How do you prove that the limit of #(3x+2)=8 # as x approaches 2 using the epsilon delta proof?

Question

Emily Ammerman · Accepted Answer

#delta(epsilon)=epsilon/3# #abs(f(x)-L)=abs(3x+2-8)=3abs(x-2)# Then if we choose #delta=epsilon/3# #abs(x-2) < delta=\ \ =>\ \ abs(f(x)-L)=3abs(x-2)<3delta=epsilon#

Paisley Johnson · Answer

undefined To prove that the limit of (3x+2) as x approaches 2 is 8 using the epsilon-delta proof, we need to show that for any given epsilon greater than 0, there exists a delta greater than 0 such that if 0 < |x - 2| < delta, then |(3x+2) - 8| < epsilon. Let's proceed with the proof: Given epsilon > 0, we need to find a suitable delta > 0. We start by manipulating the expression |(3x+2) - 8| < epsilon: |(3x+2) - 8| = |3x - 6| = 3|x - 2| Now, we want to find a delta such that if 0 < |x - 2| < delta, then 3|x - 2| < epsilon. Since we have control over the value of delta, we can choose it to be smaller than epsilon/3. This ensures that 3|x - 2| will be less than epsilon. Therefore, if we choose delta = epsilon/3, we can proceed with the proof. Now, let's assume that 0 < |x - 2| < delta = epsilon/3. From this assumption, we can deduce: |x - 2| < epsilon/3 Multiplying both sides by 3, we get: 3|x - 2| < epsilon Since 3|x - 2| is equal to |(3x+2) - 8|, we have: |(3x+2) - 8| < epsilon This completes the epsilon-delta proof, as we have shown that for any epsilon > 0, there exists a delta > 0 (specifically, delta = epsilon/3) such that if 0 < |x - 2| < delta, then |(3x+2) - 8| < epsilon.