$\epsilon-N$ for $\lim\limits_{n \to \infty} \sqrt{n^{2} +3n-3} -n = \frac{3}{2}$ First, I tried to use the triangle inequality only once to find an N:
$$
\left | \sqrt{n^2+3n-3}-n-\frac{3}{2}  \right | \leqslant \left | \sqrt{n^2+3n-3}-n \right | + \left | \frac{3}{2} \right | = \epsilon 
$$
$$
N=\left \lfloor \frac{(\epsilon -\frac{3}{2})^{2}+3}{6-2\epsilon }  \right \rfloor +1
$$
I choose epsilon to be 0.01, and N is 1, which is incorrect.
Then I manipulated the inequality again by using the triangle inequality one more time:
$$
\left | \sqrt{n^2+3n-3}-n-\frac{3}{2}  \right | \leqslant \left | \sqrt{n^2+3n-3}-n \right | + \left | \frac{3}{2} \right |
$$
$$
\leqslant \left | \sqrt{n^2+3n-3} \right | +n+  \frac{3}{2} =\epsilon 
$$
$$
N=\left \lfloor \frac{(\epsilon -\frac{3}{2})^{2}+3}{2\epsilon }  \right \rfloor +1
$$
and this time, when epsilon is 0.01, N is 2624, which is correct
I would like to know why the first approach is wrong and the second one is right, Thank you.

You have
$$\sqrt{n^2+3n-3}-n = \frac{ (\sqrt{n^2+3n-3}-n)(\sqrt{n^2+3n-3}+n)}{\sqrt{n^2+3n-3}+n}$$
$$= \frac{3n-3}{\sqrt{n^2+3n-3}+n}.$$
Divide top an bottom by $n$ to get
$$\frac{3-\frac{3}{n}}{\sqrt{1+\frac{3}{n} - \frac{3}{n^2}}+1} $$
