## University Calculus: Early Transcendentals (3rd Edition)

Apply a direct limit comparison test with $\dfrac{1}{x}$ Thus, we have $\dfrac{1}{ \sqrt {x^2-1}} \geq \dfrac{1}{x}$ This suggests that $0\leq \dfrac{1}{x}\leq \dfrac{1}{ \sqrt {x^2-1}}$; as $x \to\infty$ Now, we have $\int_{1}^{\infty} \dfrac{dx}{x}=\lim\limits_{p \to \infty} \int_{1}^{a} \dfrac{dx}{x}$ or, $\lim\limits_{p \to \infty} \int_{1}^{a} \dfrac{dx}{x}=\lim\limits_{p \to \infty} [\ln |x|]_{1}^{a} \dfrac{dx}{x}$ or, $\lim\limits_{p \to \infty} [\ln |x|]_{1}^{a} \dfrac{dx}{x}=\lim\limits_{p \to \infty} [0-\ln |a|]=\infty$ Hence, the given integral diverges by the direct limit comparison test.