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

MATH110 April 2012

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

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[show] 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.

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[show] Solution
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.