Why partial fraction decomposition requires a proper fraction on input?

Question

Is this only a desirable condition to make the job easier or a strict constraint of the method itself?

("Proper fraction" = a fraction where numerator is smaller then denominator)

Answer 1

Try to find constants $A$ and $B$ here: $$ \frac{x^{10}}{x(1-x)} = \frac{A}{x}+\frac{B}{1-x} $$ If you cannot do it, you have your reason...

(I assume "poper fraction" means; degree of numerator is less than degree of denominator.)

added
All the terms for a partial fraction decomposition go to $0$ as $x \to \infty$. So if you add them all together, still you have something at goes to $0$ as $x \to \infty$. The only rational functions with that property are the "proper" rational functions.