How do you verify #lim x->2# of #(3x-5)=1#, using the definition of the limit?

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Answer 1 · 2018-03-06T23:31:55

Evaluate the difference:

#abs(3x-5-1) = abs(3x-6) = 3abs(x-2)#

Given any number #epsilon >0# choose #delta_epsilon < epsilon/3#.
For #x in (2-delta_epsilon, 2+delta_epsilon)# we have:

#abs(x-2) < delta_epsilon#

and then:

#abs(3x-5-1) = 3abs(x-2) <3delta_epsilon < 3epsilon/3 = epsilon#

In conclusion:

#x in (2-delta_epsilon, 2+delta_epsilon) => abs(3x-5-1) < epsilon#

which proves the limit.