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

1 Answer
Dec 6, 2016

There's an interesting trick you can do called L'Hopital's rule, since #cos(pi/2) = 0#, and #pi/2 - pi/2 = 0#, giving the proper #0/0# form.

#color(blue)(lim_(x-> pi"/"2) cosx/(x - pi/2)) = lim_(x-> pi"/"2) (d/(dx)[cosx])/(d/(dx)[x - pi/2])#

#= lim_(x->pi"/"2) (-sinx)/(1)#

#= -sin(pi/2)#

#= color(blue)(-1)#

Indeed, at #pi/2 ~~ 1.57#, #f(x) = -1#:

graph{cosx/(x - pi/2) [-7.023, 7.024, -3.51, 3.513]}