$\lim \limits_{x \rightarrow \infty} \sqrt[]{x^2+2x+3}-\sqrt[]{x^2-x+5}.$ 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.

$$\sqrt{x^2+2x+3}-(x+1) = \frac{2}{x+1+\sqrt{x^2+2x+3}}=O\left(\frac{1}{x}\right)\tag{1}$$
$$\sqrt{x^2-x-5}-\left(x-\frac{1}{2}\right) = -\frac{5+\frac{1}{4}}{\left(x-\frac{1}{2}\right)+\sqrt{x^2-x-5}}=O\left(\frac{1}{x}\right)\tag{2} $$
$$\lim_{x\to +\infty}\sqrt{x^2+2x+3}-\sqrt{x^2-x-5}=\lim_{x\to +\infty}(x+1)-\left(x-\frac{1}{2}\right) = \color{red}{\frac{3}{2}}.\tag{3}$$
