Question

Evaluate the given limits of the piecewise defined functions \(f\). \(f(x)=\left\\{\begin{array}{cl}x+1 & x \leq 1 \\ x^{2}-5 & x>1\end{array}\right.\) (a) \(\lim _{x \rightarrow 1^{-}} f(x)\) (c) \(\lim _{x \rightarrow 1} f(x)\) (b) \(\lim _{x \rightarrow 1^{+}} f(x)\) (d) \(f(1)\)

Short answer

The left-hand limit is 2, right-hand limit is -4, the two-sided limit does not exist, and \( f(1) = 2 \).

Step-by-step solution

Identify the Left-Hand Limit

To evaluate \ \lim _{x \rightarrow 1^{-}} f(x) \, note that \(x < 1\). According to the piecewise function, if \(x \leq 1\), \(f(x) = x+1\). Therefore, \ \lim _{x \rightarrow 1^{-}} f(x) = \lim _{x \rightarrow 1^{-}} (x+1) = 1+1 = 2. \

Identify the Right-Hand Limit

To evaluate \ \lim _{x \rightarrow 1^{+}} f(x) \, note that \(x > 1\). According to the piecewise function, if \(x > 1\), \(f(x) = x^{2} - 5\). Therefore, \ \lim _{x \rightarrow 1^{+}} f(x) = \lim _{x \rightarrow 1^{+}} (x^{2}-5) = 1^{2} - 5 = -4. \

Evaluate the Two-Sided Limit

To find \ \lim _{x \rightarrow 1} f(x) \, check if both side limits are equal. Previously, we found \ \lim _{x \rightarrow 1^{-}} f(x) = 2 \ and \ \lim _{x \rightarrow 1^{+}} f(x) = -4. \ Since \(2 eq -4\), the two-sided limit does not exist.

Evaluate the Value of the Function at the Point

To find \(f(1)\), use the piecewise definition: when \(x \leq 1\), \(f(x) = x + 1\). Thus, \(f(1) = 1 + 1 = 2\).

Key concepts

Limits

In calculus, the concept of limits helps us understand the behavior of functions as the input approaches a certain point. When we evaluate limits, we are not interested in the value of the function at that exact point, but rather the values the function's output approaches as we get infinitely close to the point in question. In the context of a piecewise function like the one given, limits can differ depending on which piece of the function is applicable as the input approaches the specified point. For example, checking

• The left-hand limit considers inputs approaching the point from the negative side.
• The right-hand limit considers inputs approaching from the positive side.

By considering these different approaches, we gain detailed insights about the function's behavior around that point.

Continuity

Continuity is an important concept that indicates whether a function has any 'jumps' or breaks as you move along its graph. A function is considered continuous at a point if the graph can be drawn without lifting the pencil at that point. Mathematically, a function is continuous at a point if:

• The limit as the input approaches the point from the left exists.
• The limit as the input approaches from the right exists.
• These two limits are equal to each other and also equal to the function's actual value at that point.

For our piecewise function, we found that the left-hand limit as we approached 1 is 2, but the right-hand limit is -4. Because these are not equal, the function has a jump discontinuity at that point.

Left-Hand Limit

The left-hand limit of a function focuses on the behavior of the function as the input value approaches a specific point from the left or negative side. It's expressed as \(\lim_{x \to 1^-} f(x)\), meaning we look at values of x that are less than, but getting very close to, 1. For our piecewise function:

• When \(x \leq 1\), the appropriate piece is \(f(x) = x + 1\).
• So, \(\lim_{x \to 1^{-}} (x + 1) = 1 + 1 = 2\).

This calculation shows that as x approaches 1 from values less than 1, the function's output approaches 2.

Right-Hand Limit

The right-hand limit examines the function's behavior as the input approaches the specific point from the right, or positive side. It's represented as \(\lim_{x \to 1^+} f(x)\). For our piecewise function where:

• When \(x > 1\), the function follows \(f(x) = x^2 - 5\).
• So, \(\lim_{x \to 1^{+}} (x^{2} - 5) = 1^{2} - 5 = -4\).

This demonstrates that as x approaches 1 from values greater than 1, the output of the function approaches -4, which is a different result from the left-hand limit. Hence, there is an observable discontinuity.