1. According to the given limit, it is necessary to factor the expressions in numerator and denominator to solve it:
[tex]\begin{gathered} \lim _{x\to6}\frac{x^2-3x+2}{x^2-x-30} \\ \lim _{x\to6}\frac{(x-2)(x-1)}{(x-6)(x+5)} \\ \lim _{x\rightarrow6}\frac{4\cdot5}{0\cdot11} \\ \lim _{x\rightarrow6}\frac{20}{0}=\infty \end{gathered}[/tex]The limit does exist and it is infinite.
2. As the limit when x tends to 6 is infinite, it means there is a vertical asymptote in x=6.
3. The equation of the asymptote is x=6.
4. Evaluate the function in values that close to 6 for example 5.9 and 6.1 to know if it approaches negative or positive infinity from the left and right:
[tex]\begin{gathered} \lim _{x\rightarrow5.9}\frac{(5.9-2)(5.9-1)}{(5.9-6)(5.9+5)} \\ \lim _{x\rightarrow5.9}\frac{(3.9)(4.9)}{(-0.1)(10.9)}=\frac{19.11}{-1.09}=-17.53 \\ \lim _{x\rightarrow6.1}\frac{(6.1-2)(6.1-1)}{(6.1-6)(6.1+5)} \\ \lim _{x\rightarrow6.1}\frac{(4.1)(5.1)}{(0.1)(11.1)}=\frac{20.91}{1.11}=18.83 \end{gathered}[/tex]According to this, the limit approaches to negative infinity from the left (when x tends to 5.9 the limit is negative) and to positive infinity from the right (when x tends to 6.1 the limit is positive)
