Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.8 Exercises - Page 390: 19

$\displaystyle \lim_{x\rightarrow\infty} {\rm sech} x=0$

${\rm sech} x=\displaystyle \frac{1}{\cosh x}=\frac{2}{e^{x}+e^{-x}}$ When $x\rightarrow\infty,\qquad e^{x}\rightarrow\infty\quad e^{-x}\rightarrow 0$ the numerator is constant, the denominator $\rightarrow\infty$ $\displaystyle \lim_{x\rightarrow\infty} {\rm sech} x=0$