Determining Limits Using the Squeeze Theorem
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AP Calculus AB › Determining Limits Using the Squeeze Theorem
Which of the following is a correct application of the Squeeze Theorem to find $$\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right)$$?
Since $$-1 \leq \sin\left(\frac{1}{x}\right) \leq 1$$, we have $$-x^2 \leq x^2 \sin\left(\frac{1}{x}\right) \leq x^2$$
Since $$\sin\left(\frac{1}{x}\right)$$ oscillates, the limit does not exist by the Squeeze Theorem
The Squeeze Theorem cannot be applied because $$\sin\left(\frac{1}{x}\right)$$ is undefined at $$x = 0$$
Since $$0 \leq \sin\left(\frac{1}{x}\right) \leq 1$$, we have $$0 \leq x^2 \sin\left(\frac{1}{x}\right) \leq x^2$$
Explanation
Since $$-1 \leq \sin\left(\frac{1}{x}\right) \leq 1$$ for all $$x \neq 0$$, multiplying by $$x^2 \geq 0$$ gives $$-x^2 \leq x^2 \sin\left(\frac{1}{x}\right) \leq x^2$$. Since both $$\lim_{x \to 0} (-x^2) = 0$$ and $$\lim_{x \to 0} x^2 = 0$$, the Squeeze Theorem gives $$\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0$$.
What is $$\lim_{x \to 0} x \cos\left(\frac{1}{x^2}\right)$$ using the Squeeze Theorem?
The limit does not exist due to oscillation
$$1$$
$$0$$
$$-1$$
Explanation
Since $$-1 \leq \cos\left(\frac{1}{x^2}\right) \leq 1$$ for all $$x \neq 0$$, we have $$-|x| \leq x \cos\left(\frac{1}{x^2}\right) \leq |x|$$ for all $$x \neq 0$$. Since $$\lim_{x \to 0} (-|x|) = 0$$ and $$\lim_{x \to 0} |x| = 0$$, the Squeeze Theorem gives $$\lim_{x \to 0} x \cos\left(\frac{1}{x^2}\right) = 0$$.
If $$3x^2 - x^4 \leq g(x) \leq 3x^2 + x^4$$ for all $$x$$ near 0, what is $$\lim_{x \to 0} \frac{g(x)}{x^2}$$?
$$3$$
$$1$$
The limit does not exist
$$0$$
Explanation
Dividing the inequality by $$x^2 > 0$$ (for $$x$$ near but not equal to 0), we get $$3 - x^2 \leq \frac{g(x)}{x^2} \leq 3 + x^2$$. Since $$\lim_{x \to 0} (3 - x^2) = 3$$ and $$\lim_{x \to 0} (3 + x^2) = 3$$, the Squeeze Theorem gives $$\lim_{x \to 0} \frac{g(x)}{x^2} = 3$$.
If $$|f(x) - 3| \leq x^2$$ for all $$x$$ near 0, what is $$\lim_{x \to 0} f(x)$$?
$$3$$
The limit cannot be determined from this information
$$0$$
$$-3$$
Explanation
The inequality $$|f(x) - 3| \leq x^2$$ is equivalent to $$-x^2 \leq f(x) - 3 \leq x^2$$, which gives $$3 - x^2 \leq f(x) \leq 3 + x^2$$. Since $$\lim_{x \to 0} (3 - x^2) = 3$$ and $$\lim_{x \to 0} (3 + x^2) = 3$$, the Squeeze Theorem gives $$\lim_{x \to 0} f(x) = 3$$.
What is $$\lim_{x \to 0} x^3 \sin\left(\frac{2}{x}\right)$$ using the Squeeze Theorem?
$$-2$$
$$2$$
The limit does not exist due to the oscillating sine function
$$0$$
Explanation
Since $$-1 \leq \sin\left(\frac{2}{x}\right) \leq 1$$ for all $$x \neq 0$$, multiplying by $$x^3$$ gives $$-|x|^3 \leq x^3 \sin\left(\frac{2}{x}\right) \leq |x|^3$$ (accounting for the sign of $$x^3$$). Since $$\lim_{x \to 0} (-|x|^3) = 0$$ and $$\lim_{x \to 0} |x|^3 = 0$$, the Squeeze Theorem gives $$\lim_{x \to 0} x^3 \sin\left(\frac{2}{x}\right) = 0$$.
Which of the following inequalities would allow us to conclude that $$\lim_{x \to 0} f(x) = 0$$ using the Squeeze Theorem?
$$-x^4 \leq f(x) \leq x^2$$ for all $$x$$ near 0
$$x^2 - 1 \leq f(x) \leq x^2 + 1$$ for all $$x$$ near 0
$$\sin x \leq f(x) \leq \cos x$$ for all $$x$$ near 0
$$-|x| \leq f(x) \leq |x|$$ for all $$x$$ near 0
Explanation
For the Squeeze Theorem to give $$\lim_{x \to 0} f(x) = 0$$, both bounding functions must approach 0. In choice B, $$\lim_{x \to 0} (-|x|) = 0$$ and $$\lim_{x \to 0} |x| = 0$$, so the theorem applies. Choice A has bounds approaching different values, choice C has bounds both approaching 1, and choice D has bounds approaching $$\sin(0) = 0$$ and $$\cos(0) = 1$$.
What is $$\lim_{t \to 0} \frac{t^2 \sin(3t)}{t}$$ using the Squeeze Theorem?
$$0$$
$$3$$
The limit does not exist
$$1$$
Explanation
First, simplify: $$\frac{t^2 \sin(3t)}{t} = t \sin(3t)$$ for $$t \neq 0$$. Since $$-1 \leq \sin(3t) \leq 1$$, multiplying by $$t$$ gives $$-|t| \leq t \sin(3t) \leq |t|$$ (considering the sign of $$t$$). Since $$\lim_{t \to 0} (-|t|) = 0$$ and $$\lim_{t \to 0} |t| = 0$$, the Squeeze Theorem gives $$\lim_{t \to 0} t \sin(3t) = 0$$.
If $$5 - x^2 \leq h(x) \leq 5 + x^6$$ for all $$x$$ near 0, what is $$\lim_{x \to 0} h(x)$$?
The limit cannot be determined because the bounds have different powers
$$0$$
$$10$$
$$5$$
Explanation
By the Squeeze Theorem, since $$\lim_{x \to 0} (5 - x^2) = 5 - 0 = 5$$ and $$\lim_{x \to 0} (5 + x^6) = 5 + 0 = 5$$, and $$5 - x^2 \leq h(x) \leq 5 + x^6$$, we have $$\lim_{x \to 0} h(x) = 5$$. The fact that the bounds have different powers is irrelevant as long as both approach the same limit.
What is $$\lim_{x \to 0} \sqrt{x^2 + x^4} \cos\left(\frac{1}{x}\right)$$ using the Squeeze Theorem?
$$-1$$
$$0$$
$$1$$
The limit does not exist due to oscillation
Explanation
Since $$-1 \leq \cos\left(\frac{1}{x}\right) \leq 1$$ and $$\sqrt{x^2 + x^4} \geq 0$$, we have $$-\sqrt{x^2 + x^4} \leq \sqrt{x^2 + x^4} \cos\left(\frac{1}{x}\right) \leq \sqrt{x^2 + x^4}$$. Since $$\sqrt{x^2 + x^4} = |x|\sqrt{1 + x^2}$$ and $$\lim_{x \to 0} |x|\sqrt{1 + x^2} = 0 \cdot 1 = 0$$, both bounds approach 0, so by the Squeeze Theorem, the limit is 0.
If $$1 - 2|x| \leq g(x) \leq 1 + 2|x|$$ for all $$x$$ near 0, what is $$\lim_{x \to 0} g(x)$$?
$$1$$
$$2$$
$$-1$$
$$0$$
Explanation
Since $$\lim_{x \to 0} (1 - 2|x|) = 1 - 2(0) = 1$$ and $$\lim_{x \to 0} (1 + 2|x|) = 1 + 2(0) = 1$$, and $$1 - 2|x| \leq g(x) \leq 1 + 2|x|$$, the Squeeze Theorem gives $$\lim_{x \to 0} g(x) = 1$$. Both bounding functions approach the same limit of 1.