Is that infinite ?

Question

#lim x-> pi/2 = ((x-pi/2))/((1-sinx)cosx) #

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Answer 1 · 2017-03-01T14:36:09

#-oo#

Calling #y=x-pi/2# we have

# ((x-pi/2))/((1-sinx)cosx) = -y/((1-cos(y))sin(y))#

and also

#lim_( x-> pi/2) ((x-pi/2))/((1-sinx)cosx) =lim_(y->0) -y/((1-cos(y))sin(y))=# #= lim_(y->0)1/(1-cos(y))lim_(y->0)(-y/siny) = oo (-1) = -oo#