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

Question

Samantha Adkison · Accepted Answer

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.

Charles Arocha · Answer

undefined To prove that the limit of (3x - 5) as x approaches 2 is equal to 1 using the epsilon-delta proof, we need to show that for any given epsilon greater than 0, there exists a corresponding delta greater than 0 such that whenever 0 < |x - 2| < delta, then |(3x - 5) - 1| < epsilon. Let's proceed with the proof: Given epsilon > 0, we need to find a delta > 0 such that |(3x - 5) - 1| < epsilon whenever 0 < |x - 2| < delta. Now, let's manipulate the expression |(3x - 5) - 1| < epsilon: |(3x - 5) - 1| = |3x - 6| = 3|x - 2| We want to find a delta such that 3|x - 2| < epsilon whenever 0 < |x - 2| < delta. Since we have control over the expression |x - 2|, we can manipulate it to fit our needs: |x - 2| < delta Dividing both sides by 3: |x - 2|/3 < delta/3 Now, we can choose delta/3 = epsilon, which implies delta = 3epsilon. Therefore, for any given epsilon > 0, if we choose delta = 3epsilon, then whenever 0 < |x - 2| < delta, we have: 3|x - 2| < 3(3epsilon) = 9epsilon Hence, we have shown that for any epsilon > 0, there exists a delta > 0 (specifically, delta = 3epsilon) such that whenever 0 < |x - 2| < delta, we have |(3x - 5) - 1| < epsilon. This proves that the limit of (3x - 5) as x approaches 2 is equal to 1 using the epsilon-delta proof.