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My attempt:

We prove that $$\displaystyle \lim\limits_{n \rightarrow \infty} \left(\frac{23n+2}{4n+1}\right) = \frac{23}{4} $$

It is sufficient to show that for an arbitrary real number $\epsilon\gt0$, there is a $K$ such that for all $n\gt K$, $$\left| \frac{23n+2}{4n+1} - \frac{23}{4} \right| < \epsilon. $$

Note that $$ \displaystyle\left| \frac{23n+2}{4n+1} - \frac{23}{4} \right| = \left| \frac{-15}{16n+4} \right| $$ and for $ n > 1 $ $$ \displaystyle \left| \frac{-15}{16n+4} \right| = \frac{15}{16n+4} < \frac{1}{n}. $$

Suppose $ \epsilon \in \textbf{R} $ and $ \epsilon > 0 $. Consider $ K = \displaystyle \frac{1}{\epsilon} $. Allow that $ n > K $. Then $ n > \displaystyle \frac{1}{\epsilon} $. So $ \epsilon >\displaystyle \frac{1}{n} $. Thus

$$ \displaystyle\left| \frac{23n+2}{4n+1} - \frac{23}{4} \right| = \left| \frac{-15}{16n+4} \right| = \frac{15}{16n+4} < \frac{1}{n} < \epsilon. $$ Thus $$ \displaystyle \lim\limits_{n \rightarrow \infty} \left(\frac{23n+2}{4n+1}\right) = \frac{23}{4}. $$

Is this proof correct? What are some other ways of proving this? Thanks!

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    $\begingroup$ Yes correct. You can also factor $n$ in both numerator and denominator and use the fact that $\lim_{n\to\infty }\frac 1 n=0$. $\endgroup$ – user63181 Jan 8 '14 at 15:27
  • $\begingroup$ find a minimum for n to complete the answer. $\endgroup$ – Khosrotash Jan 8 '14 at 15:27
  • $\begingroup$ Very clear and correct. I don't know if you're allowed to, but you could also use l'Hospital's rule (probably overkill and not in the spirit of the question). $\endgroup$ – rookie Jan 8 '14 at 15:29
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    $\begingroup$ It is not correct. You have to define symbols before you use them. Therefore, there is a mistake on the line $$ \displaystyle \vert \frac{23n+2}{4n+1} - \frac{23}{4} \vert < \epsilon.$$ Before that you have to write for example "Choose an arbitrary positive real number $\epsilon$". $\endgroup$ – Jaakko Seppälä Jan 8 '14 at 15:31
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Your proof is basically correct, but I would encourage you to practice a bit on articulating exactly what you mean. Where you say

It is sufficient to show that $$\left| \frac{23n+2}{4n+1} - \frac{23}{4} \right| < \epsilon $$

you mean to say something like

It is sufficient to show that for all $\epsilon\gt0$, there is a $K$ such that for all $n\gt K$, $$\left| \frac{23n+2}{4n+1} - \frac{23}{4} \right| < \epsilon $$

As is, the $\epsilon$ comes out of nowhere and there's no stated restriction on $n$, so the inequality that is "sufficient" to show could be trivially true or patently false.

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  • $\begingroup$ Thanks. I implemented your suggestion into the OP. $\endgroup$ – William Muenzinger Jan 8 '14 at 16:24
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$\displaystyle\lim_{n\to \infty} \frac{n(23 + \frac{2}{n})}{n(4+\frac{1}{n})} = \cdots$

Use the fact that $\frac{\alpha}{n}$ tends to $0$ when $n$ tends to infinity, and theorems of limits of sequences.

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Is this proof correct? What are some other ways of proving this? Thanks!

Your proof is correct with the caveat that you are a bit more precise about what $\epsilon$ and $K$ mean. Another way to prove this is using l'Hôpital's rule. Let $f(n)=23n+2$, $g(n)=4n+1$, then we can see that $$ \lim_{n\rightarrow\infty} f(n) = \lim_{n\rightarrow\infty} g(n) = \infty $$

In this case, the rule applies because you have an "indeterminant form" of $\infty/\infty$. Then the rule is that

$$ \lim_{n\rightarrow\infty}\frac{f(n)}{g(n)} = \lim_{n\rightarrow\infty}\frac{f'(n)}{g'(n)} $$

All that remains is to evaluate $f'(n)=23$, $g'(n)=4$, $$ \lim_{n\rightarrow\infty}\frac{f(n)}{g(n)} = \lim_{n\rightarrow\infty}\frac{23}{4} = \frac{23}{4} $$

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