How to evaluate this limit question? (+Infinity)

$$\lim_{n\rightarrow +\infty }(\sqrt{n^2+2n+2}-\sqrt{n^2+3n-1})$$

Is the answer $-\frac{1}{2}$?

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hint: use the formula: $\sqrt{A} – \sqrt{B} = \dfrac{A-B}{\sqrt{A}+\sqrt{B}}$, and take out $n$ from the two denominators, then simplify with the $n$ on top to see the $-1/2$ appear in your answer after taking limit .

$$F=\lim_{n\rightarrow +\infty }(\sqrt{n^2+2n+2}-\sqrt{n^2+3n-1})$$

Set $n=1/h,$

$\implies\sqrt{n^2+2n+2}=\sqrt{\frac{1+2h+2h^2}{h^2}}=\dfrac{\sqrt{1+2h+2h^2}}{|h|}=\dfrac{\sqrt{1+2h+2h^2}}h$ as $h>0$


$$=\lim_{h\to0^+}\frac{(1+2h+2h^2)-(1+3h-h^2))}h \cdot\frac1{\lim_{h\to0^+}(\sqrt{1+2h+2h^2}+\sqrt{1+3h-h^2})}=\cdots$$

Another way to solve the problem (and more) is based on Taylor series. $$A=\sqrt{n^2+2n+2}-\sqrt{n^2+3n-1}=n~\Big(\sqrt{1+\frac{2n+2}{n^2}}-\sqrt{1+\frac{3n-1}{n^2}}\Big)$$ and use the fact that, for small values of $y$, $$\sqrt{1+y}=1+\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right)$$ Now, make in the first radical $y=\frac{2n+2}{n^2}$ and in the second radical $y=\frac{3n+1}{n^2}$.

So, after developments and minor simplifications, $$A=n~\Big((n+1+\frac{1}{2 n}-\frac{1}{2 n^2}+O\left(\left(\frac{1}{n}\right)^3\right))-(n+\frac{3}{2}-\frac{13}{8 n}+\frac{39}{16
n^2}+O\left(\left(\frac{1}{n}\right)^3\right))\Big)$$ $$A=-\frac{1}{2}+\frac{17}{8 n}-\frac{47}{16
n^2}+O\left(\left(\frac{1}{n}\right)^3\right)$$ This shows the limit and moreover how it is approached.

This could easily be generalized for the limit of $$A=\sqrt{n^2+an+b}-\sqrt{n^2+cn+d}$$ An identical procedure would lead to $$A=\frac{a-c}{2}+\frac{-a^2+4 b+c^2-4 d}{8 n}+\frac{a^3-4 a b-c^3+4 c d}{16
n^2}+O\left(\left(\frac{1}{n}\right)^3\right)$$ in which the role of each $a,b,c,d$ coefficients appear.

$\sqrt{n^2+2n+2}-\sqrt{n^2+3n-1} = {3 -n \over \sqrt{n^2+2n+2}+\sqrt{n^2+3n-1}}$.

Now divide above and below by $n$ and take limits.

If we divide above and below by $n$ we get
$\sqrt{n^2+2n+2}-\sqrt{n^2+3n-1} = {{3\over n} -1 \over \sqrt{1+{2 \over n}+{2 \over n^2}}+\sqrt{1+{3 \over n}-{1 \over n^2}}}$.

We have $\lim_n (1+{2 \over n}+{2 \over n^2}) = 1$, $\lim_n (1+{3 \over n}-{1 \over n^2}) = 1$ and $\lim_n ({3\over n} -1) = -1$.

Since $x \mapsto \sqrt{x}$ is continuous, we see that
$\lim_n (\sqrt{1+{2 \over n}+{2 \over n^2}}+\sqrt{1+{3 \over n}-{1 \over n^2}}) = 2$.

Since $(x,y) \mapsto {x \over y}$ is continuous (for $y \neq 0$), we have
$\lim_n{{3\over n} -1 \over \sqrt{1+{2 \over n}+{2 \over n^2}}+\sqrt{1+{3 \over n}-{1 \over n^2}}} = { -1 \over 2} = – {1 \over 2}$.