Determining Limits Using the Squeeze Theorem - AP Calculus AB
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Find $\lim_{x \to 0} x \sin(\frac{1}{x^2})$ using the Squeeze Theorem.
Find $\lim_{x \to 0} x \sin(\frac{1}{x^2})$ using the Squeeze Theorem.
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$0$. Since $-|x| \leq x \sin(\frac{1}{x^2}) \leq |x|$ and both bounds approach $0$.
$0$. Since $-|x| \leq x \sin(\frac{1}{x^2}) \leq |x|$ and both bounds approach $0$.
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Determine $\lim_{x \to 0} x^3 \cos(\frac{1}{x^3})$ using the Squeeze Theorem.
Determine $\lim_{x \to 0} x^3 \cos(\frac{1}{x^3})$ using the Squeeze Theorem.
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$0$. Since $-x^3 \leq x^3 \cos(\frac{1}{x^3}) \leq x^3$ and both bounds approach $0$.
$0$. Since $-x^3 \leq x^3 \cos(\frac{1}{x^3}) \leq x^3$ and both bounds approach $0$.
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What is $\lim_{x \to 0} x^4 \sin(\frac{1}{x^3})$?
What is $\lim_{x \to 0} x^4 \sin(\frac{1}{x^3})$?
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$0$. Since $-x^4 \leq x^4 \sin(\frac{1}{x^3}) \leq x^4$ and both bounds approach $0$.
$0$. Since $-x^4 \leq x^4 \sin(\frac{1}{x^3}) \leq x^4$ and both bounds approach $0$.
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Determine $\lim_{x \to 0} x^4 \cos(\frac{1}{x})$ using the Squeeze Theorem.
Determine $\lim_{x \to 0} x^4 \cos(\frac{1}{x})$ using the Squeeze Theorem.
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$0$. Since $-x^4 \leq x^4 \cos(\frac{1}{x}) \leq x^4$ and both bounds approach $0$.
$0$. Since $-x^4 \leq x^4 \cos(\frac{1}{x}) \leq x^4$ and both bounds approach $0$.
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Find $\lim_{x \to \infty} \frac{\sin x}{x}$ using the Squeeze Theorem.
Find $\lim_{x \to \infty} \frac{\sin x}{x}$ using the Squeeze Theorem.
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$0$. Since $-\frac{1}{x} \leq \frac{\sin x}{x} \leq \frac{1}{x}$ and both bounds approach $0$.
$0$. Since $-\frac{1}{x} \leq \frac{\sin x}{x} \leq \frac{1}{x}$ and both bounds approach $0$.
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What should $g(x)$ and $h(x)$ approach for $\lim_{x \to c}f(x)$ to exist?
What should $g(x)$ and $h(x)$ approach for $\lim_{x \to c}f(x)$ to exist?
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The same limit $L$. This ensures the squeezed function approaches the unique limit.
The same limit $L$. This ensures the squeezed function approaches the unique limit.
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Find $\lim_{x \to 0} x^2 \sin(\frac{1}{x})$ using the Squeeze Theorem.
Find $\lim_{x \to 0} x^2 \sin(\frac{1}{x})$ using the Squeeze Theorem.
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$0$. Since $-|x^2| \leq x^2 \sin(\frac{1}{x}) \leq |x^2|$ and both bounds approach $0$.
$0$. Since $-|x^2| \leq x^2 \sin(\frac{1}{x}) \leq |x^2|$ and both bounds approach $0$.
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Determine $\lim_{x \to 0} x^2 \cos(x^2)$ using the Squeeze Theorem.
Determine $\lim_{x \to 0} x^2 \cos(x^2)$ using the Squeeze Theorem.
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$0$. Since $|x^2 \cos(x^2)| \leq x^2$ and $x^2 \to 0$.
$0$. Since $|x^2 \cos(x^2)| \leq x^2$ and $x^2 \to 0$.
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Find $\lim_{x \to 0} x^3 \sin(\frac{1}{x})$ using the Squeeze Theorem.
Find $\lim_{x \to 0} x^3 \sin(\frac{1}{x})$ using the Squeeze Theorem.
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$0$. Since $-|x^3| \leq x^3 \sin(\frac{1}{x}) \leq |x^3|$ and both bounds approach $0$.
$0$. Since $-|x^3| \leq x^3 \sin(\frac{1}{x}) \leq |x^3|$ and both bounds approach $0$.
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State the Squeeze Theorem in terms of $f(x)$, $g(x)$, $h(x)$.
State the Squeeze Theorem in terms of $f(x)$, $g(x)$, $h(x)$.
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If $g(x) \leq f(x) \leq h(x)$ and $\lim_{x \to c}g(x) = \lim_{x \to c}h(x) = L$, then $\lim_{x \to c}f(x) = L$. The fundamental statement of the Squeeze Theorem with three functions.
If $g(x) \leq f(x) \leq h(x)$ and $\lim_{x \to c}g(x) = \lim_{x \to c}h(x) = L$, then $\lim_{x \to c}f(x) = L$. The fundamental statement of the Squeeze Theorem with three functions.
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What is $\lim_{x \to 0} x^3 \sin(x^2)$ using the Squeeze Theorem?
What is $\lim_{x \to 0} x^3 \sin(x^2)$ using the Squeeze Theorem?
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$0$. Since $|x^3 \sin(x^2)| \leq x^3$ and $x^3 \to 0$.
$0$. Since $|x^3 \sin(x^2)| \leq x^3$ and $x^3 \to 0$.
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Find $\lim_{x \to 0} x^2 \cos(\frac{1}{x})$ using the Squeeze Theorem.
Find $\lim_{x \to 0} x^2 \cos(\frac{1}{x})$ using the Squeeze Theorem.
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$0$. Since $-x^2 \leq x^2 \cos(\frac{1}{x}) \leq x^2$ and both bounds approach $0$.
$0$. Since $-x^2 \leq x^2 \cos(\frac{1}{x}) \leq x^2$ and both bounds approach $0$.
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Find $\lim_{x \to 0} x^5 \sin(\frac{1}{x^4})$ using the Squeeze Theorem.
Find $\lim_{x \to 0} x^5 \sin(\frac{1}{x^4})$ using the Squeeze Theorem.
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$0$. Since $-x^5 \leq x^5 \sin(\frac{1}{x^4}) \leq x^5$ and both bounds approach $0$.
$0$. Since $-x^5 \leq x^5 \sin(\frac{1}{x^4}) \leq x^5$ and both bounds approach $0$.
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What is the limit $\lim_{x \to 0} x^2 \sin(x^2)$?
What is the limit $\lim_{x \to 0} x^2 \sin(x^2)$?
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$0$. Since $|x^2 \sin(x^2)| \leq x^2$ and $x^2 \to 0$.
$0$. Since $|x^2 \sin(x^2)| \leq x^2$ and $x^2 \to 0$.
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What is $\lim_{x \to 0} x^4 \cos(\frac{1}{x^5})$?
What is $\lim_{x \to 0} x^4 \cos(\frac{1}{x^5})$?
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$0$. Since $-x^4 \leq x^4 \cos(\frac{1}{x^5}) \leq x^4$ and both bounds approach $0$.
$0$. Since $-x^4 \leq x^4 \cos(\frac{1}{x^5}) \leq x^4$ and both bounds approach $0$.
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Determine $\lim_{x \to 0} x^4 \sin(\frac{1}{x})$ using the Squeeze Theorem.
Determine $\lim_{x \to 0} x^4 \sin(\frac{1}{x})$ using the Squeeze Theorem.
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$0$. Since $-x^4 \leq x^4 \sin(\frac{1}{x}) \leq x^4$ and both bounds approach $0$.
$0$. Since $-x^4 \leq x^4 \sin(\frac{1}{x}) \leq x^4$ and both bounds approach $0$.
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What inequality must $f(x)$ satisfy in the Squeeze Theorem?
What inequality must $f(x)$ satisfy in the Squeeze Theorem?
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$g(x) \leq f(x) \leq h(x)$. This defines the sandwich relationship between the three functions.
$g(x) \leq f(x) \leq h(x)$. This defines the sandwich relationship between the three functions.
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Find $\lim_{x \to 0} x^2 \cos x$ using the Squeeze Theorem.
Find $\lim_{x \to 0} x^2 \cos x$ using the Squeeze Theorem.
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$0$. Since $|x^2 \cos x| \leq x^2$ and $x^2 \to 0$.
$0$. Since $|x^2 \cos x| \leq x^2$ and $x^2 \to 0$.
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Find $\lim_{x \to 0} x^3 \sin(\frac{2}{x})$ using the Squeeze Theorem.
Find $\lim_{x \to 0} x^3 \sin(\frac{2}{x})$ using the Squeeze Theorem.
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$0$. Since $-x^3 \leq x^3 \sin(\frac{2}{x}) \leq x^3$ and both bounds approach $0$.
$0$. Since $-x^3 \leq x^3 \sin(\frac{2}{x}) \leq x^3$ and both bounds approach $0$.
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Determine $\lim_{x \to 0} x^3 \cos(x^4)$ using the Squeeze Theorem.
Determine $\lim_{x \to 0} x^3 \cos(x^4)$ using the Squeeze Theorem.
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$0$. Since $|x^3 \cos(x^4)| \leq x^3$ and $x^3 \to 0$.
$0$. Since $|x^3 \cos(x^4)| \leq x^3$ and $x^3 \to 0$.
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What is the Squeeze Theorem used for in calculus?
What is the Squeeze Theorem used for in calculus?
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To determine the limit of a function. Forces a function between two bounds that approach the same limit.
To determine the limit of a function. Forces a function between two bounds that approach the same limit.
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What is the limit of $x \sin(\frac{1}{x})$ as $x \to 0$?
What is the limit of $x \sin(\frac{1}{x})$ as $x \to 0$?
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$0$. Bounded between $-|x|$ and $|x|$, both approaching $0$.
$0$. Bounded between $-|x|$ and $|x|$, both approaching $0$.
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What is the limit $\lim_{x \to 0} x^2 \sin(\frac{1}{x^4})$?
What is the limit $\lim_{x \to 0} x^2 \sin(\frac{1}{x^4})$?
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$0$. Since $-x^2 \leq x^2 \sin(\frac{1}{x^4}) \leq x^2$ and both bounds approach $0$.
$0$. Since $-x^2 \leq x^2 \sin(\frac{1}{x^4}) \leq x^2$ and both bounds approach $0$.
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What is $\lim_{x \to 0} x^2 \sin x$?
What is $\lim_{x \to 0} x^2 \sin x$?
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$0$. Since $|x^2 \sin x| \leq x^2$ and $x^2 \to 0$.
$0$. Since $|x^2 \sin x| \leq x^2$ and $x^2 \to 0$.
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Is $\lim_{x \to 0} x^2 \sin(x^3)$ zero?
Is $\lim_{x \to 0} x^2 \sin(x^3)$ zero?
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Yes. Since $|x^2 \sin(x^3)| \leq x^2$ and $x^2 \to 0$.
Yes. Since $|x^2 \sin(x^3)| \leq x^2$ and $x^2 \to 0$.
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Is $\lim_{x \to 0} x^2 \sin(\frac{1}{x})$ finite?
Is $\lim_{x \to 0} x^2 \sin(\frac{1}{x})$ finite?
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Yes, it is $0$. The limit equals $0$, which is a finite value.
Yes, it is $0$. The limit equals $0$, which is a finite value.
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Is $\lim_{x \to 0} x \sin x$ equal to zero?
Is $\lim_{x \to 0} x \sin x$ equal to zero?
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Yes. Since $|x \sin x| \leq |x|$ and $|x| \to 0$.
Yes. Since $|x \sin x| \leq |x|$ and $|x| \to 0$.
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Can the Squeeze Theorem be used if $\lim_{x \to c}g(x)
eq \lim_{x \to c}h(x)$?
Can the Squeeze Theorem be used if $\lim_{x \to c}g(x) eq \lim_{x \to c}h(x)$?
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No. The bounding functions must approach the same value for the theorem to work.
No. The bounding functions must approach the same value for the theorem to work.
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Determine $\lim_{x \to 0} x^5 \sin(\frac{1}{x^6})$ using the Squeeze Theorem.
Determine $\lim_{x \to 0} x^5 \sin(\frac{1}{x^6})$ using the Squeeze Theorem.
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$0$. Since $-x^5 \leq x^5 \sin(\frac{1}{x^6}) \leq x^5$ and both bounds approach $0$.
$0$. Since $-x^5 \leq x^5 \sin(\frac{1}{x^6}) \leq x^5$ and both bounds approach $0$.
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What is $\lim_{x \to 0} x \sin(\frac{2}{x})$?
What is $\lim_{x \to 0} x \sin(\frac{2}{x})$?
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$0$. Bounded between $-|x|$ and $|x|$, both approaching $0$.
$0$. Bounded between $-|x|$ and $|x|$, both approaching $0$.
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