\(\displaystyle\lim_{x\to4}\frac{x^3-2x^2-9x+4}{x^2-2x-8}\)
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 \(4^3-2\times4^2-9\times4+4=0\), which means, by the polynomial remainder theorem, that \(x-4\) is indeed a linear factor of the numerator. Dividing yields a quotient of
\(\dfrac{x^3-2x^2-9x+4}{x-4}=x^2+2x-1\)
so the limit is
\(\displaystyle\lim_{x\to4}\frac{x^2+2x-1}{x+2}=\frac{4^2+2\times4-1}{4+2}=\frac{23}6\)
