Problem 21 - Calculus Early Transcendentals

Question

Determining limits analytically Determine the following limits or state that they do not exist. a. \(\lim _{x \rightarrow 3^{+}} \frac{(x-1)(x-2)}{(x-3)}\) b. \(\lim _{x \rightarrow 3^{-}} \frac{(x-1)(x-2)}{(x-3)}\) c. \(\lim _{x \rightarrow 3} \frac{(x-1)(x-2)}{(x-3)}\)

Step-by-Step Solution

Answer

Answer: a. The limit as x approaches 3 from the right side (3+) is \(+\infty\). b. The limit as x approaches 3 from the left side (3-) is \(-\infty\). c. The overall limit when x approaches 3 does not exist (DNE).

Step 1: Identify the given function

We have the given function: $$f(x) = \frac{(x-1)(x-2)}{(x-3)}$$

Step 2: Determine the Limit as x approaches 3 from the right side (3+)

To find the limit as x approaches 3 from the right side (3+), we can plug in a number slightly greater than 3 into the function and find the value. As x approaches 3 from the right side (3+), \(f(x)\) becomes: $$\lim _{x \rightarrow 3^{+}} \frac{(x-1)(x-2)}{(x-3)}$$ Since the denominator approaches zero as x approaches 3, we need to analyze the sign of the numerator and denominator as x approaches 3 from the right: Numerator: \((x-1)(x-2)\) will be positive since both factors are positive when x > 3. Denominator: \((x-3)\) will be positive since x > 3. As x approaches 3 from the right, both numerator and denominator are positive, thus: $$\lim _{x \rightarrow 3^{+}} \frac{(x-1)(x-2)}{(x-3)} = +\infty$$

Step 3: Determine the limit as x approaches 3 from the left side (3-)

To find the limit as x approaches 3 from the left side (3-), we can plug in a number slightly smaller than 3 into the function and find the value. As x approaches 3 from the left side (3-), \(f(x)\) becomes: $$\lim _{x \rightarrow 3^{-}} \frac{(x-1)(x-2)}{(x-3)}$$ Since the denominator approaches zero as x approaches 3, we need to analyze the sign of the numerator and denominator as x approaches 3 from the left: Numerator: \((x-1)(x-2)\) will be positive since both factors are positive when x < 3. Denominator: \((x-3)\) will be negative since x < 3. As x approaches 3 from the left, the numerator is positive and denominator is negative, thus: $$\lim _{x \rightarrow 3^{-}} \frac{(x-1)(x-2)}{(x-3)} = -\infty$$

Step 4: Determine the overall limit as x approaches 3

We can see that $$\lim _{x \rightarrow 3^{+}} \frac{(x-1)(x-2)}{(x-3)} \neq \lim _{x \rightarrow 3^{-}} \frac{(x-1)(x-2)}{(x-3)}$$ Since the right-sided and left-sided limits are not equal, the overall limit when x approaches 3 does not exist: $$\lim _{x \rightarrow 3} \frac{(x-1)(x-2)}{(x-3)} = \text{DNE}$$ (DNE = Does Not Exist) So, the final answers are: a. \(\lim _{x \rightarrow 3^{+}} \frac{(x-1)(x-2)}{(x-3)} = +\infty\) b. \(\lim _{x \rightarrow 3^{-}} \frac{(x-1)(x-2)}{(x-3)} = -\infty\) c. \(\lim _{x \rightarrow 3} \frac{(x-1)(x-2)}{(x-3)} = \text{DNE}\)

Key Concepts

Analytic Techniques

Analytic techniques in calculus allow us to find limits by examining the behavior of functions. They are essential for understanding how functions behave as variables approach certain values. Specifically, in these exercises, we are looking at how a rational function behaves as \(x\) nears 3 from different directions.To use analytic techniques effectively:

• Identify the behavior of the numerator and denominator as \(x\) approaches the point of interest.
• Assess the signs of both the numerator and denominator to determine the overall sign of the limit.
• Check if the overall limit exists by comparing the left-sided and right-sided limits.

In our case, it was crucial to understand that the function involved a factor that turned the denominator to zero while the numerator remained positive. This leads to different behaviors depending on the direction from which \(x\) approaches 3.

One-Sided Limits

One-sided limits focus on evaluating a function's limit by approaching a specific value from one side only. These are crucial when the function behaves differently from the left and the right of the point.When calculating one-sided limits, we noted:

• \(\lim _{x \rightarrow 3^+}\): Looked at values slightly greater than 3, where the denominator \((x-3)\) is positive, leading to an infinite positive limit.
• \(\lim _{x \rightarrow 3^-}\): Examined values slightly less than 3, causing the denominator \((x-3)\) to be negative, resulting in a negative infinite limit.

The distinction between the right-hand and left-hand limit reveals key insights about discontinuities and behaviors at specific points. If both one-sided limits were equal, the overall limit would exist.

Infinite Limits

Infinite limits occur when a function's value increases or decreases without bound as \(x\) approaches a certain value. In our exercise, the expression becomes undefined as \(x\) nears 3 because the denominator tends to zero.To understand infinite limits better:

• The numerator \((x-1)(x-2)\) remains positive around \(x=3\).
• The sign of the denominator \((x-3)\) determines the direction of the infinity.

These cases occur often in rational functions, highlighting points where the function does not settle to a finite number. Because the right-handed and left-handed limits differ, this particular limit approaches both positive and negative infinity from those directions.

Discontinuous Functions

Discontinuous functions do not have a defined value at some points, or their limits differ based on the direction of approach. Such functions can have breaks or jumps in their graph.In our given function \(\frac{(x-1)(x-2)}{(x-3)}\), the discontinuity at \(x=3\) results because:

• The function becomes undefined as \(x\) approaches 3, resulting in non-existing general limits.
• The differing one-sided limits (\(+\infty\) from the right, \(-\infty\) from the left) confirm the discontinuity.

Understanding discontinuous functions helps in recognizing why certain limits don’t exist and enhances our ability to analyze functions for more complex calculus problems. It's crucial for students to know how limits behave at points of discontinuity to solve similar exercises effectively.