How do you find the limit of #(1/(x-1)+1/(x^2-3x+2))# as #x->1#?

2 Answers
Eddie
Feb 9, 2017

#-1#

Explanation:

#lim_(x to 1) (1/(x-1)+1/(x^2-3x+2))#

#= lim_(x to 1) (1/(x-1)+1/((x-1)(x-2)))#

if we let #x-1 = delta #

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

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

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

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

Steve M
Feb 9, 2017

# lim_(x to 1) 1/(x-1)+1/(x^2-3x+2) =-1 #

Explanation:

# lim_(x to 1) 1/(x-1)+1/(x^2-3x+2) = lim_(x to 1) 1/(x-1)+1/((x-1)(x-2)) #
# " "= lim_(x to 1) ((x-2)+1)/((x-1)(x-2)) #
# " "= lim_(x to 1) ((x-1))/((x-1)(x-2)) #
# " "= lim_(x to 1) (1)/((x-2)) #
# " "= (1)/((1-2)) #
# " "= -1 #

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