How do you determine the limit of #(x^2-2x)/(x^2-4x+4)# as x approaches 2-?

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Answer 1 · 2016-08-31T16:13:04

#-oo#

#lim_(x to 2^-) (x^2-2x)/(x^2-4x+4)#

it indeterminate so we can simplify a bit using L'Hopital's Rule

#= lim_(x to 2^-) (2x-2)/(2x-4)#

factor out the 2
#= lim_(x to 2^-) (x-1)/(x-2)#

if we now sub #x = 2 - delta# where #0 < delta " << " 1#, as it is a left sided limit then we have

#= lim_(delta to 0) (2 - delta -1)/(2 - delta-2)#

#= lim_(delta to 0) (1 - delta)/(-delta) = -oo#