Why is $x(\sqrt{x^2+1} - x )$ approaching $1/2$ when $x$ becomes infinite? Why is $$\lim_{x\to \infty}  x\left( (x^2+1)^{\frac{1}{2}} - x\right)  = \frac{1}{2}?$$
What is the right way to simplify this?
My only idea is: $$x((x²+1)^{\frac{1}{2}} - x) > x(x^{2^{0.5}}) - x^2 = 0$$
But 0 is too imprecise.

You already know that $\sqrt{x^2 + 1} \sim x$. The problem is that that approximation isn't good enough for computing this limit.
However, you've done exercises to do this exact thing: use a differential approximation and the value of $\sqrt{x^2}$ to estimate the value of $\sqrt{x^2 + 1}$. e.g. maybe you've been asked to estimate $\sqrt{81.4}$ using the knowledge of $\sqrt{81}$.
To put it back in familiar symbols, set $f(t) = \sqrt{x^2 + t}$. Then differential approximation tells us that
$$ f(1) \approx f(0) + f'(0) \cdot (1 - 0) $$
or that
$$ \sqrt{x^2 + 1} \approx x + \frac{1}{2 \sqrt{x^2}} =  x + \frac{1}{2x} $$
and now we can see that the limit probably should be $1/2$. There is one last problem, though: we need to make sure there isn't too much error in the differential approximation! And unfortunately, the description of the error term for differential approximation isn't good enough for our purposes, since it tells us what happens as $t \to 0$, not as $x \to \infty$.
While there are some tricks for doing this, I will instead give a more systematic approach: use Taylor series with remainder. Here is one form:
$$ f(1) = f(0) + f'(0) \cdot (1-0) + \frac{1}{2} f''(a)  \cdot (1-0)^2 $$
where $a$ is some unknown value in the interval $[0,1]$. So we have
$$ \sqrt{x^2 + 1} = \sqrt{x^2} + \frac{1}{2 \sqrt{x^2} } - \frac{1}{8 (x^2 + a)^{3/2}}
= x + \frac{1}{2 x } - \frac{1}{8 (x^2 + a)^{3/2}} $$
where $a \in [0,1]$ (and varies with $x$). Plugging this into our original limit now gives us something we can rigorously show converges to $1/2$.
It may help to use the fact this is a strictly increasing function of $a$ to arrange the above as
$$ x + \frac{1}{2 x } - \frac{1}{8x^3} \leq \sqrt{x^2 + 1} \leq  x + \frac{1}{2 x } - \frac{1}{8(x^2 + 1)^{3/2}}  $$

Interestingly, even though we needed a first-order estimate to solve the problem, we could have used a zeroth-order Taylor series with remainder, or equivalently the mean value theorem:
$$ f(1) = f(0) + f'(a) \cdot (1-0) = x + \frac{1}{2 \sqrt{x^2 + a}} $$
so that
$$x + \frac{1}{2\sqrt{x^2+1}} \leq \sqrt{x^2 + 1} \leq x + \frac{1}{2 x}$$
