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Domain of a logarithmic function

Consider the function t(x)=ln⁡(x2−1)t(x) = \ln(x^2 - 1).

Find the domain of t(x)=ln⁡(x2−1)t(x) = \ln(x^2 - 1). Express your answer using interval notation.

Hint

The argument of a logarithm must be strictly positive. Solve the inequality x2−1>0x^2 - 1 > 0.

Step-by-step solution

Since ln⁡(u)\ln(u) is only defined for u>0u > 0, we require x2−1>0x^2 - 1 > 0, which means x2>1x^2 > 1. This inequality holds when x>1x > 1 or x<−1x < -1.

Therefore the domain of t(x)t(x) consists of all real numbers except those in [−1,1][-1,1], giving (−∞,−1)∪(1,∞)(-\infty, -1) \cup (1, \infty).

Vertical asymptotes

Using the domain found above, identify all vertical asymptotes of t(x)=ln⁡(x2−1)t(x) = \ln(x^2 - 1) by evaluating the appropriate one-sided limits as xx approaches the boundary points of the domain.

Hint

Vertical asymptotes occur where the function's argument approaches 0+0^+ from within the domain. Check the behavior of x2−1x^2-1 as x→−1−x \to -1^- and as x→1+x \to 1^+.

Step-by-step solution

As x→1+x \to 1^+, we have x2−1→0+x^2 - 1 \to 0^+, so ln⁡(x2−1)→−∞\ln(x^2-1) \to -\infty. Similarly, as x→−1−x \to -1^-, x2−1→0+x^2 - 1 \to 0^+ as well, so ln⁡(x2−1)→−∞\ln(x^2-1) \to -\infty.

Since the function heads to −∞-\infty at both boundary values of the domain, there are vertical asymptotes at x=1x = 1 and x=−1x = -1.

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