# Prove that $\lim\limits_{n \rightarrow \infty} \left(\frac{23n+2}{4n+1}\right) = \frac{23}{4}$.

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!

• Yes correct. You can also factor $n$ in both numerator and denominator and use the fact that $\lim_{n\to\infty }\frac 1 n=0$. – user63181 Jan 8 '14 at 15:27
• find a minimum for n to complete the answer. – Khosrotash Jan 8 '14 at 15:27
• 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). – rookie Jan 8 '14 at 15:29
• 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 $\epsilon$". – Jaakko Seppälä Jan 8 '14 at 15:31

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.

• Thanks. I implemented your suggestion into the OP. – William Muenzinger Jan 8 '14 at 16:24

$\displaystyle\lim_{n\to \infty} \frac{n(23 + \frac{2}{n})}{n(4+\frac{1}{n})} = \cdots$

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

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 $f'(n)=23$, $g'(n)=4$, $$\lim_{n\rightarrow\infty}\frac{f(n)}{g(n)} = \lim_{n\rightarrow\infty}\frac{23}{4} = \frac{23}{4}$$