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

My attempt:

We prove that $\lim\limits_{n \rightarrow \infty } \dfrac{n+6}{n^2-6} =0$.

It is sufficient to show that for any $\epsilon \in\textbf{R}^+$, there exists an $K \in \textbf{R}$ such that for any $n > K$

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

Suppose $n>4$. Then $n+6 < 7n$ and $n^2 -6 > \frac{1}{2} n^2$. So

$\displaystyle \left| \frac{n+6}{n^2-6} - 0 \right| = \frac{n+6}{n^2-6 } < \frac{7n}{\frac{1}{2}n^2} = \frac{14}{n}$.

Consider $K = \max\{4,\displaystyle \frac{14}{\epsilon}\}$ and suppose $n> K$. Then $n > \displaystyle \frac{14}{\epsilon}$. This implies that $\epsilon > \displaystyle \frac{14}{n}$. Therefore

$\displaystyle \left| \frac{n+6}{n^2-6} - 0 \right| = \frac{n+6}{n^2-6 } < \frac{7n}{\frac{1}{2}n^2} = \frac{14}{n} < \epsilon$.

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

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

• Do you have to prove it using $\epsilon$-$\delta$? Jan 9, 2014 at 21:02
• Alternative proofs are free to use other methods, but that is the method I was using. Jan 9, 2014 at 21:07
• Ok, because it's much more straightforward to find the limit using some limit calculation rules. Jan 9, 2014 at 21:08
• Divide each summand by $n^2$ and you will see, applying the limit rules. Jan 9, 2014 at 21:12
• This is the first way it is taught in real analysis. More efficient methods involving things like lim x + y = lim x + lim y, and so forth come next. Jan 9, 2014 at 21:13

Personally (assuming I wasn't supposed to use sharper tools), I would tackle it by first noting that the fraction should be about $1/n$ when $n$ is very large. So pull out a factor $1/n$, and simplify the rest, and you get $$\frac{n+6}{n^2-6}=\frac1n\cdot\frac{1+6/n}{1-6/n^2}.$$ Now you just need to find a good bound for the fracton on the right, when $n$ is large. To make life really simple, you could require $n\ge6$, which implies $1+6/n\le2$ and $1-6/n^2\ge5/6$, so that $$\frac{n+6}{n^2-6}\le\frac{12}5\frac1n\qquad\text{for n\ge6}.$$ The rest is now trivial.
It is sufficient to use the fact that $\displaystyle \frac{n+6}{n^2-6}\sim_{\infty}\frac{n}{n^2}=\frac{1}{n}$ then you have the result.
Your proof is correct. Just to be pedantic, you can define $K$ directly as $14/\varepsilon$, since you precedently assumed that $n>4$.
As you've already noted in the comments, the actual way to solve limits without going through all those $\delta$-$\epsilon$ arguments is to use limits properties, such as the preservation of operations $+,\cdot$ (and, with some attention, $\div$). With these little theorems the proof is very easy: $$\{\dfrac{n+6}{n^2-6}\}_{n\to \infty}=\{\dfrac{1+6/n}{n-6/n}\}_{n\to \infty}=\dfrac{ \lim 1 + \lim 6/n}{\lim n -\lim6/n}=0$$ since all but the left bottom corner terms are bounded.
$$\frac{n+6}{n^{2}-6}=r_{n}\times\frac{1+6r_{n}}{1-6r_{n}}$$ for: $$r_{n}=\frac{1}{n}$$
If $n\rightarrow\infty$ then $r_{n}\rightarrow0$ (easy) and consequently: $$\frac{n+6}{n^{2}-6}\rightarrow0\times\frac{1+0}{1-0}=0$$