Answer:
Explanation:
To prove that the limit of (1 - 3x) as x approaches infinity is -2, we need to show that for every positive number e, there exists a positive number 8 such that whenever 0 < |x - 8| < e, then |(1 - 3x) - (-2)| < e.
Let's start by manipulating the expression |(1 - 3x) - (-2)| < e:
|(1 - 3x) + 2| < e
|3 - 3x| < e
|3(x - 1)| < e
3|x - 1| < e
From here, we need to set an upper bound for |x - 1| based on the given inequality.
We want to find 8 such that if |x - 1| < d, then 3|x - 1| < e where d is some positive number.
Assume d = e/3. This allows us to rewrite the inequality as:
|3(x - 1)| = 3|x - 1| < 3(d) = 3(e/3) = e
Therefore, if we choose 8 such that |x - 1| < e/3 (where e > 0), then 3|x - 1| < e, which satisfies |(1 - 3x) - (-2)| < e.
Hence, by definition, the limit of (1 - 3x) as x approaches infinity is -2.