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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:
$$ left|frac{n+3}{n^2 -3}-0right|=frac{n+3}{n^2 -3} $$
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 $n_0 =lceilfrac{1}{epsilon} +3rceil$, 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?

Yup, the proof is correct.

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

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 $dfrac1infty$.