It is meaningless state that $\lim\limits_{x\to \infty} \sqrt{x^2+x}-x \ne x - x$ and $\lim\limits_{x\to \infty} \sqrt{x^2+x}+x = x + x$ we should state that
$$\lim\limits_{x\to \infty} \sqrt{x^2+x}-x = \lim\limits_{x\to \infty} \frac1{2}+o(1/x)=\frac12$$
and
$$\lim\limits_{x\to \infty} \sqrt{x^2+x}+x = \lim\limits_{x\to \infty} 2x+o(1)=\infty$$
the explanation in both case is in binomial first order approximation that is
$$\sqrt{x^2+x}=x\left(1+\frac1x\right)^\frac12= x\left(1+\frac1{2x}+o\left(\frac1x\right)\right)=x+ \frac1{2}+o\left(1\right) $$
which means that for $x$ large we have
$$\sqrt{x^2+x}\sim x+ \frac1{2}$$
and therefore
$$\sqrt{x^2+x}-x \sim \frac 12$$
$$\sqrt{x^2+x}+x \sim 2x+\frac 12$$
Edit
Note that for $\frac{x}{ \sqrt{x^2+x} + x}$ it is not correct to state that the denominator is $2x$ the complete steps are
$$\frac{x}{ \sqrt{x^2+x} + x}=\frac x x \frac{1}{ \sqrt{1+1/x} + 1} \to \frac12$$
For $\sqrt{x^2+x} - x$ indeed is not correct take$ \sqrt{x^2+x}=x$ what is true is that $\sqrt{x^2+x}\sim x+\frac12$.
If we use that approximation, it works with both limits. In this particular case first order approximation works and we can use it to evaluate both limits.
Since $\frac{\pi}{2} - \arctan(x) =\arctan \left(\frac1x\right)\to 0$ we can use that
$$\left(1+ \frac{\pi}{2} - \arctan(x)\right)^x=\left[\left(1+ \arctan\left(\frac1x\right)\right)^{\frac{1}{\arctan\left(\frac1x\right)}}\right]^{x\arctan\left(\frac1x\right)}$$
and then refer to standard limits.
Or as an alternative, following your idea
$$\lim\limits_{x \to \infty}\frac{\ln(1+\frac{\pi}{2}- \arctan(x))}{\frac{1}{x}}=\lim\limits_{x \to \infty}\frac{\ln\left(1+\arctan \left(\frac1x\right)\right)}{\arctan \left(\frac1x\right)}\,\frac{\arctan \left(\frac1x\right)}{\frac1x}$$
and then conclude again by standard limits.
Best Answer
Hint: Let $t=\dfrac1x$ , and use the fact that $\displaystyle\lim_{a\to\infty}x^a=0$ when $x\in(0,1)$.