Science:Math Exam Resources/Courses/MATH110/April 2012/Question 05 (d)

MATH110 April 2012

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Question 05 (d)
Let $\displaystyle f(x)={\frac {2x^{2}}{x^{2}-1}}$ (d) Find the vertical asymptotes of ƒ.

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
We check for vertical asymptotes by taking the limit of the function at points where it is undefined or otherwise not continuous. From part (a) of this question we know that the points of interest are x = -1 and x = 1.

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

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. To find the vertical asymptotes, we will take the limit of the function at both x = -1 and x = 1. $\lim _{x\to -1}f(x)=\lim _{x\to -1}{\frac {2x^{2}}{x^{2}-1}}$ This limit is undefined since the denominator goes to zero but the numerator is nonzero. We will try taking the left and right hand limit at x = -1 $\lim _{x\to -1^{-}}{\frac {2x^{2}}{x^{2}-1}}=\lim _{x\to -1^{-}}{\frac {2x^{2}}{(x+1)(x-1)}}=\infty$ Note that as x approaches -1 from the left, the numerator of the fraction will be positive. Both terms in the denominator will be negative, yielding an overall positive number. As the denominator is decreasing, this is the same as saying that the entire fraction is increasing. Thus the limit is positive infinity. Using similar reasoning we can solve the right hand limit to get $\lim _{x\to -1^{+}}{\frac {2x^{2}}{x^{2}-1}}=-\infty$ Taking the limit at x = 1 will give us an undefined limit in the same way as before. Checking right and left hand limits gives $\lim _{x\to 1^{-}}{\frac {2x^{2}}{x^{2}-1}}=-\infty$ $\lim _{x\to 1^{+}}{\frac {2x^{2}}{x^{2}-1}}=\infty$ Thus we have vertical asymptotes at x = -1 and x = 1, where the value of the function near the asymptote depends which side of the asymptote it is approaching.

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