<s>[INST] Attempt at sequence proof $\frac{n+3}{n^2 -3}$ converges to $0$ 
Prove convergence of the following sequence: $$\frac{n+3}{n^2 -3} \rightarrow 0$$

Proof discussion:
Notice that since whenever $n>3$,  we have $n^2 -3 >0$, we also know that $n+3 >0$, so $\frac{n+3}{n^2 -3}>0$. This means we can drop the absolute value signs in: 
$$  \left|\frac{n+3}{n^2 -3}-0\right|=\frac{n+3}{n^2 -3} $$
We now notice that for $n>3$ also $n^2 -9>0$ and $n^2 -3 > n^2 -9$ so $\frac{1}{n^2 -3}< \frac{1}{n^2 -9}$ we can thus write:
$$\frac{n+3}{n^2 -3}<\frac{n+3}{n^2 -9}=\frac{(n+3)}{(n+3)(n-3)}=\frac{1}{n-3} $$ 
To be able to complete this proof we want that $\frac{1}{n-3}<\epsilon$, we write $n-3>\frac{1}{\epsilon}$ or $n> \frac{1}{\epsilon} +3$. If we pick $n_0 =\lceil\frac{1}{\epsilon} +3\rceil$, it will also be automatically larger than $3$. We can now write our proof:
Proof:
For all $\epsilon>0$, we let $n_0=\lceil{\frac{1}{\epsilon}+3 }\rceil$ then for all $n>n_0$, we know that:
$$|a_n-0|=\left|\frac{n+3}{n^2-3} \right|<\frac{n+3}{n^2-9}=\frac{1}{n-3}< \frac{1}{\frac{1}{\epsilon}+3-3}=\epsilon$$
And hence our sequence converges to $0$ $\square$.
Is my proof okay?
 [/INST] Alternatively, decompose
$$\frac{n+3}{n^2-3}=\frac1{n-\sqrt3}+(3-\sqrt3)\frac1{n^2-3}$$
and the two terms are of the form $\dfrac1\infty$.
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