Problem 16 - Calculus Early Transcendentals

Question

\(15-20\) Explain why the function is discontinuous at the given number a. Sketch the graph of the function. \(f(x)=\left\\{\begin{array}{ll}{\frac{1}{x-1}} & {\text { if } x \neq 1} \\\ {2} & {\text { if } x=1}\end{array}\right. \quad a=1\)

Step-by-Step Solution

Answer

The function is discontinuous at \(x=1\); the limit does not exist and differs from \(f(1)=2\).

Step 1: Identify the Function Definition

The function is defined piecewise. For all values except 1, it behaves as \(f(x) = \frac{1}{x-1}\). At \(x = 1\), the function instead takes the value \(f(1) = 2\). This sets up the potential for discontinuity.

Step 2: Evaluate Limits as x Approaches 1

To determine if a function is continuous at \(x=1\), evaluate the limit of \(f(x)\) as \(x\) approaches 1. Since \(f(x) = \frac{1}{x-1}\) for \(x eq 1\), the limit of \(f(x)\) as \(x\) approaches 1 does not exist because it involves division by zero (making the denominator 0 when \(x = 1\)).

Step 3: Compare Function Value and Limit

For a function to be continuous at a point "a", the limit as \(x\) approaches \(a\) must equal \(f(a)\). Since the limit as \(x\) approaches 1 does not exist (due to division by zero), it cannot equal \(f(1) = 2\), indicating that \(f(x)\) is discontinuous at \(x=1\).

Step 4: Sketch the Graph

Draw the graph by plotting \(f(x) = \frac{1}{x-1}\) for values close to 1, which results in sharp increases/decreases near \(x=1\), appearing as a vertical asymptote. Plot \(f(1) = 2\) as a distinct point separate from the curve at \(x=1\). This visualizes the discontinuity as there is no connecting path (where the rest of the graph does not meet at the point \(x=1\)).

Key Concepts

Piecewise Function

A piecewise function is a mathematical expression built from different sub-functions, each defined on a certain interval of the domain. In this case, the function \( f(x) \) is defined as

• \( f(x) = \frac{1}{x-1} \) for \( x eq 1 \)
• \( f(x) = 2 \) for \( x = 1 \)

This means the behavior of the function changes at \( x = 1 \). These functions are useful when different rules apply to distinct parts of the domain. To analyze such a function, it's imperative to consider each piece separately, ensuring that transitions between them are smooth if we want the function to be continuous throughout.

Limits

Limits help us understand the behavior of functions as they approach a certain point. For the piecewise function we're examining, the crucial point is \( x = 1 \). As we approach this point from either side, using \( f(x) = \frac{1}{x-1} \), the division by zero causes the expression to be undefined at exactly \( x = 1 \). The left-hand and right-hand limits can provide insights about the general trend near the point, but in this case, both tend towards infinity (positive or negative, depending on the direction of approach):

• As \( x \to 1^- \), \( f(x) \to -\infty \)
• As \( x \to 1^+ \), \( f(x) \to +\infty \)

Because these one-sided limits do not converge to a single value, the limit \( \lim_{x \to 1} f(x) \) does not exist.

Vertical Asymptote

A vertical asymptote appears when the value of a function tends towards infinity as the input approaches a particular point. For \( f(x) = \frac{1}{x-1} \), there's an infinite jump as \( x \) approaches 1 from either side of the number line. This is seen in the graph where there are drastic increases or decreases, suggesting an imaginary line the function races towards but never touches. The graph, thus, splits at \( x = 1 \), highlighting that \( x = 1 \) is a vertical asymptote. This kind of asymptote signals discontinuity, alerting that the function isn't well-behaved and doesn't settle into a fixed value around this point.

Continuity Conditions

For a function to be continuous, it must fulfill certain criteria at every point:

• The function has to be defined at the point.
• The limit of the function as it approaches the point from both directions exists and equals the function's value at that point.

In our problem with \( f(x) \), although \( f(1) = 2 \) is defined, the limit of \( f(x) \) as \( x \to 1 \) does not exist because of the discontinuity at the vertical asymptote. Thus, continuity conditions are not satisfied. To concretely show discontinuity, sketching the graph evidences that the function abruptly jumps or diverges, lacking a seamless transition through \( x = 1 \). This highlights why \( f(x) \) is not continuous at \( x = 1 \).