Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.8 Exercises - Page 390: 18

$\displaystyle \lim_{x\rightarrow-\infty}\tanh x=-1$

$\displaystyle \tanh x=\frac{\sinh x}{\cosh x} =\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} $ $\displaystyle \lim_{x\rightarrow-\infty}\tanh x=\lim_{x\rightarrow-\infty}\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}$ multiply both the numerator and the denominator with $e^{x}$ $=\displaystyle \lim_{x\rightarrow-\infty}\frac{e^{2x}-e^{0}}{e^{2x}+e^{0}}=\frac{-1}{1}=-1$