Science:Math Exam Resources/Courses/MATH 180/December 2017/Question 03 (a)/Solution 1

Let us first simplify the function:

$$f(x)={\dfrac {x^{2}+x-2}{x^{2}+2x-3}}={\dfrac {(x+2)(x-1)}{(x+3)(x-1)}}.$$

Note that ${\textstyle f(x)}$ is defined for all real numbers except at ${\textstyle -3}$ (where ${\textstyle x+3=0}$) and ${\textstyle 1}$ (where ${\textstyle x-1=0}$).

The candidates for vertical asymptotes are ${\textstyle x=-3}$ and ${\textstyle x=1}$. However we need to compute the limits of ${\textstyle f(x)}$ at these values to check whether each is a vertical asymptote.

For ${\textstyle x\neq -3,1}$, we can write

$$f(x)={\dfrac {x+2}{x+3}}.$$

From this we see that

$$\lim _{x\to -3^{+}}f(x)=\lim _{x\to -3^{+}}{\dfrac {x+2}{x+3}}=-\infty ,\quad \quad \lim _{x\to -3^{-}}f(x)=\lim _{x\to -3^{-}}{\dfrac {x+2}{x+3}}=\infty$$

and

$$\lim _{x\to 1}f(x)=\lim _{x\to 1}{\dfrac {x+2}{x+3}}={\dfrac {\lim \limits _{x\to 1}(x+2)}{\lim \limits _{x\to 1}(x+3)}}={\dfrac {1+2}{1+3}}={\dfrac {3}{4}}.$$

Therefore, ${\textstyle x=-3}$ is a vertical asymptote, while ${\textstyle x=1}$ is not one.

Answer: The (only) vertical asymptote is ${\textstyle {\color {blue}x=-3}}$.