Using algebraic limit theorem Is it possible to provide an example of functions so that neither one individually has a limit as x approaches 1 BUT their product DOES have a limit as x approaches 1

Yes, it is possible to find two functions, neither of which has a limit as \( x \) approaches 1, but their product does have a limit as \( x \) approaches 1. Here's an example to illustrate this concept:

### Example Functions
Let:

1. \( f(x) = \sin\left(\frac{1}{x-1}\right) \)
2. \( g(x) = x - 1 \)

### Behavior of Each Function Individually:
- The function \( f(x) = \sin\left(\frac{1}{x-1}\right) \) oscillates infinitely as \( x \) approaches 1 because \( \frac{1}{x-1} \) goes to \( \pm \infty \), causing \( \sin\left(\frac{1}{x-1}\right) \) to oscillate without settling at any particular value. Therefore, \( f(x) \) does not have a limit as \( x \) approaches 1.

- The function \( g(x) = x - 1 \) goes to 0 as \( x \) approaches 1, but it does not have a specific non-zero limit at 1. Hence, \( g(x) \) does not have a non-trivial limit as \( x \) approaches 1.

### Product of the Functions:
Now consider the product of these functions:

\[
h(x) = f(x) \cdot g(x) = \sin\left(\frac{1}{x-1}\right) \cdot (x - 1)
\]

### Evaluating the Limit of the Product:
Let's find the limit of \( h(x) \) as \( x \) approaches 1:

\[
\lim_{x \to 1} \sin\left(\frac{1}{x-1}\right) \cdot (x - 1)
\]

Using the substitution \( y = \frac{1}{x-1} \), as \( x \to 1 \), \( y \to \infty \). Therefore, \( x - 1 = \frac{1}{y} \). The expression becomes:

\[
\lim_{y \to \infty} \sin(y) \cdot \frac{1}{y}
\]

Using the fact that \( -1 \leq \sin(y) \leq 1 \), we can write:

\[
-\frac{1}{y} \leq \sin(y) \cdot \frac{1}{y} \leq \frac{1}{y}
\]

As \( y \to \infty \), both \( \frac{1}{y} \) and \( -\frac{1}{y} \) approach 0. By the Squeeze Theorem:

\[
\lim_{y \to \infty} \sin(y) \cdot \frac{1}{y} = 0
\]

Thus, we have:

\[
\lim_{x \to 1} \sin\left(\frac{1}{x-1}\right) \cdot (x - 1) = 0
\]

### Conclusion:
- Neither \( f(x) = \sin\left(\frac{1}{x-1}\right) \) nor \( g(x) = x - 1 \) has a limit as \( x \) approaches 1.
- However, their product \( h(x) = \sin\left(\frac{1}{x-1}\right) \cdot (x - 1) \) does have a limit, and it is 0.

Would you like more details or have any further questions?

### Additional Questions:

1. How does the Squeeze Theorem work, and why is it applicable here?
2. Can you provide another example where a product has a limit, but neither factor does?
3. What is the role of oscillatory behavior in determining limits?
4. How does substitution help in evaluating limits involving oscillatory functions?
5. What are the general conditions under which the product of two functions has a limit?

### Tip:
When evaluating limits involving products of functions, consider using substitutions and the Squeeze Theorem to manage oscillatory behavior effectively.

Explore an example where two functions, neither having a limit as x approaches 1 individually, demonstrate a product that does have a limit. Understand the behavior of oscillatory functions and the application of the Squeeze Theorem in evaluating limits.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation