Problem 94 - Single Variable Calculus

Question

Show by means of an example that \(\lim _{x \rightarrow a a}[f(x) g(x)]\) may exist even though neither \(\lim _{x \rightarrow a} f(x)\) nor \(\lim _{x \rightarrow a} g(x)\) exists. Does this example contradict the Product Law of limits?

Step-by-Step Solution

Answer

The example functions are: \(f(x) = \begin{cases} a & \text{if}\ x = a \\[1em] 1 & \text{if}\ x \neq a \end{cases}\) \(g(x) = \begin{cases} a & \text{if}\ x = a \\[1em] \dfrac{1}{x-a} & \text{if}\ x \neq a \end{cases}\) The limit of their product, \(\lim_{x \rightarrow a} h(x)\), doesn't exist. However, the individual limits \(\lim_{x \rightarrow a} f(x)\) and \(\lim_{x \rightarrow a} g(x)\) exist. This example contradicts the Product Law of limits, but it's important to note that the functions f(x) and g(x) do not satisfy conditions for the Product Law to hold, such as continuity and differentiability at point a, so the example still follows the general rules for the Product Law of limits.

Step 1: Define the product function h(x)

Let's define a function, h(x), as the product of f(x) and g(x): \(h(x) = f(x)g(x)\)

Step 2: Determine h(x)

Now, calculate h(x) as the product of f(x) and g(x): For x ≠ a, we have: \(h(x) = f(x)g(x) = (1) * \left(\dfrac{1}{x-a}\right) = \dfrac{1}{x-a}\) For x = a, we have: \(h(x) = f(x)g(x) = (a) * (a) = a^2\) Next, let's find the limit of h(x) as x approaches a.

Step 3: Find the limit of h(x)

As x approaches a: \(\lim_{x \rightarrow a} h(x) = \lim_{x \rightarrow a} \frac{1}{x-a} = \frac{1}{a-a} = \frac{1}{0}\) Since the limit of h(x) as x approaches a is undefined, we can conclude that the limit of the product of f(x) and g(x) does not exist. Now, let's examine the limits of f(x) and g(x) individually:

Step 4: Find the limit of f(x) and g(x)

\(\lim_{x \rightarrow a} f(x) = f(a) = a\) \(\lim_{x \rightarrow a} g(x) = g(a) = a\) Since the limits of f(x) and g(x) exist when x approaches a, the Product Law of limits states that the limit of the product of these functions should also exist. In our example, the limit of the product of f(x) and g(x) does not exist while the individual limits of f(x) and g(x) exist. This means that our example contradicts the Product Law of limits. However, it is important to note that the Product Law of limits holds true for functions that satisfy certain conditions, such as continuity and differentiability at the point a. Our functions f(x) and g(x) do not satisfy these conditions, and that is why our example presents a scenario where the Product Law of limits does not apply.

Key Concepts

Limits of Functions

In calculus, the concept of limits is fundamental. A limit is the value that a function approaches as the input, or x-value, gets indefinitely close to a particular point. This point is often represented as \(a\) in mathematical exercises.

• The limit of a function \(f(x)\) as \(x\) approaches \(a\) is written as \(\lim_{x \rightarrow a} f(x)\).
• If the function approaches a specific value, we say that the limit exists. Otherwise, it does not.

Understanding limits helps us analyze the behavior of functions near points where they might not be defined or are discontinuous. For instance, when a function takes the form \(\dfrac{1}{x-a}\), it becomes difficult to determine its limit as \(x\) approaches \(a\), since division by zero is undefined.
This undefined nature is a common obstacle in limit evaluations, and it drives the necessity to understand the conditions under which a limit can exist or not.

Continuity

Continuity of a function at a point essentially means that the graph of the function is unbroken at that point. In mathematical terms, a function \(f(x)\) is said to be continuous at \(x = a\) if:

• \(\lim_{x \rightarrow a} f(x)\) exists,
• \(f(a)\) is defined, and
• \(\lim_{x \rightarrow a} f(x) = f(a)\)

Continuity ensures the smooth progression of a graph without any jumps or gaps. It is crucial when larger ideas like the Product Law of limits come into play because continuity checks ensure that the behavior of functions is predictable around points of interest.
Functions that are not continuous might behave unexpectedly, making it possible for their limits to not exist even if under normal conditions we expect them to. In the example exercise, continuity conditions are not met, which contributes to the contradictory situation with the Product Law of limits.

Product Law of Limits

The Product Law of limits is a valuable tool in calculus that allows us to simplify the process of finding limits of products of functions. The law states:

• If \(\lim_{x \rightarrow a} f(x)\) and \(\lim_{x \rightarrow a} g(x)\) exist, then \(\lim_{x \rightarrow a} [f(x)g(x)] = \lim_{x \rightarrow a} f(x) \cdot \lim_{x \rightarrow a} g(x)\).

This rule gives us a shortcut by allowing us to evaluate the limits of the individual functions and then use their product to find the limit of the composite function.
However, this law does require that both functions be continuous around the limit point \(a\). When these conditions aren't met, such as in the provided exercise where \(g(x) = \dfrac{1}{x-a}\) becomes problematic, the rule doesn't apply as expected.
This is crucial to note because in real-world scenarios, unpredicted behavior in the functions can lead to situations where the Product Law of limits cannot be utilized, highlighting the importance of verifying continuity before applying this helpful rule.