## Thomas' Calculus 13th Edition

$a.$ Using the alternative definition for $|x|,\ \quad |x|=\sqrt{x^{2}}$ $|\displaystyle \frac{1}{b}|=\sqrt{(\frac{1}{b})^{2}}=\sqrt{\frac{1^{2}}{b^{2}}}=\frac{1}{\sqrt{b^{2}}}=\frac{1}{|b|}$ $b.$ $|\displaystyle \frac{a}{b}|=\sqrt{(\frac{a}{b})^{2}}=\sqrt{\frac{a^{2}}{b^{2}}}=\frac{\sqrt{a^{2}}}{\sqrt{b^{2}}}=\frac{|a|}{|b|}$