$\epsilon-N$ 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.
 A: 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
$$\frac{3-\frac{3}{n}}{\sqrt{1+\frac{3}{n} - \frac{3}{n^2}}+1} $$
A: $$ \lim_{n->\infty}(\sqrt{n^{2}+3n-3} - n) = \lim_{n->\infty}\frac{3n-3}{\sqrt{n^{2}+3n-3} + n} = \lim_{n->\infty}\frac{3-\frac{3}{n}}{\sqrt{1+\frac{3}{n}-\frac{3}{n^2}} + 1} = \frac{3}{2}. $$
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 $$1 = \frac{\sqrt{n^{2}+3n-3} + n}{\sqrt{n^{2}+3n-3} + n}.$$
A: You cannot “choose” $\varepsilon$, forget it. You want to show that, for every $\varepsilon>0$, the inequality
$$
\Bigl|{\textstyle\sqrt{n^{2} +3n-3}} -n - \frac{3}{2}\Bigr|<\varepsilon
$$
is satisfied for all $n$ greater that some integer $N$ (depending on $\varepsilon$).
Your idea of applying the triangle inequality might be good: indeed, if you are able to see that
$$
\bigl|{\textstyle\sqrt{n^{2} +3n-3}} -n\bigr|+\frac{3}{2}<\varepsilon
$$
is satisfied for $n>N$, then also the wanted inequality would be satisfied. Unfortunately this turns out not to be a good strategy, because when $\varepsilon<3/2$, the inequality is satisfied for no value of $n$.
The required inequality is equivalent to
$$
\textstyle\bigl|\sqrt{4n^2+12n-12}-2n-3\bigr|<2\varepsilon
$$
and under the square root you can complete the square:
$$
4n^2+12n-12=4n^2+12n+9-21=(2n+3)^2-21
$$
and you can set temporarily $2n+3=m$, so you have
$$
\bigl|\textstyle\sqrt{m^2-21}-m\bigr|<2\varepsilon
$$
Since $\sqrt{m^2-21}<m$, this can be simplified to
$$
\textstyle m-\sqrt{m^2-21}<2\varepsilon
$$
hence to
$$
\textstyle m<\sqrt{m^2-21}+2\varepsilon
$$
that we can square to find
$$
\textstyle 0<-21+4\varepsilon^2+4\varepsilon\sqrt{m^2-21}
$$
so
$$
\textstyle 4\varepsilon\sqrt{m^2-21}>21-4\varepsilon^2
$$
which is certainly valid when
$$
m^2-21>\frac{(21-4\varepsilon)^2}{16\varepsilon^2}
$$
hence for
$$
m>\sqrt{21+\frac{(21-4\varepsilon)^2}{16\varepsilon^2}}
$$
Getting back to $n$, we find
$$
n>\frac{1}{2}\biggl(\sqrt{21+\frac{(21-4\varepsilon)^2}{16\varepsilon^2}}-3\biggr)
$$
and we're done.
A: 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.
