What is the limit of #7/(4(x-1)^2)# as x approaches 1?

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Answer 1 · 2018-04-16T16:58:04

Look below

First, rewrite this as #lim_(x->1) 7/(4(x-1)^2#

now factor #(x-1)^2# = #(x-1)(x-1)# = #x^2-2x+1#

#\frac{7}{4x^2-2x+1}#

now substitute x #-># 1

#\frac{7}{4(1)^2 -2(1)+1#

#7/3#

#therefore lim_(x->1) 7/(4(x-1)^2) = 7/6#