Squeeze Theorem - Calculus 1

1. If $2 - x^4 \le f(x) \le 2 + \cos(x)$ for all $x$, what is $\lim_{x \to 0} f(x)$?
   Answer: The Squeeze Theorem is not applicable. To use the Squeeze Theorem, the limits of the upper and lower bounding functions must be equal. Let's find these limits as $x \to 0$. The limit of the lower bound is $\lim_{x \to 0} (2 - x^4) = 2 - 0 = 2$. The limit of the upper bound is $\lim_{x \to 0} (2 + \cos(x)) = 2 + \cos(0) = 2 + 1 = 3$. Since the limits of the bounding functions are different (2 and 3), the Squeeze Theorem cannot be applied to determine $\lim_{x \to 0} f(x)$. The given information is insufficient to find the limit. A student might incorrectly average the limits or assume the theorem works even if the limits differ.
2. To evaluate $\lim_{x \to \infty} \frac{e^{-x}}{2-\cos(x)}$ using the Squeeze Theorem, which of the following inequalities provides the correct bounding functions, $g(x)$ and $h(x)$, for $f(x) = \frac{e^{-x}}{2-\cos(x)}$?
   Answer: $\frac{e^{-x}}{3} \le f(x) \le e^{-x}$ First, we must bound the denominator, $2-\cos(x)$. We know that $-1 \le \cos(x) \le 1$. Multiplying by -1 reverses the inequalities: $1 \ge -\cos(x) \ge -1$. Adding 2 to all parts gives $2+1 \ge 2-\cos(x) \ge 2-1$, which is $3 \ge 2-\cos(x) \ge 1$. Since the denominator is always positive, we can take the reciprocal, which reverses the inequalities again: $\frac{1}{3} \le \frac{1}{2-\cos(x)} \le 1$. The numerator $e^{-x}$ is positive for all real $x$. Multiplying the inequality by $e^{-x}$ gives the correct bounds: $\frac{e^{-x}}{3} \le \frac{e^{-x}}{2-\cos(x)} \le e^{-x}$.
3. If $\lim_{x \to c} g(x) = \lim_{x \to c} h(x) = L$ and $g(x) < f(x) < h(x)$ for all $x$ in an open interval containing $c$ (except possibly at $c$), what can be said about $\lim_{x \to c} f(x)$?
   Answer: The limit is $L$. The Squeeze Theorem holds even if the inequalities are strict (i.e., $g(x) < f(x) < h(x)$). The conclusion is the same. As $x$ gets arbitrarily close to $c$, the values of $g(x)$ and $h(x)$ get arbitrarily close to $L$. Since $f(x)$ is trapped between them, its value must also get arbitrarily close to $L$. Therefore, $\lim_{x \to c} f(x) = L$. The strictness of the inequality does not affect the limit value.
4. For which of the following limits is the Squeeze Theorem the most direct method of evaluation among standard Calculus 1 techniques?
   Answer: $\lim_{x \to 0} (e^x-1) \cos\left(\frac{1}{x}\right)$ Let's analyze each option. (A) is typically solved using L'Hôpital's Rule or the known limit $\lim_{x \to 0} \frac{\sin(x)}{x}=1$. (B) is most directly solved by dividing the numerator and denominator by $x^4$. (D) is solved by factoring the numerator and simplifying. (C) is a classic Squeeze Theorem problem. The function $\cos(1/x)$ is bounded between -1 and 1, and the factor $(e^x-1)$ approaches 0 as $x \to 0$. This '0 times bounded' structure is the canonical setup for the Squeeze Theorem, making it the most direct method. The inequality is $-(e^x-1) \le (e^x-1) \cos(1/x) \le (e^x-1)$ for $x>0$, and the limits of the bounds are both 0.
5. Evaluate $\lim_{x \to 0} \frac{x^2}{4 - \cos^2(1/x)}$.
   Answer: $0$ We can use the Squeeze Theorem by first establishing bounds for the denominator. We know that $0 \le \cos^2(1/x) \le 1$. Multiplying by $-1$ reverses the inequality: $-1 \le -\cos^2(1/x) \le 0$. Adding $4$ to all parts gives $3 \le 4 - \cos^2(1/x) \le 4$. Since the denominator is always positive, we can take the reciprocal, which reverses the inequality again: $\frac{1}{4} \le \frac{1}{4 - \cos^2(1/x)} \le \frac{1}{3}$. Finally, we multiply the entire inequality by $x^2$. Since $x^2 \ge 0$, the inequality direction is preserved: $\frac{x^2}{4} \le \frac{x^2}{4 - \cos^2(1/x)} \le \frac{x^2}{3}$. Taking the limit as $x \to 0$, we have $\lim_{x \to 0} \frac{x^2}{4} = 0$ and $\lim_{x \to 0} \frac{x^2}{3} = 0$. By the Squeeze Theorem, the limit of the original function is $0$.
6. Evaluate the limit: $$\lim_{x \to 1} (x-1)^2 \sin\left(\frac{\pi}{x-1}\right)$$
   Answer: 0 The Squeeze Theorem is applicable here. We know that the sine function is bounded between -1 and 1. So, for any $x \neq 1$, we have $-1 \le \sin\left(\frac{\pi}{x-1}\right) \le 1$. Since $(x-1)^2$ is always non-negative, we can multiply the inequality by it without changing the direction of the inequality signs: $-(x-1)^2 \le (x-1)^2 \sin\left(\frac{\pi}{x-1}\right) \le (x-1)^2$. Now, we take the limit of the bounding functions as $x \to 1$: $\lim_{x \to 1} -(x-1)^2 = 0$ and $\lim_{x \to 1} (x-1)^2 = 0$. Since the function is squeezed between two functions that both approach 0, its limit must also be 0.
7. Let $f(x) = x^2$ if $x$ is a rational number, and $f(x) = -x^2$ if $x$ is an irrational number. What is $\lim_{x \to 0} f(x)$?
   Answer: 0 This is a classic application of the Squeeze Theorem. For any real number $x$, either $f(x) = x^2$ or $f(x) = -x^2$. In all cases, the value of $f(x)$ is bounded by $-x^2$ and $x^2$. That is, for all $x$, we have the inequality $-x^2 \le f(x) \le x^2$. We can find the limit of the bounding functions as $x \to 0$: $\lim_{x \to 0} (-x^2) = 0$ and $\lim_{x \to 0} (x^2) = 0$. Since $f(x)$ is squeezed between two functions that both approach 0, the limit of $f(x)$ as $x \to 0$ must also be 0.
8. Evaluate the limit: $$\lim_{x \to \infty} \frac{3x^2 - \cos(2x)}{x^2 + \sin(2x)}$$
   Answer: 3 To evaluate this limit at infinity, we can divide the numerator and the denominator by the highest power of $x$, which is $x^2$. This gives $\lim_{x \to \infty} \frac{3 - \frac{\cos(2x)}{x^2}}{1 + \frac{\sin(2x)}{x^2}}$. We need to evaluate the limits of $\frac{\cos(2x)}{x^2}$ and $\frac{\sin(2x)}{x^2}$ as $x \to \infty$. For $\frac{\cos(2x)}{x^2}$, we use the Squeeze Theorem. Since $-1 \le \cos(2x) \le 1$, we have $\frac{-1}{x^2} \le \frac{\cos(2x)}{x^2} \le \frac{1}{x^2}$. As $x \to \infty$, both $\frac{-1}{x^2}$ and $\frac{1}{x^2}$ approach 0. Thus, $\lim_{x \to \infty} \frac{\cos(2x)}{x^2} = 0$. Similarly, $\lim_{x \to \infty} \frac{\sin(2x)}{x^2} = 0$. Substituting these results back into the main expression, we get $\frac{3 - 0}{1 + 0} = 3$.
9. Suppose that for $x$ in $(-1, 1)$, we have $g(x) \le f(x) \le h(x)$, where $\lim_{x \to 0} g(x) = 0$ and $\lim_{x \to 0} h(x) = 0$. What additional condition is necessary to prove that $f(x)$ is continuous at $x=0$?
   Answer: $f(0)=0$. The Squeeze Theorem, with the given conditions, guarantees that $\lim_{x \to 0} f(x) = 0$. For a function to be continuous at a point $c$, three conditions must be met: 1) $f(c)$ is defined, 2) $\lim_{x \to c} f(x)$ exists, and 3) $\lim_{x \to c} f(x) = f(c)$. We have established that the limit exists and is 0. To satisfy the third condition for continuity at $x=0$, we must have $f(0)$ equal to this limit. Therefore, the necessary additional condition is $f(0)=0$. Option A is not sufficient, as $f(0)$ could be defined but not equal to 0. Option C is about the bounding functions, not $f(x)$ itself. Option D is incorrect because the limit existing does not guarantee continuity.
10. Let $g(x) = -x^2$ and $h(x) = x^2$. The Squeeze Theorem can be used to show $\lim_{x \to 0} f(x) = 0$ for which of the following functions $f(x)$?
    Answer: $f(x) = x^2 \cos(x)$ For the Squeeze Theorem to apply with $g(x) = -x^2$ and $h(x) = x^2$, we need $-x^2 \leq f(x) \leq x^2$ near $x = 0$. Let's check each option: A) $f(x) = x\sin(1/x)$: We have $|x\sin(1/x)| \leq |x|$, so $-|x| \leq f(x) \leq |x|$. For small $x$, $|x| > x^2$, so the bounds $\pm x^2$ are too restrictive. B) $f(x) = x^3\cos(1/x)$: We have $|f(x)| \leq |x^3|$, and for $|x| < 1$, we have $|x^3| < x^2$, so this works. C) $f(x) = x^{1/3}\sin(x)$: For small positive $x$, $x^{1/3} > x^2$, so the bounds are too restrictive. D) $f(x) = x^2\cos(x)$: Since $|\cos(x)| \leq 1$, we have $|f(x)| \leq x^2$, which gives $-x^2 \leq f(x) \leq x^2$. This satisfies the required inequality.
11. Evaluate the limit: $$\lim_{x \to \infty} (2 + \frac{\cos(x^2)}{x+1})$$
    Answer: 2 We can evaluate this limit by considering its parts. The limit of a sum is the sum of the limits, provided they exist. So, $\lim_{x \to \infty} (2 + \frac{\cos(x^2)}{x+1}) = \lim_{x \to \infty} 2 + \lim_{x \to \infty} \frac{\cos(x^2)}{x+1}$. The first limit is clearly 2. For the second limit, we use the Squeeze Theorem. We know $-1 \le \cos(x^2) \le 1$. For $x > -1$, we can divide by $x+1$ (which is positive) to get $\frac{-1}{x+1} \le \frac{\cos(x^2)}{x+1} \le \frac{1}{x+1}$. As $x \to \infty$, both $\frac{-1}{x+1}$ and $\frac{1}{x+1}$ approach 0. So, by the Squeeze Theorem, $\lim_{x \to \infty} \frac{\cos(x^2)}{x+1} = 0$. Therefore, the original limit is $2 + 0 = 2$.
12. Let $f, g,$ and $h$ be functions defined for all real numbers. Suppose that for all $x \neq 2$, we have $g(x) \le f(x) \le h(x)$. If $\lim_{x \to 2} g(x) = -1$ and $\lim_{x \to 2} h(x) = 1$, what can be concluded about $\lim_{x \to 2} f(x)$?
    Answer: The Squeeze Theorem does not provide sufficient information to determine the limit. The Squeeze Theorem requires that the limits of the lower and upper bounding functions are equal. In this case, $\lim_{x \to 2} g(x) = -1$ and $\lim_{x \to 2} h(x) = 1$. Since these limits are not equal, the conditions for the Squeeze Theorem are not met. Therefore, we cannot draw any conclusion about $\lim_{x \to 2} f(x)$. The limit of $f(x)$ may or may not exist. For example, if $f(x) = 0$, the limit is 0. If $f(x) = (x-2) \sin(1/(x-2))$, the limit is 0. If $f(x) = \sin(\pi/(x-2))$, the limit does not exist. The theorem is inconclusive.
13. Evaluate $\lim_{x \to \infty} \frac{3x + \cos(x)}{x - \sin(x)}$.
    Answer: $3$ To evaluate this limit at infinity, we can divide the numerator and the denominator by the highest power of $x$, which is $x$. This gives $\lim_{x \to \infty} \frac{3 + \frac{\cos(x)}{x}}{1 - \frac{\sin(x)}{x}}$. We can evaluate the limits of the terms $\frac{\cos(x)}{x}$ and $\frac{\sin(x)}{x}$ using the Squeeze Theorem. For $\frac{\cos(x)}{x}$, we have $-1 \le \cos(x) \le 1$, so for $x>0$, $-\frac{1}{x} \le \frac{\cos(x)}{x} \le \frac{1}{x}$. As $x \to \infty$, both bounds approach $0$, so $\lim_{x \to \infty} \frac{\cos(x)}{x} = 0$. Similarly, $\lim_{x \to \infty} \frac{\sin(x)}{x} = 0$. Substituting these values back into the main expression gives $\frac{3 + 0}{1 - 0} = 3$.
14. Evaluate the limit: $$\lim_{x \to 0} (x^2 + x^4) \sin(\pi/x)$$
    Answer: 0 The Squeeze Theorem can be applied. The term $\sin(\pi/x)$ is bounded between -1 and 1. The term $(x^2 + x^4)$ is non-negative and approaches 0 as $x \to 0$. We can establish the inequality: $-1 \le \sin(\pi/x) \le 1$. Multiplying by $(x^2+x^4)$ (which is $\ge 0$) gives $-(x^2+x^4) \le (x^2+x^4)\sin(\pi/x) \le (x^2+x^4)$. Now we take the limits of the bounding functions: $\lim_{x \to 0} -(x^2+x^4) = 0$ and $\lim_{x \to 0} (x^2+x^4) = 0$. By the Squeeze Theorem, the limit of the original function must be 0.
15. Suppose $f(x)$ is a function such that $4x - 4 \le f(x) \le x^2$ for all $x \ge 0$. What is $\lim_{x \to 2} f(x)$?
    Answer: 4 We are asked to find the limit of $f(x)$ as $x \to 2$. We can use the Squeeze Theorem. Let's evaluate the limits of the bounding functions as $x \to 2$. The lower bound is $g(x) = 4x - 4$, and its limit is $\lim_{x \to 2} (4x - 4) = 4(2) - 4 = 8 - 4 = 4$. The upper bound is $h(x) = x^2$, and its limit is $\lim_{x \to 2} x^2 = 2^2 = 4$. Since $\lim_{x \to 2} g(x) = \lim_{x \to 2} h(x) = 4$, and $f(x)$ is between $g(x)$ and $h(x)$ in the neighborhood of 2, the Squeeze Theorem guarantees that $\lim_{x \to 2} f(x) = 4$.
16. Evaluate the limit: $$\lim_{x \to \infty} \frac{\ln(x) + \sin(x)}{\ln(x)}$$
    Answer: 1 We can rewrite the expression as $\lim_{x \to \infty} \left( \frac{\ln(x)}{\ln(x)} + \frac{\sin(x)}{\ln(x)} \right) = \lim_{x \to \infty} \left( 1 + \frac{\sin(x)}{\ln(x)} \right)$. To find the limit of the second term, $\frac{\sin(x)}{\ln(x)}$, we use the Squeeze Theorem. We know $-1 \le \sin(x) \le 1$. For $x > 1$, $\ln(x) > 0$, so we can divide by $\ln(x)$ without changing the inequality direction: $\frac{-1}{\ln(x)} \le \frac{\sin(x)}{\ln(x)} \le \frac{1}{\ln(x)}$. As $x \to \infty$, $\ln(x) \to \infty$, so both $\frac{-1}{\ln(x)}$ and $\frac{1}{\ln(x)}$ approach 0. By the Squeeze Theorem, $\lim_{x \to \infty} \frac{\sin(x)}{\ln(x)} = 0$. Therefore, the original limit is $1 + 0 = 1$.
17. Evaluate the limit: $\lim_{x \to \infty} e^{-3x} \sin(x)$.
    Answer: $0$ The function $\sin(x)$ oscillates between $-1$ and $1$. This suggests using the Squeeze Theorem. We have the inequality $-1 \le \sin(x) \le 1$ for all $x$. Since $e^{-3x}$ is always positive, we can multiply the inequality by $e^{-3x}$ without changing the direction of the inequality signs: $-e^{-3x} \le e^{-3x} \sin(x) \le e^{-3x}$. Now we take the limit of the bounding functions as $x \to \infty$. $\lim_{x \to \infty} (-e^{-3x}) = 0$ and $\lim_{x \to \infty} e^{-3x} = 0$. Since the function is squeezed between two functions that both approach $0$, its limit must also be $0$.
18. Let $f(x) = \frac{\arctan(x^3)}{x^2}$. What is $\lim_{x \to \infty} f(x)$?
    Answer: $0$ When evaluating limits at infinity involving rational functions with trigonometric components, you need to consider the dominant behavior of each part as the variable grows large. For $f(x) = \frac{\arctan(x^3)}{x^2}$, let's analyze what happens as $x \to \infty$. The key insight is understanding how $\arctan(x^3)$ behaves for large values of $x$. As $x$ approaches infinity, $x^3$ also approaches infinity, and $\arctan(x^3)$ approaches $\frac{\pi}{2}$ (since the arctangent function has horizontal asymptotes at $\pm\frac{\pi}{2}$). So we have: $\lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{\arctan(x^3)}{x^2} = \lim_{x \to \infty} \frac{\pi/2}{x^2} = \frac{\pi/2}{\infty} = 0$ This confirms answer (D) 0 is correct. (A) is wrong because the limit does exist and equals zero. Both the numerator and denominator have well-defined behavior as $x \to \infty$. (B) 1 might tempt students who incorrectly think $\arctan(x^3)$ grows like $x^2$, but arctangent is bounded. (C) π/2 catches students who focus only on the numerator's limit without considering that it's being divided by $x^2$, which grows without bound. Strategy tip: When you see arctangent in a rational function, remember that $\arctan$ is always bounded between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$, so any polynomial in the denominator will typically drive the overall limit to zero.
19. Evaluate the limit: $$\lim_{x \to 0} \frac{x^3}{2 + \sin^2(1/x)}$$
    Answer: 0 We can use the Squeeze Theorem. First, let's establish bounds for the denominator. We know $0 \le \sin^2(1/x) \le 1$. Adding 2 to all parts gives $2 \le 2 + \sin^2(1/x) \le 3$. Taking the reciprocal reverses the inequalities: $\frac{1}{3} \le \frac{1}{2 + \sin^2(1/x)} \le \frac{1}{2}$. Now we multiply by $x^3$. We must consider $x>0$ and $x<0$. For $x>0$, $\frac{x^3}{3} \le \frac{x^3}{2 + \sin^2(1/x)} \le \frac{x^3}{2}$. As $x \to 0^+$, both bounds approach 0. For $x<0$, $x^3$ is negative, so multiplying reverses the inequalities: $\frac{x^3}{3} \ge \frac{x^3}{2 + \sin^2(1/x)} \ge \frac{x^3}{2}$. As \(x \to 0^-\, both bounds approach 0. Since both one-sided limits are 0, the limit is 0.
20. Let $f$ and $g$ be functions such that $\lim_{x \to 1} f(x) = 0$ and $|g(x)| \le 7$ for all real numbers $x$. What is $\lim_{x \to 1} f(x)g(x)$?
    Answer: $0$ We are given that $|g(x)| \le 7$, which means $-7 \le g(x) \le 7$. This allows us to bound the expression $|f(x)g(x)|$. We have $0 \le |f(x)g(x)| = |f(x)||g(x)| \le 7|f(x)|$. We can apply the Squeeze Theorem to the function $|f(x)g(x)|$. We have $\lim_{x \to 1} 0 = 0$ and $\lim_{x \to 1} 7|f(x)| = 7|\lim_{x \to 1} f(x)| = 7|0| = 0$. By the Squeeze Theorem, $\lim_{x \to 1} |f(x)g(x)| = 0$. A property of limits states that if $\lim_{x \to c} |h(x)| = 0$, then $\lim_{x \to c} h(x) = 0$. Therefore, $\lim_{x \to 1} f(x)g(x) = 0$.

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