How do you determine the limit of #(x-pi/2)tan(x)# as x approaches pi/2?

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Answer 1 · 2018-05-03T17:46:21

#lim_(xrarr(pi)/2)(x-(pi)/2)tanx=-1#

#lim_(xrarr(pi)/2)(x-(pi)/2)tanx#

#(x-(pi)/2)tanx#

#=# #(x-(pi)/2)sinx/cosx#

#(xsinx-(πsinx)/2)/cosx#

So we need to calculate this limit

#lim_(xrarrπ/2)(xsinx-(πsinx)/2)/cosx=_(DLH)^((0/0))#

#lim_(xrarrπ/2)((xsinx-(πsinx)/2)')/((cosx)'# #=#

#-lim_(xrarrπ/2)(sinx+xcosx-(πcosx)/2)/sinx# #=#

#-1#

because #lim_(xrarrπ/2)sinx=1# ,

#lim_(xrarrπ/2)cosx=0#

Some graphical help enter image source here

Answer 2 · 2018-05-03T17:54:13

For an algebraic solution, please see below.

#(x-pi/2)tanx = (x-pi/2)sinx/cosx#

# = (x-pi/2)sinx/sin(pi/2-x)#

# = (-(pi/2-x))/sin(pi/2-x) sinx#

Take limit as #xrarrpi/2# using #lim_(trarr0)t/sint = 1# to get

#lim_(xrarrpi/2)(x-pi/2)tanx = -1#