Attempt at sequence proof $\frac{n+3}{n^2 -3}$ converges to $0$

Question

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?

Answer 1

Yup, the proof is correct.

Also, "we write $n-3>\color{blue}{\frac1{\epsilon}}$"

Answer 2

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 $\dfrac1\infty$.