Problem 82 - Precalculus with Limits

Question

CONSUMER AWARENESS The cost of sending a package overnight is \(\$ 15\) for the first pound and \(\$ 1.30\) for each additional pound or portion of a pound. A plastic mailing bag can hold up to 3 pounds. The cost \(f(x)\) of sending a package in a plastic mailing bag is given by $$f(x)=\left\\{\begin{array}{ll}{15.00,} & {0 < x \leq 1} \\ {16.30,} & {1 < x \leq 2} \\ {17.60,} & {2 < x \leq 3}\end{array}\right.$$ where \(x\) represents the weight of the package \((\mathrm{in}\) pounds). Show that the limit of \(f\) as \(x \rightarrow 1\) does not exist.

Step-by-Step Solution

Answer

The limit of the function as \(t \rightarrow 1\) does not exist, since the left limit (\$15) and right limit (\$16.30) at \(t = 1\) are not equal.

Step 1: Computation of left-hand limit

The left-hand limit as \(t \rightarrow 1\) is computed by taking the limit of the function as \(t\) approaches 1 from values less than 1. In this case, if \(0 < t \leq 1\), then the function \(f(t)\) equals \$15. Therefore, the left limit at 1 is \$15.

Step 2: Computation of right-hand limit

The right-hand limit as \(t \rightarrow 1\) is computed by taking the limit of the function as \(t\) approaches 1 from values greater than 1. In this case, if \(1 < t \leq 2\), then the function \(f(t)\) equals \$16.30. Therefore, the right limit at 1 is \$16.30.

Step 3: Comparison of left and right-hand limit

The limit of a function at a given point exists if and only if the left-hand limit and the right-hand limit at that point are equal. As obtained from the above steps, the left limit at \(t = 1\) is \$15 and the right limit at \(t = 1\) is \$16.30. Since these two values do not match, the limit of \(f(t)\) as \(t \rightarrow 1\) does not exist.