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Chapter 1: Problem 34

Find the limit \(L\). Then use the \(\varepsilon-\delta\) definition to prove that the limit is \(L\). $$ \lim _{x \rightarrow-3}(2 x+5) $$

Short Answer

Expert verified
The limit \(L\) of the function as \(x\) approaches -3 is -1. The limit can be demonstrated using the 蔚-未 proof: for any 蔚 > 0, there is a 未 = 蔚/2 such that if \(0 < |x + 3| < 未\), then \( |2x + 6| = 2|x + 3| < 2未 = 蔚\).

Step by step solution

01

Identify and Solve the Limit

The limit of the function \(f(x) = 2x + 5\) as \(x\) approaches -3 can be found by simple substitution: \(f(x) = 2(-3) + 5 = -1\). So the limit \(L = -1\).
02

Implement the 蔚-未 Definition

Now that the limit \(L = -1\) is determined, the task is to find a suitable value for 未 for any given value of 蔚. If \(0 < |x+3| < 未\), then we want to ensure that |(2x+5) - (-1)| < 蔚.\n\nStart by simplifying the expression to the left: |(2x+5) - (-1)| = |2x + 6| = 2|x + 3|. Given that \(0 < |x + 3| < 未\), we can say \(2|x + 3| < 2未\).\n\nTo fulfill the condition \(2|x + 3| < 蔚\), we choose 未 = 蔚/2. Then for any 蔚 > 0, there is a 未 = 蔚/2 such that if \(0 < |x+3| < 未\), then \( |2x + 6| = 2|x + 3| < 2未 = 蔚\). This proves the limit using the 蔚-未 definition.

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