Question

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

Short answer

(a) 2, (b) 0, (c) Does not exist, (d) 1.

Step-by-step solution

Analyze the function for \(x < 1\)

For the function \(f(x)=x+1\) when \(x < 1\), we need to inspect how the function behaves as \(x\) approaches 1 from the left (i.e., \(x \to 1^-\)). As \(x\) gets closer to 1 from the left, the value of \(f(x)\) approaches \(1+1=2\).

Evaluate \(\lim_{x \to 1^-} f(x)\)

As we determined in Step 1, since \(f(x) = x + 1\) for \(x < 1\), the limit of \(f(x)\) as \(x\) approaches 1 from the left is 2. Therefore, \(\lim_{x \to 1^-} f(x) = 2\).

Analyze the function for \(x > 1\)

For \(x > 1\), the function is given by \(f(x) = x-1\). As \(x\) approaches 1 from the right (i.e., \(x \to 1^+\)), the value of \(f(x)\) approaches \(1-1=0\).

Evaluate \(\lim_{x \to 1^+} f(x)\)

Based on Step 3, the limit of \(f(x)\) as \(x\) approaches 1 from the right is 0. Thus, \(\lim_{x \to 1^+} f(x) = 0\).

Evaluate \(\lim_{x \to 1} f(x)\)

The overall limit exists only if the limits from both directions are equal. Since \(\lim_{x \to 1^-} f(x) = 2\) and \(\lim_{x \to 1^+} f(x) = 0\), \(\lim_{x \to 1} f(x)\) does not exist because the left-hand and right-hand limits are not equal.

Determine \(f(1)\)

The value of the function specifically at \(x=1\) is defined as \(f(x) = 1\). Hence, \(f(1) = 1\).

Key concepts

Left-hand Limit

The concept of the left-hand limit is essential in understanding how a function behaves as it approaches a particular point from the left side. In mathematical terms, we denote this as \( \lim_{x \to c^-} f(x) \), meaning that \( x \) approaches the point \( c \) from values less than \( c \).

To calculate the left-hand limit, we use the piece of the function defined for \( x < c \). For instance, in our given exercise involving a piecewise function, the function is \( f(x) = x + 1 \) for \( x < 1 \). As \( x \) gets closer to 1 from the left, \( f(x) \) approaches \( 1 + 1 = 2 \).

Thus, the left-hand limit for this function as \( x \to 1^- \) is 2. This demonstrates how we evaluate the behavior of piecewise functions using the specific definition applicable to the domain of interest.

Right-hand Limit

The right-hand limit investigates the function's behavior as \( x \) approaches a certain point from the right, i.e., from values greater than the point. This is expressed as \( \lim_{x \to c^+} f(x) \).

In our piecewise function example, the segment applicable for \( x > 1 \) is \( f(x) = x - 1 \). As \( x \) gets nearer to 1 from the right, \( f(x) \) tends to \( 1 - 1 = 0 \).

Consequently, the right-hand limit of the function as \( x \to 1^+ \) is 0. By analyzing the right-hand limit, you can determine how the function progresses from the immediate right of the specified point.

Continuity at a Point

Continuity at a point is a fundamental concept ensuring that a function behaves predictably at that point without any discontinuities. A function is continuous at a point \( c \) if the following three conditions are satisfied:

• The function \( f(x) \) is defined at \( x = c \).
• The left-hand limit \( \lim_{x \to c^-} f(x) \) and right-hand limit \( \lim_{x \to c^+} f(x) \) both exist.
• The left-hand and right-hand limits are equal, and both are equal to \( f(c) \).

In our exercise, continuing the function at \( x = 1 \) would mean that \( f(1) = 1 \), yet the two-sided limit \( \lim_{x \to 1} f(x) \) does not exist since \( \lim_{x \to 1^-} f(x) = 2 \) and \( \lim_{x \to 1^+} f(x) = 0 \) are not equal. This absence of equality implies a discontinuity at \( x = 1 \) for this function.

Understanding these conditions will help you recognize points of discontinuity effectively in various mathematical contexts.