$\epsilon-N$ 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.

First, notice that both of the methods you followed are incorrect because the sum is larger than 3/2. Hence it can't be less than epsilon (for values small enough). I propose a different approach.$$$$
Notice that for all $n \geq 1$ :
$$|\sqrt{n^2+3n-3}-n-\frac{3}{2}| = -\sqrt{n^2+3n-3}+n+\frac{3}{2}$$
Now we're looking for N such that for all $n \geq N$
$$|\sqrt{n^2+3n-3}-n-\frac{3}{2}| \lt \epsilon$$
Using the formula above we find that:
$$\frac{(\frac{3}{2} - \epsilon)^2+3}{2\epsilon} \lt n$$
Thus, we take:
$$N = \pmb\lfloor{}\frac{(\epsilon - \frac{3}{2})^2+3}{2\epsilon}\pmb\rfloor{} + 1$$
which happens to be the exact same answer you got in the second approach you followed. That's why you got a correct answer for N although the method is wrong.
