Science:Math Exam Resources/Courses/MATH110/April 2017/Question 04 (b) (ii)

MATH110 April 2017

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[hide] Question 04 (b) (ii)
(b) Consider the function $g(x)={\frac {\ln x}{x^{2}-1}}$ (ii) Find all vertical asymptotes of $g$ .

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[show] Hint
Recall that a line $x=a$ is a vertical asymptote of $g$ if either $\lim _{x\to a^{-}}g(x)$ or $\lim _{x\to a^{+}}g(x)$ are infinite.

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[show] Solution 1
Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Recall that $x=a$ is a vertical asymptote of $g$ if either one side of the limits, as $x\to a^{+}$ or $x\to a^{-}$ , of $g$ is infinite. When $g$ is given as a fraction form, a one side limit is infinite, then either the denominator goes to 0 or the numerator goes to infinity. Considering $g(x)={\frac {\ln x}{x^{2}-1}}$ , the numerator approaches to $-\infty$ , as $x\to 0+$ , while the denominator $x^{2}-1=(x+1)(x-1)$ goes to zero as $x\to \pm 1$ . This shows that the possible candidates for vertical asymptotes are $x=0,1$ and $x=-1$ . From part (a), we know that $g(x)$ is not defined for $x$ near $-1$ , and hence the limit of $g$ as $x\to -1$ doesn't exists. This implies that $x=-1$ is not a vertical asymptote. Let's consider $x=0$ . Since the domain of $g$ is $(0,1)\cup (1,\infty )$ (from part (a)), we consider the limit as $x\to 0+$ . Using $\lim _{x\to 0+}x^{2}-1=0-1=-1$ , we have $\lim _{x\to 0+}g(x)=\lim _{x\to 0+}{\frac {\ln x}{x^{2}-1}}=\lim _{x\to 0+}\ln x\cdot \lim _{x\to 0+}{\frac {1}{x^{2}-1}}=-\infty \cdot (-1)=+\infty$ . This implies that $x=0$ is a vertical asymptote of $g$ . Finally, let's think about the last candidate $x=1$ . Since the direct substitution gives ${\frac {\ln 1}{1^{2}-1}}={\frac {0}{0}}$ , we use l'Hospital's rule to evaluate the limit; $\lim _{x\to 1}g(x)=\lim _{x\to 1}{\frac {\ln x}{x^{2}-1}}=\lim _{x\to 1}{\frac {(\ln x)'}{(x^{2}-1)'}}=\lim _{x\to 1}{\frac {1}{x}}\cdot {\frac {1}{2x}}={\frac {1}{2}},$ . which means that $x=1$ is also not a vertical asymptote. To summarize, the only vertical asymptote of $g$ is $\color {blue}x=0$ .