<s>[INST] $\epsilon-N$ 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.
 [/INST] 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.
</s>