Problem 25 - Calculus: Early Transcendentals

Question

Determining limits analytically Determine the following limits. a. \(\lim _{z \rightarrow 3^{+}} \frac{(z-1)(z-2)}{(z-3)}\) b. \(\lim _{z \rightarrow 3^{-}} \frac{(z-1)(z-2)}{(z-3)}\) c. \(\lim _{z \rightarrow 3} \frac{(z-1)(z-2)}{(z-3)}\)

Step-by-Step Solution

Answer

Answer: The one-sided limits are as follows: a. \(\lim _{z \rightarrow 3^{+}} \frac{(z-1)(z-2)}{(z-3)} = \infty\) b. \(\lim _{z \rightarrow 3^{-}} \frac{(z-1)(z-2)}{(z-3)} = -\infty\) The general limit does not exist: c. \(\lim _{z \rightarrow 3} \frac{(z-1)(z-2)}{(z-3)}\) = Does not exist

Step 1: Simplify the function

Starting with the given function \(\frac{(z-1)(z-2)}{(z-3)}\), notice that it is a rational function with a hole at z=3 since the numerator and the denominator have the same degree.

Step 2: Analyze the one-sided limits

For (a) and (b), we will examine the behavior of the function as z approaches 3 from the right (3+) and from the left (3-), respectively. a. \(\lim _{z \rightarrow 3^{+}} \frac{(z-1)(z-2)}{(z-3)}\) To determine this limit, observe the behavior of the function as z approaches 3 from the right: As z approaches 3 from the right, (z-1) and (z-2) are both positive. The expression (z-3) is positive as well since we are approaching 3 from the right. Therefore, the ratio \(\frac{(z-1)(z-2)}{(z-3)}\) will also be positive. b. \(\lim _{z \rightarrow 3^{-}} \frac{(z-1)(z-2)}{(z-3)}\) To determine this limit, observe the behavior of the function as z approaches 3 from the left: As z approaches 3 from the left, (z-1) and (z-2) are both positive. The expression (z-3), however, is negative since we are approaching 3 from the left. Therefore, the ratio \(\frac{(z-1)(z-2)}{(z-3)}\) will be negative.

Step 3: Evaluate the one-sided limits by plugging z into the simplified function

Now we can evaluate the limits for both (a) and (b): a. \(\lim _{z \rightarrow 3^{+}} \frac{(z-1)(z-2)}{(z-3)} = \frac{(3-1)(3-2)}{(3-3)} = \frac{(2)(1)}{(0)} = \infty\) b. \(\lim _{z \rightarrow 3^{-}} \frac{(z-1)(z-2)}{(z-3)} = \frac{(3-1)(3-2)}{(3-3)} = \frac{(2)(1)}{(0)} = -\infty\)

Step 4: Evaluate the general limit

Finally, we can evaluate the general limit (c): c. \(\lim _{z \rightarrow 3} \frac{(z-1)(z-2)}{(z-3)}\) Since the one-sided limits are not the same, the general limit does not exist: \(\lim _{z \rightarrow 3} \frac{(z-1)(z-2)}{(z-3)}\) = Does not exist

Step 5: Results

We have successfully analyzed and calculated the given limits: a. \(\lim _{z \rightarrow 3^{+}} \frac{(z-1)(z-2)}{(z-3)} = \infty\) b. \(\lim _{z \rightarrow 3^{-}} \frac{(z-1)(z-2)}{(z-3)} = -\infty\) c. \(\lim _{z \rightarrow 3} \frac{(z-1)(z-2)}{(z-3)}\) = Does not exist

Key Concepts

One-Sided Limits

When studying how functions behave near a certain point, it's crucial to consider one-sided limits. These limits assess the behavior of a function as the input approaches a specific value from one side only - either from the left (left-hand limit) or the right (right-hand limit). Understanding this concept is essential because it can influence whether a general limit exists at that point. For instance, if one-sided limits don't match, as in the exercise where the limits as z approaches 3 from the left and right are infinite and negative infinite respectively, the general limit at that point is deemed nonexistent. This difference in one-sided limits indicates a discontinuity or a break in the graph of the function at that point.

Rational Functions

A rational function is defined as the ratio of two polynomials. The behavior of these functions is quite interesting, especially near values that set the denominator to zero, known as asymptotes. In the exercise, the function (z-1)(z-2)/(z-3) is undefined when z equals 3 because the denominator becomes zero, creating a hole in the graph. Rational functions often have restrictions on their domains where they are not defined, leading to interesting limits and often resulting in infinite limits as seen in the exercise.

It's also noteworthy to remember that rational functions can be simplified by factoring and reducing common terms, however, in cases where factors in the denominator are not cancelable, they create points of discontinuity.

Limits of Rational Functions

The limits of rational functions can sometimes be calculated by simple substitution if the function is continuous at the point of interest. However, when approaching values that make the denominator zero, the situation becomes more nuanced. The one-sided limits must be analyzed separately to understand how the function behaves near the point of discontinuity. In the given exercise, the substitution directly leads to an undefined expression, forcing us to look at the behavior of the numerator and denominator separately as z approaches 3. The signs of the resulting quantities determine whether the limit trends towards positive infinity or negative infinity. These infinite results are a key characteristic of rational functions' behavior near their vertical asymptotes.

Infinite Limits

Infinite limits occur when the value of a function increases or decreases without bound as the input approaches a certain point. This is typically visualized on a graph as a vertical asymptote, which the function approaches but never actually reaches or crosses. In the exercise, both one-sided limits as z approaches 3 result in the function growing without bound in opposite directions (positive or negative infinity). Even though infinity is not a real number and we cannot say the function 'equals' infinity, in calculus, we use it to describe this unbounded behavior. Consequently, when dealing with infinite limits, it's fundamental to analyze the sign and direction in which the function is heading to properly understand the function's behavior near those points of interest.