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Here's my proof for the fact that the sequence $\frac{n+6}{n^2-7}$ converges to $0$. Can someone verify it?

Let $\epsilon > 0$. Pick N = $max\{2, \frac{7}{\epsilon}\}$.

Then, if $n > N$, $\left|\frac{n+6}{n^2-6}\right| = \frac{n+6}{n^2-6}$, since $n > 2$.

Now, $\frac{n+6}{n^2 - 6} \leq \frac{7n}{n^2} = \frac{7}{n} \leq \epsilon$.

This completes the proof.

Ryan Stiles
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1 Answers1

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You proof is correct. (There is a minor typo, the denominator should be $n^2-7$ instead of $n^2-6$)

hrkrshnn
  • 6,287