Science:Math Exam Resources/Courses/MATH110/April 2016/Question 05 (c)/Solution 1

Since the denominator $x^2$ is zero exactly when $x=0$, it is the only candidate for the vertical asymptote.

As $x$ approaches to $0$, the numerator $x-1$ approaches to $-1$, while the denominator $x^2>0$ approaches 0. This gives

$\underset{x\to 0}{\lim }f(x)=\underset{x\to 0}{\lim }\frac{x-1}{x^2}=-\mathrm{\infty }$.

In other words, $f(x)$ has exactly one vertical asymptote, which is $x=0$.

On the other hand, we have

$\underset{x\to -\mathrm{\infty }}{\lim }f(x)=\underset{x\to -\mathrm{\infty }}{\lim }\frac{x-1}{x^2}=\underset{x\to -\mathrm{\infty }}{\lim }\frac{1-1/x}{x}=0$

and

$\underset{x\to \mathrm{\infty }}{\lim }f(x)=\underset{x\to \mathrm{\infty }}{\lim }\frac{x-1}{x^2}=\underset{x\to \mathrm{\infty }}{\lim }\frac{1-1/x}{x}=0$

It follows that it has $y=0$ as its horizontal asymptote.

Answer, the asymptotes are: $x=0,y=0$.