$\lim \limits_{x \rightarrow \infty} \sqrt[]{x^2+2x+3}-\sqrt[]{x^2-x+5}.$ I am trying to formally evaluate the following limit:
$$\lim \limits_{x \rightarrow \infty} \sqrt[]{x^2+2x+3}-\sqrt[]{x^2-x+5}.$$
Empirically, the limit seems to be converging to $1.5$, although I am not sure how to formally prove this.  I had one idea thus far: it appears the constant terms within the square roots do not matter, so I rewrote the limit as
$$\lim \limits_{x \rightarrow \infty} \sqrt[]{x^2+2x-3}-\sqrt[]{x^2-x-6}
=\lim \limits_{x \rightarrow \infty} \sqrt[]{(x+3)(x-1)}-\sqrt[]{(x+2)(x-3)}.$$
In each square root, there are two factors.  I assumed that in the limit the geometric mean of the two factors (i.e. square root of the product) is equal to their arithmetic mean.  This is a step I am not quite certain of, and if it is true I would like to prove it. However, as I found it led to the correct answer, since
$$\lim \limits_{x \rightarrow \infty} \frac{(x+3)+(x-1)}{2}-\frac{(x+2)+(x-3)}{2}=\lim \limits_{x \rightarrow \infty} (x+1)-(x-0.5)=1.5.$$
I am not sure if the result was coincidental, but if not, I would like some help formalizing each of my steps.  I would also like to hear of alternate approaches that do not alter the original problem in the way I did.
 A: $$\lim \limits_{x \rightarrow \infty} \sqrt[]{x^2+2x+3}-\sqrt[]{x^2-x+5}=\frac{3x-2}{\sqrt{x^2+2x+3}+\sqrt{x^2-x+5}}\xrightarrow[x\to\infty]{}\frac3{1+1}=\frac32$$
A: A trick I find very useful is to do $x=1/t$ (and we can restrict to $x>0$), so the limit becomes
$$
\lim_{t\to0^+}
\left(\sqrt{\frac{1}{t^2}+\frac{2}{t}+3}-
\sqrt{\frac{1}{t^2}-\frac{1}{t}+5}\right)
=
\lim_{t\to0^+}
\frac{\sqrt{1+2t+3t^2}-\sqrt{1-t+5t^2}}{t}
$$
Now, with a simple Taylor expansion, the numerator becomes
$$
\left(1+\frac{1}{2}(2t+3t^2)+o(t)\right)-
\left(1+\frac{1}{2}(-t+5t^2)+o(t)\right)=
t+\frac{1}{2}{t}+o(t)
$$
so we have
$$
\lim_{t\to0^+}\frac{\frac{3}{2}t+o(t)}{t}=\frac{3}{2}
$$
Alternatively, the limit is the derivative at $0$ of
$$
f(t)=\sqrt{1+2t+3t^2}-\sqrt{1-t+5t^2}
$$
Since
$$
f'(t)=\frac{2+6t}{2\sqrt{1+2t+3t^2}}-\frac{-1+10t}{2\sqrt{1-t+5t^2}}
$$
we have $f'(0)=\frac{3}{2}$.
Or you can “rationalize”:
$$
\lim_{t\to0^+}
\lim_{t\to0^+}
\frac{\sqrt{1+2t+3t^2}-\sqrt{1-t+5t^2}}{t}
=
\lim_{t\to0^+}
\frac{(1+2t+3t^2)-(1-t+5t^2)}{t(\sqrt{1+2t+3t^2}+\sqrt{1-t+5t^2})}
$$
A: $$\sqrt{x^2+2x+3}-(x+1) = \frac{2}{x+1+\sqrt{x^2+2x+3}}=O\left(\frac{1}{x}\right)\tag{1}$$
$$\sqrt{x^2-x-5}-\left(x-\frac{1}{2}\right) = -\frac{5+\frac{1}{4}}{\left(x-\frac{1}{2}\right)+\sqrt{x^2-x-5}}=O\left(\frac{1}{x}\right)\tag{2} $$
$$\lim_{x\to +\infty}\sqrt{x^2+2x+3}-\sqrt{x^2-x-5}=\lim_{x\to +\infty}(x+1)-\left(x-\frac{1}{2}\right) = \color{red}{\frac{3}{2}}.\tag{3}$$
A: Your finding is excellent, the constant terms do not matter.
$$\sqrt{x^2+2x+a}-\sqrt{x^2-x+b}=\frac{(x^2+2x+a)-(x^2-x+b)}{\sqrt{x^2+2x+a}+\sqrt{x^2-x+b}}\approx\frac{3x+a-b}{2x}$$ and for large $x$, $a-b$ is neglectible.
Then why not set $a=1,b=1/4$ to form perfect squares so that
$$\sqrt{x^2+2x+1}-\sqrt{x^2-x+\frac14}=(x+1)-(x-\frac12)=\frac32.$$
