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

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

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

On the other hand, we have

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

and

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

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

Answer, the asymptotes are: $\color {blue}x=0,y=0$.