The correct value of δ when ε = 0.001 for the limit statement limx→7(2x−6)=8 is δ = 0.0005.
How to solve
Epsilon-delta definition of a limit: A limit statement limx→aL=L means that for any positive number ε, there exists a positive number δ such that whenever 0 < |x - a| < δ, we have |f(x) - L| < ε.
Apply the definition to the given problem:
We want to find δ such that whenever 0 < |x - 7| < δ, we have |2x - 6 - 8| = |2x - 14| < ε = 0.001.
Solve the inequality: |2x - 14| < 0.001 Since the absolute value of an expression is its distance from zero, we can write this inequality as: -2(0.001) < 2x - 14 < 2(0.001)
Simplifying, we get: -13.998 < 2x < 14.002
Dividing everything by 2, we obtain: -6.999 < x < 7.001
Choose δ: The above inequality tells us that if we choose any value of δ less than 0.001 (e.g., 0.0005), then any x such that 0 < |x - 7| < δ will satisfy the condition |2x - 14| < ε = 0.001.
Therefore, the correct value of δ when ε = 0.001 is δ = 0.0005.