Answer
$+\infty$
Work Step by Step
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
$\displaystyle \lim_{x\rightarrow 2^{+}}\frac{x^{2}-2x-8}{x^{2}-5x+6}=+\infty$
