How do you use the epsilon delta definition to prove that the limit of #2x-4=6# as #x->1#?

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Answer 1 · 2017-05-09T13:11:26

It cannot be proven. It is false.

#lim_(xrarr1)(2x-4) = -2# and #-2 != 6#

Proof that #lim_(xrarr1)(2x-4) = -2#

Given #epsilon > 0#, choose #delta = epsilon/2# (Note that #delta > 0#.)

Then for any #x# with #0 < abs (x-1) < delta#,

#abs((2x-4)-(-2)) = abs(2x-2)#

# = 2abs(x-1)#

# < 2delta#

# = 2epsilon/2#

# = epsilon#

That is: #abs(2x-4)-(-2)) < epsilon#.

We have shown that:

For any #epsilon > 0#, there is a #delta > 0# such that

for all #x# with #0 < abs(x-1) < delta# we have #abs((2x-4)-(-2)) < epsilon#.

So, by the definition of limit, #lim_(xrarr3)(2x-4) = -2#.