How do you prove that the limit of #x-2 = 2# as x approaches 3 using the epsilon delta proof?

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Answer 1 · 2016-04-19T13:38:05

You can't prove it because it is false. #lim_(xrarr3)(x - 2) != 2#

You can prove thar #lim_(xrarr3)(x-2) = 1#

Proof

Let #epsilon > 0# be given. Choose #delta = epsilon#

Now for every #x# with #0 < abs(x-3) < delta#, we get

#abs(f(x)-L) = abs((x-2)-1) = abs(x-3) < delta = epsilon#.

We have shown that
for any #epsilon > 0#, there is a delta such that #0 < abs(x-2) < delta# implies #abs((x-2)-1) < epsilon#

Therefore, by the definition of limit, #lim_(xrarr3)(x-2) = 1#

Also, using a similar argument
we could prove that #lim_(xrarr3)(x-1) = 2#.

Sketch: again choose #delta = epsilon#

Then show that #0 < abs(x-3) < delta# implies #abs((x-1)-2) < epsilon#