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My initial question was: why partial fractions do require proper fractions as input?

My uni professor guided me through saying that I should think about is what happens at the limits when x goes to infinity. All our partial fraction terms have higher powers of x in the denominator and so go to zero. If the input has a higher power of x in the numerator than the denominator then this will go to infinity as x goes to infinity.

I think, I have almost built the intuition. However, I am still not sure: if x goes to infinity, does it imply that we won't be able to find A,B,C (etc.) to do the partial fractions properly?

Karina
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  • To do the decomposition properly in this case, you have first make the Euclidean division of the numerator by the denominator, write the corresponding equality, and split the fraction to obtain the base case. – Bernard Oct 14 '20 at 14:54
  • I don't think there's any such requirement. I was taught that, for example this would be "partial fractions" $\frac{x^2}{x+1}=x-1+\frac{1}{x+1}$. – ancient mathematician Oct 14 '20 at 14:56
  • It's a little bit confusing. However, my question is rather simple: if x goes to infinity, does it imply that we won't be able to find A,B,C (etc.) to do the partial fractions properly? Is it the case? – Karina Oct 14 '20 at 15:05
  • Partial fraction decomposition algorithms (e.g. here) work fine for any fraction - proper or not. You seem to be considering some application of partial fractions that requires proper fractions, but it is not at all clear what that may be from what little you wrote. – Bill Dubuque Oct 14 '20 at 23:59

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