\( \displaystyle\lim_{x\to4}\frac{x^3-2x^2-9x+4}{x^2-2x-8} \)
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
\( \displaystyle\lim_{x\to4}\frac{x^2+2x-1}{x+2}=\frac{4^2+2\times4-1}{4+2}=\frac{23}6 \)