Question

What is limit of f(x) = (x^3-2x^2-9x+4)/(x^2-2x-8) as x approaches 4

Answer 1

`\displaystyle\lim_(x\to4)(x^3-2x^2-9x+4)/(x^2-2x-8)`

First notice that
`x^2-2x-8=(x-4)(x+2)`. If
`x-4` is not a factor of the numerator, then there is a non-removable discontinuity at
`x=4`, and a removable discontinuity otherwise.

You have
`4^3-2*4^2-9*4+4=0`, which means, by the polynomial remainder theorem, that
`x-4` is indeed a linear factor of the numerator. Dividing yields a quotient of


`(x^3-2x^2-9x+4)/(x-4)=x^2+2x-1`

so the limit is


`\displaystyle\lim_(x\to4)(x^2+2x-1)/(x+2)=(4^2+2*4-1)/(4+2)=\frac{23}6`