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Explain the core idea behind the Squeeze Theorem.
If a function is trapped between two other functions that converge to the same limit, it must also converge to that limit.
When is the Squeeze Theorem most useful?
When dealing with oscillating functions or functions where direct substitution yields an indeterminate form.
Why are continuous functions important for the Squeeze Theorem?
Continuous functions allow us to evaluate limits by direct substitution, which is often needed to verify the bounding functions.
How does the Squeeze Theorem relate to inequalities?
The Squeeze Theorem relies on establishing inequalities to bound the function of interest between two other functions.
Explain the role of bounding functions in the Squeeze Theorem.
Bounding functions, $f(x)$ and $h(x)$, 'squeeze' the function $g(x)$ between them, allowing us to determine $\lim_{x \to a} g(x)$ if $\lim_{x \to a} f(x) = \lim_{x \to a} h(x)$.
What is the importance of checking the limits of the bounding functions?
The Squeeze Theorem can only be applied if the limits of the bounding functions are equal at the point of interest.
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 $\lim_{x \to a} g(x) = L$.
Steps to apply the Squeeze Theorem.
1. 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.
1. 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 $g(x) \leq k(x) \leq h(x)$?
1. Find $\lim_{x \to a} g(x)$ and $\lim_{x \to a} h(x)$. 2. If both limits are equal to $L$, then $\lim_{x \to a} k(x) = L$.