How do you show the limit does not exist #lim_(x->4)(x-4)/(x^2-8x+16)# ?
1 Answer
Aug 4, 2014
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#oo# Solution
#=lim_(x→4)((x-4)/(x^2-8x+16))# , plugging the limit we get#0/0#
#=lim_(x→4)((x-4)/(x^2-4x-4x+16))#
#=lim_(x→4)((x-4)/(x(x-4)-4(x-4)))#
#=lim_(x→4)(x-4)/((x-4)^2)#
#=lim_(x→4)1/((x-4))# Now applying the limit, we get
#=1/0=oo# , which implies limit does not exist
