How do you prove that the limit # (x^2 - 4x + 5) = 1# as x approaches 2 using the formal definition of a limit?

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Answer 1 · 2017-02-06T20:31:38

see below

Recall that #lim_(x→a)⁡f(x)=L# ,

if for every number ε>0 there is a number δ>0 such that

|f(x)-L|<ε whenever 0<|x-a|<δ

So from #lim_(x->2) x^2-4x+5=1, L=1 and a=2#

#abs(f(x)-L) < epsilon if 0 < abs(x-a) < delta#

#abs(x^2 -4x +5 -1) < epsilon if 0 < abs (x-2) < delta#

#abs(x^2 -4x +4) < epsilon #

#abs((x-2)^2 ) < epsilon#

#abs(x-2 )^2 < epsilon#

#abs(x-2) < sqrt epsilon#

#:. delta = sqrt epsilon#

Verify:

#abs(x^2 -4x +4) #

#=abs((x-2)^2 )#

#=abs(x-2 )^2#

#=(sqrt epsilon )^2#

#= epsilon#

Hence,

#lim_(x->2) x^2-4x+5=1#