Chapter 6 - Exponential, Logarithmic, And Inverse Trigonometric Functions - 6.5 L'Hopital's Rule; Indeterminate Forms - Exercises Set 6.5 - Page 448: 43

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$$\eqalign{ & \mathop {\lim }\limits_{x \to \pi /{2^ - }} {\left( {\tan x} \right)^{\left( {\pi /2} \right) - x}} \cr & {\text{Evaluating}} \cr & \mathop {\lim }\limits_{x \to \pi /{2^ - }} {\left( {\tan x} \right)^{\left( {\pi /2} \right) - x}} = {\left( {\tan \frac{\pi }{2}} \right)^{\left( {\pi /2} \right) - \pi /2}} = {\infty ^0} \cr & {\text{This limit has the form }}{\infty ^0}.{\text{ }} \cr & {\left( {\tan x} \right)^{\left( {\pi /2} \right) - x}} = {e^{\left( {\frac{\pi }{2} - x} \right)\ln \left( {\tan x} \right)}},{\text{ then}} \cr & \mathop {\lim }\limits_{x \to \pi /{2^ - }} {\left( {\tan x} \right)^{\left( {\pi /2} \right) - x}} = \mathop {\lim }\limits_{x \to \pi /{2^ - }} {e^{\left( {\frac{\pi }{2} - x} \right)\ln \left( {\tan x} \right)}} = {e^{\mathop {\lim }\limits_{x \to \pi /{2^ - }} \left( {\frac{\pi }{2} - x} \right)\ln \left( {\tan x} \right)}} \cr & {\text{The first step is to evaluate }} \cr & L = \mathop {\lim }\limits_{x \to \pi /{2^ - }} \left( {\frac{\pi }{2} - x} \right)\ln \left( {\tan x} \right) = \mathop {\lim }\limits_{x \to \pi /{2^ - }} \frac{{\ln \left( {\tan x} \right)}}{{\frac{1}{{\left( {\pi /2 - x} \right)}}}} = \frac{\infty }{\infty } \cr & {\text{Using the L'Hopital's rule}} \cr & L = \mathop {\lim }\limits_{x \to \pi /{2^ - }} \frac{{\ln \left( {\tan x} \right)}}{{\frac{1}{{\left( {\pi /2 - x} \right)}}}} = \mathop {\lim }\limits_{x \to \pi /{2^ - }} \frac{{\frac{{{{\sec }^2}x}}{{\tan x}}}}{{\frac{1}{{{{\left( {\pi /2 - x} \right)}^2}}}}} \cr & L = \mathop {\lim }\limits_{x \to \pi /{2^ - }} \frac{{{{\left( {\pi /2 - x} \right)}^2}}}{4}\left( {\frac{1}{{\sin x}}} \right)\left( {\frac{1}{{\cos x}}} \right) \cr & L = \mathop {\lim }\limits_{x \to \pi /{2^ - }} \frac{{\pi /2 - x}}{{\sin x}}\mathop {\lim }\limits_{x \to \pi /{2^ - }} \frac{{\pi /2 - x}}{{\cos x}} \cr & L = \mathop {\lim }\limits_{x \to \pi /{2^ - }} \frac{{\pi /2 - x}}{{\sin x}}\mathop {\lim }\limits_{x \to \pi /{2^ - }} \frac{{\pi /2 - x}}{{\cos x}} \cr & L = \mathop {\lim }\limits_{x \to \pi /{2^ - }} \frac{{\pi /2 - x}}{{\sin x}}\mathop {\lim }\limits_{x \to \pi /{2^ - }} \frac{{ - 1}}{{\sin x}} \cr & {\text{Evaluating}} \cr & L = \left( {\frac{{\pi /2 - \pi /2}}{{\sin \pi /2}}} \right)\left( {\frac{{ - 1}}{{\sin \pi /2}}} \right) \cr & L = 0\left( { - 1} \right) \cr & L = 0 \cr & {\text{Therefore,}} \cr & \mathop {\lim }\limits_{x \to \pi /{2^ - }} {\left( {\tan x} \right)^{\left( {\pi /2} \right) - x}} = \mathop {\lim }\limits_{x \to \pi /{2^ - }} {e^{\left( {\frac{\pi }{2} - x} \right)\ln \left( {\tan x} \right)}} = {e^{\mathop {\lim }\limits_{x \to \pi /{2^ - }} \left( {\frac{\pi }{2} - x} \right)\ln \left( {\tan x} \right)}} \cr & \mathop {\lim }\limits_{x \to \pi /{2^ - }} {\left( {\tan x} \right)^{\left( {\pi /2} \right) - x}} = {e^0} \cr & \mathop {\lim }\limits_{x \to \pi /{2^ - }} {\left( {\tan x} \right)^{\left( {\pi /2} \right) - x}} = 1 \cr & \cr} $$