Really nice one!
Making your life easier would be to get rid of #cosx# and #(0/0)#
A good idea would be to set #color(blue)x-color(blue)π/color(blue)2color(blue)=color(blue)u#
The process of thinking behind this is to create #cos(π/2+...)# which equals to #-sin(...)# which we can work with a lot easier!
#lim_(xrarrπ/2)(1/cosx-1/(π-2x))#
#x-π/2=u#
#x->π/2#
#u->0#
#=lim_(urarr0)[1/cos(π/(2)+u)-1/(π-2(π/2+u))]# #=#
#=lim_(urarr0)(-1/sinu-1/(π-2*π/2-2u))# #=#
#= lim_(urarr0)(-1/sinu-1/(π-π-2u))# #=#
#= lim_(urarr0)(-1/sinu+1/(2u))# #=#
#= lim_(urarr0)(1/(2u)-1/sinu)# #=#
#= lim_(urarr0)((sinu-2u)/(2u*sinu))#
Now here is the interesting part. I am lead to #(0/0)# once again. I can work with Rules De L' Hospital but i want to make the solution more "approachable" for everyone.
A thought now would be to create #lim_(urarr0)sinu/u# which i know equals to #1# . As you can tell though we will have problem with the denominator.
Let's see
#u->0# so that means #u!=0#
so we have #lim_(urarr0)((sinu/u-(2u)/u)/(2u*sinu/u))# #=#
#=# #lim_(urarr0)((sinu/u-2)/(2sinu))#----> #(-1/0)#
#=?#
Now we need to explore what happens near #0# for #y=sinx#
graph{sinx [-10, 10, -5, 5]}
- When #x->0^+# , #x>0# and #sinx>0#
then the #lim# leads to #-(1/0^+)# so #-(+oo)#
, #-oo#
- When #x->0^-# , #x<0# and #sinx<0#
then the #lim# leads to #-(1/0^-)# so #-(-oo)#
, #+oo#
Eventually , #lim_(urarr0)((sinu/u-2)/(2sinu))#
#=-oo# ,
when #x->0^+#
and
#+oo#
when #x->0^-#
