How do you find the limit of #[(pi)tan(x)-(2x)/(cos(x))]# as x approaches pi/2?

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Answer 1 · 2016-06-03T11:27:27

#lim_{x->pi/2}pi tan(x)-(2x)/cos(x) = 2#

#pi tan(x)-(2x)/cos(x) equiv (pi sin(x)-2x)/cos(x)#

Calculating the series expansion of
#pi sin(x)-2x# and #cos(x)# in the vicinity of #pi/2# we get

#pi sin(x)-2x = -2 (x - pi/2) - 1/2 pi(x - pi/2)^2 +cdots#
#cos(x) =-(x - pi/2) + 1/6 (x - pi/2)^3 + cdots#

then #lim_{x->pi/2}pi tan(x)-(2x)/cos(x) = 2#

Answer 2 · 2016-06-03T13:01:25

2

#LHS =lim_{x->pi/2}[pi tan(x)-(2x)/cos(x)] #

Let #" "x=pi/2-z#

#:.x->pi/2 => z->0#

Hence the given expression becomes

#lim_{z->0}[pi tan(pi/2-z)-(2(pi/2-z))/cos(pi/2-z)]#

#=lim_{z->0}[pi cot(z))-(pi-2z)/sin(z)]#

#=lim_{z->0}[(pi cos(z))/sin(z)-pi/sin(z)+2z/sin(z)]#

#=lim_{z->0}[pi /sin(z)(cos(z)-1)+2z/sin(z)]#

#=lim_{z->0}[-(2pisin^2(z/2)) /(2sin(z/2)cos(z/2))+2z/sin(z)]#

#=lim_{z->0}[-pitan(z/2)+2z/sin(z)]#

#=[-pilim_{z->0}tan(z/2)+2lim_{z->0}z/sin(z)]#

#=-pixx0+2xx1=2#