Editing Partial fraction decomposition (section)

===Example 5 (limit method)===

[[Limit (mathematics)|Limits]] can be used to find a partial fraction decomposition.<ref>{{cite book|last=Bluman|first=George W.| title=Problem Book for First Year Calculus|year=1984|publisher=Springer-Verlag|location=New York|pages=250–251}}</ref> Consider the following example:

<math display="block"> \frac{1}{x^3 - 1}</math>

First, factor the denominator which determines the decomposition:

<math display="block"> \frac{1}{x^3 - 1} = \frac{1}{(x - 1)(x^2 + x + 1)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + x + 1}.</math>

Multiplying everything by <math>x-1</math>, and taking the limit when <math>x \to 1</math>, we get

<math display="block">\lim_{x \to 1} \left((x-1)\left ( \frac{A}{x-1} + \frac{Bx + C}{x^2 + x + 1} \right )\right) = \lim_{x \to 1} A + \lim_{x \to 1}\frac{(x-1)(Bx + C)}{x^2 + x + 1} =A.</math>

On the other hand,

<math display="block">\lim_{x \to 1} \frac{(x-1)}{(x - 1)(x^2 + x + 1)} =  \lim_{x \to 1}\frac{1}{x^2 + x + 1} = \frac13,</math>

and thus:

<math display="block">A = \frac{1}{3}.</math>

Multiplying by {{math|''x''}} and taking the limit when <math>x \to \infty</math>, we have

<math display="block">\lim_{x \to \infty} x\left( \frac{A}{x-1} + \frac{Bx + C}{x^2 + x + 1} \right )= \lim_{x \to \infty} \frac{Ax}{x-1} + \lim_{x \to \infty} \frac{Bx^2+Cx}{x^2+x+1}= A+B,</math>

and

<math display="block">\lim_{x \to \infty} \frac{x}{(x - 1)(x^2 + x + 1)} =0.</math>

This implies {{math|1=''A'' + ''B'' = 0}} and so <math>B = -\frac{1}{3}</math>.

For {{math|1=''x'' = 0}}, we get  <math>-1 = -A + C,</math> and thus  <math>C = -\tfrac{2}{3}</math>.

Putting everything together, we get the decomposition

<math display="block">\frac{1}{x^3 -1} = \frac{1}{3} \left( \frac{1}{x - 1} + \frac{-x -2}{x^2 + x + 1} \right ).</math>