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

Question

Christopher Allers · Accepted Answer

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

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

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

#x->(pi)/2# so #cosx!=0#

#=# #(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

Emilia Aguilar · Answer

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#

Isabella Ackerman · Answer

undefined To determine the limit of (x - π/2)tan(x) as x approaches π/2, we can use the concept of L'Hôpital's Rule. By applying this rule, we differentiate both the numerator and denominator separately with respect to x.

Differentiating the numerator (x - π/2) gives us 1, and differentiating the denominator tan(x) gives us sec^2(x).

Now, we substitute x = π/2 into the differentiated numerator and denominator. We get 1/sec^2(π/2), which simplifies to 1/cos^2(π/2).

Since cos(π/2) equals 0, the denominator becomes 0, and the limit is undefined.