{\epsilon}$ ? $\endgroup$
    – whitenoise
    Feb 5 '20 at 4:34
  • $\begingroup$ @whitenoise Yes, you could do that too. $\endgroup$ Feb 5 '20 at 4:49

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