Sure! I'll give you a detailed example where we need to use the Squeeze Theorem and the cosine function.

Problem Example:

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

Step-by-Step Solution:

1. State the limit expression: We are tasked with finding: $\lim_{x \to 0} x^2 \cos \left( \frac{1}{x} \right)$

2. Recognize the form: Notice that as $x \to 0$, the term $\cos \left( \frac{1}{x} \right)$ oscillates between $-1$ and $1$ infinitely fast. However, the function $x^2$ approaches $0$, and the behavior of $x^2 \cos \left( \frac{1}{x} \right)$ is influenced by this product.

3. Squeeze Theorem Setup: To apply the Squeeze Theorem, we first need to bound the function $x^2 \cos \left( \frac{1}{x} \right)$ between two simpler functions.

   Recall that for all real numbers $t$, $-1 \leq \cos(t) \leq 1$ So for any $x \neq 0$, we have: $-1 \leq \cos \left( \frac{1}{x} \right) \leq 1$

4. Multiply by $x^2$: Multiply the entire inequality by $x^2$ (which is non-negative for all $x$) to get bounds on $x^2 \cos \left( \frac{1}{x} \right)$: $-x^2 \leq x^2 \cos \left( \frac{1}{x} \right) \leq x^2$

5. Apply the Squeeze Theorem: Now, observe the behavior of the bounding functions as $x \to 0$. We know that: $\lim_{x \to 0} -x^2 = 0 \quad \text{and} \quad \lim_{x \to 0} x^2 = 0$ Since both the lower bound $-x^2$ and the upper bound $x^2$ approach $0$ as $x \to 0$, by the Squeeze Theorem, we can conclude that: $\lim_{x \to 0} x^2 \cos \left( \frac{1}{x} \right) = 0$

Final Answer:

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

Summary of Steps:

1. Recognize the problem involves an oscillating function ($\cos \left( \frac{1}{x} \right)$) and that $\cos(t)$ is bounded by $-1$ and $1$.
2. Multiply by $x^2$ to set up inequalities.
3. Apply the Squeeze Theorem, showing both bounding functions