Question

Evaluate the given limits of the piecewise defined functions \(f\). \(f(x)=\left\\{\begin{array}{cc}x^{2}-1 & x<-1 \\ x^{3}+1 & -1 \leq x \leq 1 \\\ x^{2}+1 & x>1\end{array}\right.\) (a) \(\lim _{x \rightarrow-1^{-}} f(x)\) (e) \(\lim _{x \rightarrow 1^{-}} f(x)\) (b) \(\lim _{x \rightarrow-1^{+}} f(x)\) (f) \(\lim _{x \rightarrow 1^{+}} f(x)\) (c) \(\lim _{x \rightarrow-1} f(x)\) (g) \(\lim _{x \rightarrow 1} f(x)\) (d) \(f(-1)\) (h) \(f(1)\)

Short answer

\( \lim _{x \rightarrow -1} f(x) = 0 \); \( \lim _{x \rightarrow 1} f(x) = 2 \); \( f(-1) = 0 \); \( f(1) = 2 \).

Step-by-step solution

Evaluate the left-hand limit at x = -1

For \( \lim _{x \rightarrow-1^{-}} f(x) \), consider the piece where \( x < -1 \). Here, \( f(x) = x^2 - 1 \). As \( x \rightarrow -1 \), the value becomes \((-1)^2 - 1 = 0\). Thus, \( \lim _{x \rightarrow-1^{-}} f(x) = 0 \).

Evaluate the right-hand limit at x = -1

For \( \lim _{x \rightarrow-1^{+}} f(x) \), consider the piece where \( -1 \leq x \leq 1 \). Here, \( f(x) = x^3 + 1 \). Evaluate this as \( x \rightarrow -1 \) from the right, which gives \((-1)^3 + 1 = 0\). Thus, \( \lim _{x \rightarrow-1^{+}} f(x) = 0 \).

Evaluate the two-sided limit at x = -1

Since both the left-hand and right-hand limits at \( x = -1 \) are equal, \( \lim _{x \rightarrow-1} f(x) = 0 \).

Evaluate f(-1)

For \( f(-1) \), the value of \( x \) falls in the piece \(-1 \leq x \leq 1\), which gives \( f(x) = x^3 + 1 \). Thus, \( f(-1) = (-1)^3 + 1 = 0\).

Evaluate the left-hand limit at x = 1

For \( \lim _{x \rightarrow 1^{-}} f(x) \), consider the piece where \( -1 \leq x \leq 1 \). Here, \( f(x) = x^3 + 1 \). As \( x \rightarrow 1 \) from the left, evaluate as \( 1^3 + 1 = 2 \). Thus, \( \lim _{x \rightarrow 1^{-}} f(x) = 2 \).

Evaluate the right-hand limit at x = 1

For \( \lim _{x \rightarrow 1^{+}} f(x) \), consider the piece where \( x > 1 \). Here, \( f(x) = x^2 + 1 \). As \( x \rightarrow 1 \), evaluate as \( 1^2 + 1 = 2 \). Thus, \( \lim _{x \rightarrow 1^{+}} f(x) = 2 \).

Evaluate the two-sided limit at x = 1

Since both the left-hand and right-hand limits at \( x = 1 \) are equal, \( \lim _{x \rightarrow 1} f(x) = 2 \).

Evaluate f(1)

For \( f(1) \), the value of \( x \) is exactly \( 1 \), falling in the piece \(-1 \leq x \leq 1\), which gives \( f(x) = x^3 + 1 \). Thus, \( f(1) = 1^3 + 1 = 2\).

Key concepts

Piecewise Functions

Piecewise functions are mathematical expressions defined by multiple sub-functions, each with its own domain. They are like a set of rules that tell you which formula to use depending on the value of the variable. In our exercise, the piecewise function \( f(x) \) is defined as follows:

• For \( x < -1 \), \( f(x) = x^2 - 1 \)
• For \( -1 \leq x \leq 1 \), \( f(x) = x^3 + 1 \)
• For \( x > 1 \), \( f(x) = x^2 + 1 \)

To evaluate limits or find specific function values, you must determine which piece of the function applies to the given \( x \) value or limit approach. This requires careful observation of the domain specifications of each piece of the function.

Continuity

Continuity at a point means that the function is smooth and unbroken at that point. For a function to be continuous at a point \( c \), it must satisfy three conditions:

• The function \( f(x) \) is defined at \( x = c \).
• The two-sided limit \( \lim_{x \to c} f(x) \) exists.
• \( \lim_{x \to c} f(x) = f(c) \).

In our specific exercise, we examine two points, -1 and 1, to check for continuity. For instance, at \( x = -1 \), the left-hand limit, the right-hand limit, and \( f(-1) \) all equal 0, confirming continuity. Similarly, at \( x = 1 \), both the limit values and \( f(1) \) are equal to 2, indicating continuity at that point as well.

Left-Hand Limit

The left-hand limit refers to the value the function approaches as \( x \) approaches the point of interest from the left. Using the example from our exercise, let's consider the left-hand limit at \( x = -1 \):To find \( \lim_{x \to -1^-} f(x) \) (indicating approaching \(-1\) from the "negative" side), we use the piece where \( x < -1 \), which is \( f(x) = x^2 - 1 \). As \( x \) gets closer to \(-1\) from the left, we calculate \((-1)^2 - 1 = 0\).Similarly, for \( x = 1 \), the left-hand limit involves the piece for \(-1 \leq x \leq 1\), so \( f(x) = x^3 + 1 \). As \( x \to 1^- \), this evaluates to \(1^3 + 1 = 2\). Left-hand limits are crucial in analyzing discontinuities or ensuring the function behaves correctly at specific points.

Right-Hand Limit

Conversely, the right-hand limit is the value the function approaches as \( x \) moves toward the point from the right side. When analyzing the right-hand limit for our exercise, consider \( x = -1 \):For \( \lim_{x \to -1^+} f(x) \) (approaching \(-1\) from the "positive" side), the relevant piece is where \(-1 \leq x \leq 1\), i.e., \( f(x) = x^3 + 1 \). Evaluating at \( x \to -1^+ \) gives \((-1)^3 + 1 = 0\).At \( x = 1 \), for the right-hand limit, \( x > 1 \) applies, so \( f(x) = x^2 + 1 \). As \( x \to 1^+ \), this calculates to \(1^2 + 1 = 2\). These limits help determine the behavior of the function as it approaches critical points from different directions. Understanding both left-hand and right-hand limits is essential for checking overall continuity and the precise behavior of piecewise functions.