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

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