To determine the real values of $x$ such that the triangle with sides 5, 8, and $x$ is obtuse, we need to apply the properties of an obtuse triangle. In an obtuse triangle, the square of the longest side is greater than the sum of the squares of the other two sides.

Assuming $x$ is the longest side, the condition for obtuseness is:
$$x^2 > 5^2 + 8^2$$
$$x^2 > 25 + 64$$
$$x^2 > 89$$
So, $x > \sqrt{89}$.

Next, assuming 8 is the longest side, the condition for obtuseness becomes:
$$8^2 > 5^2 + x^2$$
$$64 > 25 + x^2$$
$$64 - 25 > x^2$$
$$39 > x^2$$
So, $x < \sqrt{39}$.

Lastly, we need to ensure that $x$ also satisfies the triangle inequality conditions:
1. $x + 5 > 8 \Rightarrow x > 3$
2. $x + 8 > 5$ which is always true for $x > 3$.
3. $5 + 8 > x \Rightarrow x < 13$

Thus, combining all these conditions, we have:
- $3 < x < \sqrt{39}$
- $x > \sqrt{89}$
- $x < 13$

Therefore, the values of $x$ such that the triangle is obtuse are:
$$\boxed{(3, \sqrt{39}) \cup (\sqrt{89}, 13)}$$