How do you prove that the limit of #(3x − 5) = 1# as x approaches 2 using the epsilon delta proof?

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Answer 1 · 2017-02-16T20:53:24

GIven any #epsilon > 0# we choose #delta_epsilon < epsilon /3#

We have to prove that given any #epsilon > 0# we can find a #delta_epsilon > 0# such that:

#abs(f(x) -1) < epsilon# for #x in (2-delta_epsilon , 2+ delta_epsilon )#

We evaluate the difference:

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

and we can see that:

#abs(f(x) -1) < epsilon <=> abs (x-2)< epsilon/3#

So, given #epsilon >0# we can choose #delta_epsilon < epsilon/3# and we have:

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

which proves the point.