Problem about solving infinity limit with square root (I)
$$\lim_{x \to \infty } \, \left(\sqrt{x^2+x}-\sqrt{x^2-x}\right)=$$
$$\lim_{x \to \infty } \, \left(x\sqrt{1+1/x}-x\sqrt{1-1/x}\right)=$$
$$\lim_{x \to \infty } \, \left(x\sqrt{1}-x\sqrt{1}\right)=\lim_{x \to \infty } \, \left(x-x\right)=0$$
(II)
$$\lim_{x \to \infty } \, \left(\sqrt{x^2+x}-\sqrt{x^2-x}\right)=$$
$$\lim_{x \to \infty } \, \left(\left(\sqrt{x^2+x}-\sqrt{x^2-x}\right)*\frac{\left(\sqrt{x^2+x}+\sqrt{x^2-x}\right)}{\left(\sqrt{x^2+x}+\sqrt{x^2-x}\right)}\right)=$$
$$\lim_{x \to \infty } \, \frac{2x}{\left(\sqrt{x^2+x}+\sqrt{x^2-x}\right)}=$$
$$\lim_{x \to \infty } \, \frac{2x}{\left(x\sqrt{1+1/x}+x\sqrt{1-1/x}\right)}=$$
$$\lim_{x \to \infty } \, \frac{2x}{\left(x\sqrt{1}+x\sqrt{1}\right)}=\lim_{x \to \infty } \, \frac{2x}{2x}=1$$
I found these two ways to evaluate this limit. I know the answer is 1. The first one is surely wrong. The question is: why? What is wrong there?

By the same erroneous logic you'd conclude $\rm\:2\: =\: x(1+1/x)-x(1-1/x)\to\: 0$
In your example $\rm\:\displaystyle \sqrt{1+\frac{1}x}-\sqrt{1-\frac{1}x}\ =\ \frac{1}x + \frac{1}{8\: x^3} +\: \cdots\:$ which makes the error clear.
If the dominant terms in a sum of series cancel, then you need to look at subsequent terms.
