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

Question

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

Christopher Agresta · Accepted Answer

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#.

Emilia Aguilar · Answer

undefined To use the epsilon-delta definition to prove that the limit of 2x-4 is 6 as x approaches 1, we need to show that for any given epsilon greater than 0, there exists a delta greater than 0 such that if 0 < |x - 1| < delta, then |(2x - 4) - 6| < epsilon. Let's proceed with the proof: Given epsilon > 0, we need to find a suitable delta > 0. | (2x - 4) - 6 | = |2x - 10| = 2|x - 5| To ensure that |(2x - 4) - 6| < epsilon, we can set 2|x - 5| < epsilon. Dividing both sides by 2, we have |x - 5| < epsilon/2. Now, we can choose delta = epsilon/2. If 0 < |x - 1| < delta, then |x - 5| < epsilon/2. Multiplying both sides by 2, we get 2|x - 5| < epsilon. Since 2|x - 5| = |(2x - 4) - 6|, we have shown that for any epsilon > 0, there exists a delta > 0 such that if 0 < |x - 1| < delta, then |(2x - 4) - 6| < epsilon. Therefore, by the epsilon-delta definition, the limit of 2x-4 is 6 as x approaches 1.

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

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