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I'm still working on $\epsilon-N$ proof. If you don't mind is it possible for us to give restriction on the value of $N$ as illustrated by this example:

Say after some manipulation of the limit definition I've reached this:

$\frac{1}{9n-6} < \epsilon$

Then I noticed, if $n > 1$, it is the case that $\frac{1}{3n} > \frac{1}{9n-6}$. Hence I can simplify the process of choosing $N$ by simply setting. $\frac{1}{3\epsilon} < n$. But again, there's the first restriction that $n > 1$. So given any positive $\epsilon$, our $N$ has to be at least greater than 1.

Is this additional restriction fine? I'm rather sure that it should be fine because the definition doesn't say anything about that, so long that we can still find an appropriate $N$ for each $\epsilon$.

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  • $\begingroup$ You are correct. You only have to produce an $N$, so it is fine if you need, e.g., $N > \frac{1}{3\epsilon} $ and $N > 1$. If you want to use symbols you could write $N > \max(1, \frac{1}{3\epsilon})$. $\endgroup$ Jul 30, 2014 at 12:30
  • $\begingroup$ Ok thanks guys. $\endgroup$
    – vTx
    Jul 30, 2014 at 13:02

1 Answer 1

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It is common to accumulate conditions on $N$ as your proof proceeds. You are perfectly fine if, when you write down your final proof, you can state an exact value for $N$ and proceed deductively. In what you wrote, you might begin with, "Let $\epsilon > 0$, and choose $N > \max\{1, \frac1{3\epsilon}\}$."

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