Problem 69 - Calculus: Graphical, Numerical, Algebraic

Question

In Exercises \(67 - 70\) , use the following function. \(f ( x ) = \left\\{ \begin{array} { l l } { 2 - x , } & { x \leq 1 } \\ { \frac { x } { 2 } + 1 , } & { x > 1 } \end{array} \right.\) Multiple Choice What is the value of \(\lim _ { x \rightarrow 1 } f ( x ) ?\) \(( \mathrm { A } ) 5 / 2 \quad ( \mathrm { B } ) 3 / 2\) (C) 1 (D) 0 (E) does not exist

Step-by-Step Solution

Answer

(E) does not exist

Step 1: Analyze the function pieces

As the given function is a piecewise function, the first step would be to decide which part of the function should be used for \(x = 1\). We can see that \(x = 1\) lies in the first part of the function \(2 - x\). This is because the first piece of the function is defined for \(x \leq 1\). So we will use the equation \(2 - x\) for \(x = 1\).

Step 2: Substitute the limit value

The next step would be to substitute \(x = 1\) into the function piece \(2 - x\). By doing this, we get \(2 - 1 = 1\).

Step 3: Confirming continuity of function

We also need to examine the value of the function at \(x > 1\) to ensure its continuity. Again substituting the value \(x = 1\) into the second equation, \(\frac{x}{2} + 1\), we get \(\frac{1}{2} + 1 = 1.5\) which is not equal to 1. Hence, the given function is not continuous at \(x = 1\).

Step 4: Conclusion

From Step 2 and Step 3, we can conclude that the limit \( \lim _ { x \rightarrow 1 } f ( x )\) does not exist, because the function does not approach the same value from both sides of 1.

Key Concepts

Piecewise Functions

Piecewise functions are mathematical expressions defined by multiple sub-functions. Each sub-function applies to a specific interval of the overall domain. This means that the formula used to calculate the function changes depending on which interval or piece the input belongs to.

For example, in our exercise, we have a piecewise function given by:

• First piece: if \( x \leq 1\), use the formula \( f(x) = 2 - x \).
• Second piece: if \( x > 1 \), use the formula \( f(x) = \frac{x}{2} + 1 \).

The challenge often lies in evaluating these functions at points where definitions switch from one piece to another, such as \( x = 1 \) in this case. To properly analyze the behavior, especially at boundary points, we can assess limits and continuity to understand how the function behaves when approaching such critical points.

Continuity

Continuity in a function implies that there are no abrupt changes, jumps, or breaks in its graph. A function is considered continuous at a certain point if, as you approach that point from either direction, the function values come together and are equal to the function's value at that point.

When dealing with a piecewise function, continuity becomes crucial at the points where the function pieces meet. To ensure continuity at a point "c", the following must hold:

• The limit from the left \( \lim_{x \to c^-} f(x) \) must equal the limit from the right \( \lim_{x \to c^+} f(x) \).
• Both limits must equal \( f(c) \), the function's actual value at \( c \).

In our function, we see that while the piece for \( x \leq 1 \) matches the actual value at \( x = 1 \), the piece for \( x > 1 \) does not. Thus, the function is not continuous at \( x = 1 \), showcasing a 'jump' — a key indicator that a function piecewise name may not be continuous at a boundary point.

Evaluating Limits

Evaluating limits involves determining the value that a function approaches as its input approaches a certain point. This is especially important for piecewise functions, where transitioning between pieces may introduce discrepancies.

In the given function, to evaluate the limit as \( x \to 1 \), you analyze each piece separately:

• For \( x \leq 1 \), using \( f(x) = 2 - x \), substituting \( x = 1 \) yields \( 1 \).
• For \( x > 1 \), using \( f(x) = \frac{x}{2} + 1 \), substituting \( x = 1 \) gives \( 1.5 \).

Since these values are not equal, the limit \( \lim_{x \to 1} f(x) \) does not exist. This happens because the function does not approach a single, unified value as \( x \) nears 1 from both directions. Understanding limits helps us make sense of behaviors in piecewise functions, especially at transition points, by determining whether a smooth transition occurs or if there exists a break or jump.