Chapter 6 - Applications of Integration - 6.10 Hyperbolic Functions - 6.10 Exercises - Page 505: 84

$$ - 1$$

$$\eqalign{ & \mathop {\lim }\limits_{x \to \infty } \frac{{1 - \coth x}}{{1 - \tanh x}} \cr & {\text{evaluating the limit}} \cr & \mathop {\lim }\limits_{x \to \infty } \frac{{1 - \coth x}}{{1 - \tanh x}} = \frac{{1 - \coth \left( \infty \right)}}{{1 - \tanh \left( \infty \right)}} \cr & {\text{simplifying}} \cr & \mathop {\lim }\limits_{x \to \infty } \frac{{1 - \coth x}}{{1 - \tanh x}} = \frac{{1 - 1}}{{1 - 1}} = \frac{0}{0} \cr & {\text{using the L'hopital's Rule }} \cr & \mathop {\lim }\limits_{x \to \infty } \frac{{1 - \coth x}}{{1 - \tanh x}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{d}{{dx}}\left[ {1 - \coth x} \right]}}{{\frac{d}{{dx}}\left[ {1 - \tanh x} \right]}} \cr & {\text{solving derivatives}} \cr & \mathop {\lim }\limits_{x \to \infty } \frac{{1 - \coth x}}{{1 - \tanh x}} = \mathop {\lim }\limits_{x \to \infty } \frac{{ - \left( { - {{\operatorname{csch} }^2}x} \right)}}{{ - \left( {{{\operatorname{sech} }^2}x} \right)}} \cr & = - \mathop {\lim }\limits_{x \to \infty } \frac{{{{\operatorname{csch} }^2}x}}{{{{\operatorname{sech} }^2}x}} \cr & or \cr & = - \mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{\operatorname{csch} x}}{{\operatorname{sech} x}}} \right)^2} \cr & = - \mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{\cosh x}}{{\sinh x}}} \right)^2} \cr & = - \mathop {\lim }\limits_{x \to \infty } {\left( {cothx} \right)^2} \cr & {\text{evaluate the limit}} \cr & = - {\left( {\coth \infty } \right)^2} \cr & = - {\left( 1 \right)^2} \cr & = - 1 \cr & {\text{then}} \cr & \mathop {\lim }\limits_{x \to \infty } \frac{{1 - \coth x}}{{1 - \tanh x}} = - 1 \cr} $$