1
$\begingroup$

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

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

$\endgroup$
1
  • $\begingroup$ 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
0
$\begingroup$

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

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

Then easily when $n→∞$, the limit value will approach to $0/1 = 0$

$\endgroup$
0
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.