Problem 67 - 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 \(\lim _ { x \rightarrow 1 ^ { - } } f ( x ) ?\) \(\begin{array} { l l l l l } { \text { (A) } 5 / 2 } & { \text { (B) } 3 / 2 } & { \text { (C) } 1 } & { \text { (D) } 0 } & { \text { (E) does not exist } } \end{array}\)

Step-by-Step Solution

Answer

The value of \( \lim _ { x \rightarrow 1 ^ { - } } f ( x ) \) is \( 1 \), which corresponds to option (C) in the multiple-choice options.

Step 1: Identifying the relevant section of the function

The task is to find the left-hand limit as x approaches 1. This means we will use the part of the piece-wise function defined for \( x \leq 1 \). In this case, that part is \( 2 - x \).

Step 2: Substitute the limit value

Substitute \( x = 1 \) into this function to get \( 2 - 1 \).

Step 3: Calculate the limit

Calculate the result to give a final answer of \( 1 \).

Key Concepts

Limits in Piecewise Functions

When dealing with piecewise functions, limits help us understand behavior as we approach a certain point. A piecewise function shows different behaviors based on the input value.
For our function, \[f(x) = \left\{ \begin{array}{ll} 2-x, & x \leq 1 \ \frac{x}{2}+1, & x > 1 \end{array} \right.\]limits are essential for determining how the function behaves as we near specific input values.
With piecewise functions, we carefully evaluate both parts around the input of interest, in this case, around 1. The key takeaway here is that the limit at a point, if it exists, must be the same from all directions as you near that point.

Understanding the Left-Hand Limit

The left-hand limit explores what occurs to a function as we approach a specific value from the left (or less than that value). For a piecewise function like ours, you choose the part relevant for values smaller than or equal to the point.
For our exercise, the left-hand limit of \(x\) approaching 1 is found using the equation \(2 - x\).
By substituting \(x = 1\) directly into this equation, the expected outcome is straightforward.As noted, substituting \(x=1\) gives \(2-1\), which simplifies to 1. This result is the left-hand limit as \(x\) approaches 1 from the left-hand side.

Evaluating Piecewise Functions

Evaluating piecewise functions, especially near boundaries, requires attention to each separate function section. Since each part of the function may behave differently, evaluating means substituting the specific values into the correct segment of the function.
It's crucial to note which condition or part of the piecewise function applies to your point of interest.
For example,

• If \(x \leq 1\), use the expression \(2-x\).
• If \(x > 1\), use \(\frac{x}{2} + 1\).

Understanding these conditions ensures the correct part of the function is evaluated, as seen in determining a specific limit or function value.