How do you evaluate #lim x->pi# of #(sinx)/(x-pi)# without using L'Hopital's Rule?

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Answer 1 · 2016-07-23T16:43:48

-1

#lim_(x to pi) (sinx)/(x-pi)#

let #z = x - pi, x = z + pi#

#=lim_(z to 0) (sin(z + pi))/(z)#

#=lim_(z to 0) (sinz cos pi + sin pi cos z)/(z)#

#= lim_(z to 0) (-sinz )/(z) = -1#

#=- lim_(z to 0) (sinz )/(z) = -1#

as #(sinz )/(z)|_(z to 0 ) = 1# is a well know limit