Determining Limits Using the Squeeze Theorem - AP Calculus BC

Evaluate the limit of $x^3 \sin(\frac{1}{x})$ as $x \to 0$ using the Squeeze Theorem.

0. Bounded by $-|x^3| \leq x^3 \sin(\frac{1}{x}) \leq |x^3|$ , both approach 0.
Find the limit: $x^2 \text{cos}(x)$ as $x \to 0$ using the Squeeze Theorem.

0. Since $|\cos(x)| \leq 1$ , we have $-x^2 \leq x^2\cos(x) \leq x^2$ .
Does the Squeeze Theorem apply if $f(x)$ does not converge to $g(x)$ ?

No, $f(x)$ and $g(x)$ must converge to the same limit. The theorem requires both bounding functions have identical limits.
Find the limit of $x^2 \text{cos}(\frac{1}{x^2})$ as $x \to 0$ using the Squeeze Theorem.

0. Bounded by $-x^2 \leq x^2\cos(\frac{1}{x^2}) \leq x^2$ , both approach 0.
What is the limit of $x^4 \cos(\frac{1}{x^3})$ as $x \to 0$ using the Squeeze Theorem?

0. Bounded by $-x^4 \leq x^4 \cos(\frac{1}{x^3}) \leq x^4$ , both approach 0.
What is the Squeeze Theorem used for in calculus?

Determining limits of functions trapped between two other functions. Used when a function is bounded between two converging functions.
Determine the limit: $x^2 \text{sin}(\frac{1}{x^3})$ as $x \to 0$ using the Squeeze Theorem.

0. Bounded by $-x^2 \leq x^2\sin(\frac{1}{x^3}) \leq x^2$ , both approach 0.
Determine the limit: $x^3 \text{sin}(\frac{1}{x^2})$ as $x \to 0$ using the Squeeze Theorem.

0. Bounded by $-|x^3| \leq x^3\sin(\frac{1}{x^2}) \leq |x^3|$ , both approach 0.
Does the Squeeze Theorem apply if $f(x)$ is not continuous?

Yes, continuity is not required. Continuity is not required for the Squeeze Theorem to work.
Can the Squeeze Theorem be used for bounded functions?

Yes, if they are squeezed between converging functions. Yes, bounded functions can be squeezed if appropriate bounds converge.
What is the limit of $x^3 \text{cos}(\frac{1}{x})$ as $x \to 0$ using the Squeeze Theorem?

0. Bounded by $-|x^3| \leq x^3\cos(\frac{1}{x}) \leq |x^3|$ , both approach 0.
Evaluate the limit of $x^3 \text{sin}(\frac{1}{x})$ as $x \to 0$ using the Squeeze Theorem.

0. Bounded by $-|x^3| \leq x^3\sin(\frac{1}{x}) \leq |x^3|$ , both approach 0.
Evaluate the limit of $x^4 \text{sin}(\frac{1}{x})$ as $x \to 0$ using the Squeeze Theorem.

0. Bounded by $-x^4 \leq x^4\sin(\frac{1}{x}) \leq x^4$ , both approach 0.
When is the Squeeze Theorem not applicable?

When outer functions do not converge to the same limit. Fails when bounding functions don't converge to the same value.
Evaluate the limit of $x \text{sin}(\frac{1}{x})$ as $x \to 0$ using the Squeeze Theorem.

0. Bounded by $-|x| \leq x\sin(\frac{1}{x}) \leq |x|$ , both approach 0.
Which condition is critical for applying the Squeeze Theorem?

The outer functions must converge to the same limit. Without equal limits, the theorem cannot determine the middle function's limit.
Does the Squeeze Theorem apply to oscillating functions?

Yes, if they are bounded by converging functions. Perfect application when oscillating functions are properly bounded.
Determine the limit: $x^2 \sin(\frac{1}{x^3})$ as $x \to 0$ using the Squeeze Theorem.

0. Bounded by $-x^2 \leq x^2\sin(\frac{1}{x^3}) \leq x^2$ , both approach 0.
Does the Squeeze Theorem require the same limit from both sides?

Yes, $f(x)$ and $g(x)$ must converge to the same limit. Critical condition: both outer functions must approach identical limits.
What is the limit of $x^5 \text{cos}(\frac{1}{x^3})$ as $x \to 0$ using the Squeeze Theorem?

0. Bounded by $-x^5 \leq x^5\cos(\frac{1}{x^3}) \leq x^5$ , both approach 0.
Does the Squeeze Theorem require continuity of functions?

No, continuity is not required. The theorem only requires the inequality near the limit point.
What is the limit of $x^5 \text{sin}(\frac{1}{x^4})$ as $x \to 0$ using the Squeeze Theorem?

0. Bounded by $-x^5 \leq x^5\sin(\frac{1}{x^4}) \leq x^5$ , both approach 0.
Determine the limit: $x^3 \cos\left(\frac{1}{x^2}\right)$ as $x \to 0$ using the Squeeze Theorem.

0. Bounded by $-|x^3| \leq x^3 \cos\left(\frac{1}{x^2}\right) \leq |x^3|$ , both approach 0.
What is the limit of $x^2 \text{sin}(x)$ as $x \to 0$ using the Squeeze Theorem?

0. Since $|\sin(x)| \leq 1$ , we have $-x^2 \leq x^2\sin(x) \leq x^2$ .
What must be true of the inequalities in the Squeeze Theorem?

They must hold for all $x$ near $c$ except possibly at $c$ . Must be satisfied in a neighborhood around the limit point.
Can the Squeeze Theorem be used if the middle function is undefined at a point?

Yes, it can still be used. The theorem works regardless of the middle function's definition.
Determine the limit: $x^3 \text{cos}(\frac{1}{x^2})$ as $x \to 0$ using the Squeeze Theorem.

0. Bounded by $-|x^3| \leq x^3\cos(\frac{1}{x^2}) \leq |x^3|$ , both approach 0.
Find the limit of $x^2 \text{sin}(\frac{1}{x})$ as $x \to 0$ using the Squeeze Theorem.

0. Since $-|x^2| \leq x^2\sin(\frac{1}{x}) \leq |x^2|$ and both bounds approach 0.
What is the limit of $x^6 \text{sin}(\frac{1}{x^5})$ as $x \to 0$ using the Squeeze Theorem?

0. Bounded by $-x^6 \leq x^6\sin(\frac{1}{x^5}) \leq x^6$ , both approach 0.
Determine the limit: $x^2 \text{cos}(\frac{1}{x^4})$ as $x \to 0$ using the Squeeze Theorem.

0. Bounded by $-x^2 \leq x^2\cos(\frac{1}{x^4}) \leq x^2$ , both approach 0.
Determine $\text{lim}_{x \to 0} x^2 \text{cos}(\frac{1}{x})$ using the Squeeze Theorem.

0. Bounded by $-x^2 \leq x^2\cos(\frac{1}{x}) \leq x^2$ , both approach 0.
What is a necessary condition for using the Squeeze Theorem?

Function is squeezed between two converging functions. The middle function must be trapped between two converging bounds.
Determine the limit: $x^2 \cos(\frac{1}{x^4})$ as $x \to 0$ using the Squeeze Theorem.

0. Bounded by $-x^2 \leq x^2\cos(\frac{1}{x^4}) \leq x^2$ , both approach 0.
What must be true of $f(x)$ and $g(x)$ in the Squeeze Theorem?

Both must converge to the same limit $L$ at $x=c$ . Essential requirement for the theorem to guarantee the middle function's limit.

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