Evaluate the following limit and use the ϵ, δ-definition of the limit of a function to prove your result: lim x→0 x^2 cos 1/x

To solve the limit

\[
\lim_{x \to 0} x^2 \cos\left(\frac{1}{x}\right),
\]

we first need to evaluate this limit intuitively and then provide a formal proof using the **ϵ, δ-definition of the limit**.

### 1. **Evaluation of the Limit**

We are asked to evaluate:

\[
\lim_{x \to 0} x^2 \cos\left(\frac{1}{x}\right).
\]

Let's analyze this step by step:

- The cosine function, \( \cos\left(\frac{1}{x}\right) \), oscillates between -1 and 1 for all values of \( x \neq 0 \), since \( \cos \theta \) is always bounded, i.e.,

  \[
  -1 \leq \cos\left(\frac{1}{x}\right) \leq 1.
  \]

- Now, consider the factor \( x^2 \). As \( x \to 0 \), \( x^2 \) gets smaller and smaller (since \( x^2 \) approaches 0 as \( x \) approaches 0).
  
Thus, we can write:

\[
-x^2 \leq x^2 \cos\left(\frac{1}{x}\right) \leq x^2.
\]

As \( x^2 \) approaches 0, both bounds \( -x^2 \) and \( x^2 \) approach 0. By the **squeeze theorem**, we conclude that:

\[
\lim_{x \to 0} x^2 \cos\left(\frac{1}{x}\right) = 0.
\]

### 2. **ϵ, δ-definition of the Limit**

To formally prove this result using the ϵ, δ-definition of a limit, we need to show that for every \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that whenever \( 0 < |x| < \delta \), the inequality

\[
|x^2 \cos\left(\frac{1}{x}\right) - 0| < \epsilon
\]

holds.

#### Proof:

1. We know that:

   \[
   |x^2 \cos\left(\frac{1}{x}\right)| \leq |x^2|.
   \]

   Since \( |\cos\left(\frac{1}{x}\right)| \leq 1 \), it follows that:

   \[
   |x^2 \cos\left(\frac{1}{x}\right)| \leq |x^2|.
   \]

2. Now, to make \( |x^2| < \epsilon \), we choose \( \delta = \sqrt{\epsilon} \). Hence, for \( 0 < |x| < \delta \), we have:

   \[
   |x^2| < \epsilon.
   \]

3. Since \( |x^2 \cos\left(\frac{1}{x}\right)| \leq |x^2| \), we also have:

   \[
   |x^2 \cos\left(\frac{1}{x}\right)| < \epsilon.
   \]

Thus, by the ϵ, δ-definition of the limit, we have proven that:

\[
\lim_{x \to 0} x^2 \cos\left(\frac{1}{x}\right) = 0.
\]

### Conclusion:

The limit is indeed 0, and we have proven this rigorously using the **ϵ, δ-definition of the limit**.

---

Would you like more details or have any questions about this process?

Here are some related questions that might interest you:

1. How does the Squeeze Theorem work in limits?
2. What are the key differences between the ϵ, δ-definition of a limit and the intuitive approach to limits?
3. How can we apply the ϵ, δ-definition of the limit to discontinuous functions?
4. Can oscillating functions like \( \cos(1/x) \) be continuous?
5. What happens to limits at infinity with functions involving trigonometric oscillations?

**Tip**: Always try to apply the Squeeze Theorem when you are dealing with functions that oscillate but are bounded.

Evaluate the following limit and use the ϵ, δ-definition of the limit of a function to prove your result: lim x→0 x^2 cos(1/x)

Discover how to evaluate the limit of x^2 cos(1/x) as x approaches 0, using the squeeze theorem and a formal proof via the ϵ, δ-definition of limits. This detailed solution is perfect for university-level calculus students, covering key concepts in limits and oscillating functions.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation