Problem 54 - Calculus and its applications

Question

Is the function given by $$ F(x)=\left\\{\begin{array}{ll} \frac{1}{3} x+4, & \text { for } x \leq 3 \\ 2 x-5, & \text { for } x>3 \end{array}\right. $$ continuous at \(x=3 ?\) Why or why not?

Step-by-Step Solution

Answer

No, the function is not continuous at \(x=3\) because the left and right limits are not equal.

Step 1: Evaluate Left-Hand Limit at \(x=3\)

To check for continuity at \(x=3\), we first find the left-hand limit, \(\lim_{{x \to 3^-}} F(x)\). Since \(x \leq 3\) uses the function \(\frac{1}{3}x + 4\), the left-hand limit is \(\lim_{{x \to 3}} \frac{1}{3}x + 4 = \frac{1}{3}(3) + 4 = 1 + 4 = 5\).

Step 2: Evaluate Right-Hand Limit at \(x=3\)

Next, we calculate the right-hand limit, \(\lim_{{x \to 3^+}} F(x)\). For \(x > 3\), the function used is \(2x - 5\), so the right-hand limit is \(\lim_{{x \to 3}} 2x - 5 = 2(3) - 5 = 6 - 5 = 1\).

Step 3: Compare Left-Hand and Right-Hand Limits

For a function to be continuous at a point, both the left-hand and right-hand limits must be equal, and they must also equal the function's value at that point. We found that the left-hand limit is 5 and the right-hand limit is 1. Since these are not equal, \(F(x)\) is not continuous at \(x=3\).

Step 4: Evaluate \(F(3)\)

Finally, we check the value of \(F(3)\). From the function definition \(x \leq 3\), \(F(3) = \frac{1}{3}(3) + 4 = 5\). However, since the limits are not equal, this confirms the function is not continuous.

Key Concepts

Left-Hand Limit

When we analyze the continuity of a piecewise function like \[ F(x) = \begin{cases} \frac{1}{3} x + 4, & \text{if } x \leq 3 \ 2x - 5, & \text{if } x > 3 \end{cases} \]one essential part is understanding the left-hand limit at a particular point. For our example, this point is \(x=3\).

• The left-hand limit is denoted as \( \lim_{{x \to 3^-}} F(x) \), which means we are approaching 3 from values less than 3.
• For \(x \leq 3\), the relevant part of the function is \(\frac{1}{3}x + 4\).

To compute the left-hand limit, substitute 3 into this function:\[ \lim_{{x \to 3}} \frac{1}{3}x + 4 = \frac{1}{3}(3) + 4 = 1 + 4 = 5. \]The left-hand limit evaluates to 5. This value represents the behavior of the function from the left side of 3, which is crucial in determining continuity.

Right-Hand Limit

Next, we consider the right-hand limit of the function near the same point, \(x=3\). For the continuity check, this means calculating how the function behaves as we approach 3 from numbers greater than 3.

• The right-hand limit is notated as \( \lim_{{x \to 3^+}} F(x) \).
• In our piecewise function, for \(x > 3\) the expression \(2x - 5\) is applicable.

To find this limit, we substitute 3 into the expression for the right-hand side:\[ \lim_{{x \to 3}} 2x - 5 = 2(3) - 5 = 6 - 5 = 1. \]This right-hand limit of 1 differs from our previously found left-hand limit of 5. When these two limits are not equal, the function shows a break or jump at that point.

Function Evaluation

Function evaluation involves substituting a specific value into the function to find its value at a given point. For a function to be continuous at a point \(x=a\), there are a few conditions that must be met:

• The left-hand limit and right-hand limit must both exist and be equal: in our case, previously calculated as 5 and 1, respectively.
• The function must also be defined at \(x=3\).

In our function, the value is determined using the part of the piecewise function where \(x=3\) fits, which is \(\frac{1}{3}x + 4\):\[ F(3) = \frac{1}{3}(3) + 4 = 5. \]Although \(F(3)\) itself is 5, because our left-hand limit at \(x=3\) is 5 and the right-hand limit is 1, these values mismatch. Therefore, despite the function being defined and evaluable at \(x=3\), the function is not continuous at this point.