Triple Limit

$\displaystyle \lim_{x \to 2} \frac {x-2}{x^3-8} = \lim_{x \to 2} \frac {x-2}{(x-2)(x^2+2x+4)} = \lim_{x \to 2} \frac 1{x^2+2x+4} = \frac 1{12}$ $\implies a = 12$

$\displaystyle \lim_{x \to 2} \frac {x^2-5}{4x+10} = - \frac 1{18}$ $\implies b = 18$

$\displaystyle \lim_{x \to \frac ba} \frac {x^3-3x^2}{x^2-5x+6} = \lim_{x \to \frac {18}{12}} \frac {x^2(x-3)}{(x-2)(x-3)} = \lim_{x \to \frac 32} \frac {x^2}{x-2} = \frac {\frac 94}{\frac 32 - 2} = \frac 9{6-8} = - \frac 92$ $\implies c+d+20 = 9+2+20 = \boxed{31}$

• Taking the limit of the numerator and denominator independently yields the indeterminate form $\frac{0}{0}$ . Thus, by L'Hôpital's rule, $\lim_{x\to 2} \frac{x-2}{x^3-8} = \lim_{x\to2}\frac{1}{3x^2} = \frac{1}{12}$ So $a=12$ .
• We can take the limit by evaluating the expression at $x=2$ . $\lim_{x\to 2}\frac{x^2-5}{4x+10} = \frac{-1}{18}$ So $b=18$ .
• We can again evaluate the expression at $x={}^b\!/\!_a = {}^3\!/\!_2$ . $\lim_{x\to^3\!/\!_2}\frac{x^3-3x^2}{x^2-5x+6} = \lim_{x\to^3\!/\!_2}\frac{x^2(x-3)}{(x-3)(x-2)} = \frac{\left(^3\!/\!_2\right)^2}{^3\!/\!_2-2} = -\frac{9}{2}$ Thus $c=9$ , $d=2$ , and $c+d+20 = 9+2+20 = \boxed{31}$ .