Calculus 8th Edition

$+\infty$
Factoring, numerator : find two factors of $-8$ that add to $-2$ ($-4$ and $2)$ denominator : find two factors of $6$ that add to $-5$ ($-3$ and $-2)$ $\displaystyle \lim_{x\rightarrow 2^{+}}\frac{x^{2}-2x-8}{x^{2}-5x+6}=\lim_{x\rightarrow 2^{+}}\frac{(x-4)(x+2)}{(x-3)(x-2)}$ As $x\rightarrow 2^{+}$ x is positive, approaching 2, slightly greater than 2, $(x-4)$ is negative, approaching $-2$, $(x+2)$ is positive, approaching $4$ $(x-3)$ is negative, approaching $-1$, $(x-2)$ is positive, approaching $0$ The numerator approaches$-8,$ the denominator approaches $0$ from the negative side, so $\displaystyle \lim_{x\rightarrow 2^{+}}\frac{x^{2}-2x-8}{x^{2}-5x+6}=+\infty$