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

MATH110 April 2016

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Question 05 (c)
Let $f(x)={\frac {x-1}{x^{2}}}.$ (c) Find all the vertical and horizontal asymptotes of $f$, if they exist.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
To find vertical asymptotes, use the fact that when $f$ is the ratio of two polynomials, the only candidates for vertical asymptotes are the values of $x$ for which the denominator is zero. To find horizontal asymptotes, determine the behaviour of $f(x)$ when $x$ approaches to $\pm \infty$.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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: $\color {blue}x=0,y=0$.