Science:Math Exam Resources/Courses/MATH102/December 2016/Question A 03/Solution 1

The function is not defined at $x=0$; it has a vertical asymptote at 0, because

$\lim _{x\to 0+}f(x)=0+\lim _{x\to 0+}{\frac {1}{x}}=+\infty$ and $\lim _{x\to 0-}f(x)=0+\lim _{x\to 0-}{\frac {1}{x}}=-\infty$,

so the first choice is out.

Now we check the behaviour of the function when $x\to \infty$.

First, note that $f$ has NO horizontal asymptote because $\lim _{x\to \infty }f(x)=\infty$, so the bottom left choice is out too. However, we see that $\lim _{x\to \infty }f(x)-x=\lim _{x\to \infty }{\frac {1}{x}}=0$, this means that as $x\to \infty$ the function approaches to the line $y=x$, in other words it has a slant asymptote at $y=x$.

Answer: Top Right graph