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.

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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.