<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] $$ \lim_{n->\infty}(\sqrt{n^{2}+3n-3} - n) = \lim_{n->\infty}\frac{3n-3}{\sqrt{n^{2}+3n-3} + n} = \lim_{n->\infty}\frac{3-\frac{3}{n}}{\sqrt{1+\frac{3}{n}-\frac{3}{n^2}} + 1} = \frac{3}{2}. $$
Notice that in the last limit as $n$ goes to infinity, the numerator goes to $3$ and the denominator goes to $2$. The second limit is the first one, but the argument of it is multiplied by $$1 = \frac{\sqrt{n^{2}+3n-3} + n}{\sqrt{n^{2}+3n-3} + n}.$$
</s>