What does the graph of $x\cos(\frac{1}{x})$ squeezed between $-|x|$ and $|x|$ tell us?

It visually confirms that as $x$ approaches 0, $x\cos(\frac{1}{x})$ is forced to approach 0 as well.

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What does the graph of $x\cos(\frac{1}{x})$ squeezed between $-|x|$ and $|x|$ tell us?

It visually confirms that as $x$ approaches 0, $x\cos(\frac{1}{x})$ is forced to approach 0 as well.

Steps to apply the Squeeze Theorem.

Identify the function. 2. Find bounding functions. 3. Show bounding functions have the same limit. 4. Conclude the limit of the original function.

How to find the $\lim_{x \to 0} x\cos(\frac{1}{x})$ using the Squeeze Theorem.

Recognize $-1 \leq \cos(\frac{1}{x}) \leq 1$ . 2. Multiply by $x$ : $-|x| \leq x\cos(\frac{1}{x}) \leq |x|$ . 3. $\lim_{x \to 0} -|x| = 0$ and $\lim_{x \to 0} |x| = 0$ . 4. Therefore, $\lim_{x \to 0} x\cos(\frac{1}{x}) = 0$ .

How do you handle the sign of x when multiplying inequalities?

Use absolute values to ensure inequalities hold for both positive and negative x values.

How to apply the Squeeze Theorem when given $g(x) \leq k(x) \leq h(x)$ ?

Find $\lim_{x \to a} g(x)$ and $\lim_{x \to a} h(x)$ . 2. If both limits are equal to $L$ , then $\lim_{x \to a} k(x) = L$ .

State the Squeeze Theorem.

If $f(x) \leq g(x) \leq h(x)$ for all $x$ in an interval containing $a$ (except possibly at $a$ itself) and $\lim_{x \to a} f(x) = \lim_{x \to a} h(x) = L$ , then $\lim_{x \to a} g(x) = L$ .

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What does the graph of $x\cos(\frac{1}{x})$ squeezed between $-|x|$ and $|x|$ tell us?

It visually confirms that as $x$ approaches 0, $x\cos(\frac{1}{x})$ is forced to approach 0 as well.

Steps to apply the Squeeze Theorem.

Identify the function. 2. Find bounding functions. 3. Show bounding functions have the same limit. 4. Conclude the limit of the original function.

How to find the $\lim_{x \to 0} x\cos(\frac{1}{x})$ using the Squeeze Theorem.

Recognize $-1 \leq \cos(\frac{1}{x}) \leq 1$ . 2. Multiply by $x$ : $-|x| \leq x\cos(\frac{1}{x}) \leq |x|$ . 3. $\lim_{x \to 0} -|x| = 0$ and $\lim_{x \to 0} |x| = 0$ . 4. Therefore, $\lim_{x \to 0} x\cos(\frac{1}{x}) = 0$ .

How do you handle the sign of x when multiplying inequalities?

Use absolute values to ensure inequalities hold for both positive and negative x values.

How to apply the Squeeze Theorem when given $g(x) \leq k(x) \leq h(x)$ ?

Find $\lim_{x \to a} g(x)$ and $\lim_{x \to a} h(x)$ . 2. If both limits are equal to $L$ , then $\lim_{x \to a} k(x) = L$ .

State the Squeeze Theorem.

If $f(x) \leq g(x) \leq h(x)$ for all $x$ in an interval containing $a$ (except possibly at $a$ itself) and $\lim_{x \to a} f(x) = \lim_{x \to a} h(x) = L$ , then $\lim_{x \to a} g(x) = L$ .