Science:Math Exam Resources/Courses/MATH110/April 2011/Question 08 (b)

MATH110 April 2011

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Question 08 (b)
Let $\displaystyle f(x)=\ln \left(4-x^{2}\right)$ Find all the vertical and horizontal asymptotes of ƒ, if there are any.

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
Remember that we find asymptotes using limits. Horizontal asymptotes are found by taking the limit of the function at positive and negative infinity. Vertical asymptotes are found by taking the limit of the function at points where the function is not defined - specifically endpoints of the domain.

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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. To find vertical asymptotes, we will take the limit of the function at $x=2$ and $x=-2$. Note that we are taking one-sided limits because the function is not defined on both sides of the point. $\lim _{x\to -2^{+}}\ln(4-x^{2})$ $\lim _{x\to 2^{-}}\ln(4-x^{2})$ As $x$ goes to -2 (from the right) or 2 (from the left), the expression $4-x^{2}\rightarrow 0^{+}$. So using $y=4-x^{2}$ we can rewrite both of these limits as $\lim _{y\to 0^{+}}\ln(y)$ which is simply $-\infty$. So the function has vertical asymptotes at both $x=-2$ and $x=2$. Since the function is only defined in the interval $(-2,2)$, which is bound on both sides, the function does not have horizontal asymptotes.

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