SOLUTION: Let f(x) = (x^2 + 7x - 6)/(x-1)(x-2)(x+1) (i) Express f(x) in partial fractions. (ii) Show that, when x is sufficiently small for x^4 and higher powers to be neglected, f(x) = -3

You should have grouped the factors in the denominator

 f(x) = (x^2 + 7x - 6)/[(x-1)(x-2)(x+1)]



Multiply through by (x-1)(x-2)(x+1)



Substituting x=2 will cause two of the terms on the right to become 0, 
then you can solve for B = -2

Substituting x=-1 will also cause two of the terms on the right to become 0, 
then you can solve for C = 4

Substituting x=1 will also cause two of the terms on the right to become 0, 
then you can solve for A = -1

  

Those terms look similar to the sum of an infinite geometric series:



We can get the first term in that form:



We can get the second term in that form:



The third term needs to have first term 1 in the denominator:



So we have:



Using the above equation for  on each term:



Simplifying that gives









Simplifying and omitting terms in x4 and higher we have:



Edwin