Question

question 5
find the vertical asymptotes for the function
$f(x)=(x - 1)\ln(x^{2}-1)$
$x=-1$ only
$x = 0$ only
$x=-1$ and $x = 1$
this function does not have any vertical asymptotes
$x = 0$ and $x = 1$

Explanation:

Step1: Analyze the domain of the logarithmic function

For the function \(y = \ln(u)\), the argument \(u>0\). Here \(u=x^{2}-1\), so \(x^{2}-1=(x - 1)(x + 1)>0\). The solutions of the inequality \((x - 1)(x + 1)>0\) are \(x<-1\) or \(x>1\). Also, we need to consider the behavior of the function near the points where the domain is restricted.

Step2: Check the limit as \(x\to - 1\)

We use the fact that \(\lim_{x\to a}f(x)g(x)\) where \(f(x)\to c
eq0\) and \(g(x)\to\pm\infty\). Let \(f(x)=x - 1\) and \(g(x)=\ln(x^{2}-1)\). As \(x\to - 1\), \(f(x)\to-2\) and \(\lim_{x\to - 1^{+}}\ln(x^{2}-1)=-\infty\) (since \(x^{2}-1\to0^{+}\) as \(x\to - 1^{+}\)). So \(\lim_{x\to - 1^{+}}(x - 1)\ln(x^{2}-1)=\infty\)

Step3: Check the limit as \(x\to1\)

We can rewrite the function \(f(x)=(x - 1)\ln(x^{2}-1)=(x - 1)\ln((x - 1)(x + 1))=(x - 1)\ln(x - 1)+(x - 1)\ln(x + 1)\). Let \(t=x - 1\), then \(\lim_{x\to1}(x - 1)\ln(x^{2}-1)=\lim_{t\to0}t\ln(t)+ \lim_{t\to0}t\ln(t + 2)\). Using the fact that \(\lim_{u\to0^{+}}u\ln u=\lim_{u\to0^{+}}\frac{\ln u}{\frac{1}{u}}\), by L - H rule \(\lim_{u\to0^{+}}\frac{\frac{1}{u}}{-\frac{1}{u^{2}}}=\lim_{u\to0^{+}}(-u)=0\). So \(\lim_{x\to1}(x - 1)\ln(x^{2}-1)=0\)

Answer:

\(x=-1\) only