Answer:To prove the given limit statements using the epsilon-delta definition of a limit, we need to show that for any ε (epsilon) greater than 0, we can find a δ (delta) greater than 0 such that if 0 < |x - c| < δ, then |f(x) - L| < ε, where L is the given limit.
Let's start with the first limit statement:
**(a) lim (5x + 1) = 21**
We need to prove that for any ε > 0, there exists a δ > 0 such that if 0 < |x| < δ, then |(5x + 1) - 21| < ε.
Let's work through the proof:
Given ε > 0, we want to find δ > 0 such that |(5x + 1) - 21| < ε whenever 0 < |x| < δ.
We have |(5x + 1) - 21| = |5x - 20| = 5|x - 4|.
We want this expression to be less than ε. So, we need to make sure that 5|x - 4| < ε.
To do this, we can choose δ = ε / 5.
Now, let's analyze the inequality 0 < |x - c| < δ:
0 < |x - 4| < ε / 5.
Since ε > 0, ε / 5 > 0 as well. Therefore, if |x - 4| < ε / 5, it implies that 5|x - 4| < ε.
This completes the proof for the first limit statement. We've shown that for any ε > 0, we can find a δ > 0 (specifically, δ = ε / 5) such that if 0 < |x - 4| < δ, then |(5x + 1) - 21| < ε. This satisfies the epsilon-delta definition of the limit lim (5x + 1) = 21.
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Now, let's move on to the second limit statement:
**(b) lim (x² - 9) = -8**
We need to prove that for any ε > 0, there exists a δ > 0 such that if 0 < |x - c| < δ, then |(x² - 9) - (-8)| < ε.
Let's work through the proof:
Given ε > 0, we want to find δ > 0 such that |(x² - 9) - (-8)| < ε whenever 0 < |x - c| < δ.
We have |(x² - 9) - (-8)| = |x² + 1|.
We want this expression to be less than ε. So, we need to make sure that |x² + 1| < ε.
However, for any value of x, we have |x² + 1| ≥ 1, since x² is always non-negative.
Therefore, we can choose δ = ε.
Now, let's analyze the inequality 0 < |x - c| < δ:
0 < |x - c| < ε.
Since δ = ε, it directly implies that |x² + 1| < ε.
This completes the proof for the second limit statement. We've shown that for any ε > 0, we can choose δ = ε such that if 0 < |x - c| < δ, then |(x² - 9) - (-8)| < ε. This satisfies the epsilon-delta definition of the limit lim (x² - 9) = -8.
These proofs demonstrate that the given limit statements hold true based on the epsilon-delta definition of a limit.
Explanation: