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📝 Squeeze theorem for limits (18 MCQs)

📖 From Calculus • 2. Limits and Continuity an Introduction • 18 questions available

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What is Squeeze theorem for limits?

Definition:
The Squeeze Theorem states that if $g(x) \le f(x) \le h(x)$ for all $x$ near $a$ (except possibly at $a$ ) and $\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L$ , then $\lim_{x \to a} f(x) = L$ . This theorem is used to evaluate limits of functions that are difficult to compute directly by squeezing" them between two simpler functions with known limits.

Example:
Find $\lim_{x \to 0} x^2 \sin(1/x)$ .
Solution: $-x^2 \le x^2 \sin(1/x) \le x^2$ , and $\lim x^2 = 0$ , so by Squeeze Theorem, limit is $0$ .

Reason:
The Squeeze Theorem is invaluable for proving limits of oscillating functions and for establishing fundamental limits like $\lim_{x \to 0} \frac{\sin x}{x}$ , which are central to calculus and analysis."

Easy

Medium

Hard

📝 All Squeeze theorem for limits MCQs

Q1. What does the Squeezing Theorem assert about the limit of a function f when it is bounded between two functions g and h that share the same limit L as x approaches c?

A.If g(x) ≤ f(x) ≤ h(x) and lim g = lim h = L, then lim f = L. ✅

B.If g(x) ≤ f(x) ≤ h(x) and lim g = lim h = L, then lim f may differ from L.

C.If g(x) ≤ f(x) ≤ h(x), then lim f exists regardless of limits of g and h.

D.If g(x) ≤ f(x) ≤ h(x) and lim g = lim h = L, then lim f does not exist.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The theorem requires three conditions: g and h bound f, g and h have the same limit L at c, and the inequalities hold on a punctured neighborhood of c. When these are satisfied, any function f squeezed between them must share the same limit L, guaranteeing lim f = L.

Q2. Using the Squeezing Theorem, evaluate $\displaystyle\lim_{x\to 0}\frac{\sin x}{x}$ .

A.0

B.∞

C.does not exist

D.1 ✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: By the inequality $\cos x \le \frac{\sin x}{x} \le 1$ for $x$ near 0, and knowing $\lim_{x\to0}\cos x=1$ , the squeeze forces $\lim_{x\to0}\frac{\sin x}{x}=1$ . The other options contradict this established bound.

Q3. Compare the limits $\displaystyle\lim_{x\to0}\frac{1-\cos x}{x^2}$ and $\displaystyle\lim_{x\to0}\frac{1-\cos x}{x}$ .

A.Both limits are 0

B.Both limits are

\tfrac12

C.First limit is

\tfrac12

and second limit is 0 ✅

D.First limit is 0 and second limit is

\tfrac12

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Using $1-\cos x=2\sin^2(x/2)$ and $|\sin(x/2)|\le|x|/2$ , we obtain $0\le1-\cos x\le x^2/2$ . Dividing by $x^2$ gives a squeeze to $\tfrac12$ ; dividing by $x$ yields a squeeze to 0. Hence the first limit equals $\tfrac12$ and the second equals 0.

Q4. If the inequalities $g(x)\le f(x)\le h(x)$ hold only for $x>c$ , what can be concluded about the right‑hand limit of f at c?

A.The limit does not exist

B.The limit equals L ✅

C.The limit equals the left‑hand limit

D.No conclusion can be drawn

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: When the bounding functions satisfy the squeeze on a right‑hand neighborhood of c and both have the same right‑hand limit L, the squeeze theorem guarantees $\lim_{x\to c^+}f(x)=L$ . The left‑hand behavior is irrelevant for this conclusion.

Q5. Which condition is NOT required for the Squeezing Theorem to hold?

A.The functions are defined on an interval containing c

B.The inequalities must hold at c ✅

C.Both bounding functions have the same limit at c

D.The function f is bounded on the interval

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The theorem explicitly allows the inequalities to fail at the point c itself; it only requires them on a punctured neighborhood. Therefore, demanding the inequalities at c is unnecessary, while the other three conditions are essential.

Q6. Determine $\displaystyle\lim_{x\to0} x\sin\!\left(\frac{1}{x}\right)$ using the Squeeze Theorem.

A.1

B.-1

C.does not exist

D.0 ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: Since $-1\le\sin(1/x)\le1$ for all non‑zero x, multiplying by $|x|$ yields $-|x|\le x\sin(1/x)\le|x|$ . As $x\to0$ , both bounds approach 0, so by squeezing the limit of the product is 0.

Q7. Which inequality is essential to prove $\displaystyle\lim_{x\to0}\frac{\sin x}{x}=1$ via the Squeeze Theorem?

-|x|\le\sin x\le|x|

leading to limit 1

\cos x\le\frac{\sin x}{x}\le1

✅

-1\le\sin x\le1

leading to limit 0

D.No inequality works for this limit

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The classic proof uses the geometric inequality $\cos x\le\frac{\sin x}{x}\le1$ for $0<x<\pi/2$ . Since $\lim_{x\to0}\cos x=1$ , the squeeze forces $\lim_{x\to0}\frac{\sin x}{x}=1$ . The other options either give weaker bounds or incorrect conclusions.

Q8. Suppose $-|x|\le f(x)\le|x|$ for all x. What can be inferred about $\displaystyle\lim_{x\to0}f(x)$ ?

A.0

B.1

C.does not exist ✅

D.cannot be determined

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Both bounding functions $-|x|$ and $|x|$ have limit 0 as $x\to0$ . By the squeeze theorem, any function trapped between them must also tend to 0, guaranteeing $\lim_{x\to0}f(x)=0$ .

Q9. Which of the following functions can be shown to have limit 0 at 0 by the Squeeze Theorem?

x^{2}\sin\!\left(\tfrac{1}{x}\right)

✅

\sin x

\tfrac{1}{x}

e^{x}

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: For $f(x)=x^{2}\sin(1/x)$ , we have $-x^{2}\le f(x)\le x^{2}$ . Since $\lim_{x\to0}(\pm x^{2})=0$ , the squeeze forces $\lim_{x\to0}f(x)=0$ . The other functions either lack suitable bounds or have non‑zero limits.

Q10. Find $\displaystyle\lim_{x\to0}\frac{x^{2}}{\sqrt{x^{2}+1}}$ using appropriate inequalities.

A.0 ✅

B.1

C.-1

D.does not exist

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Because $\sqrt{x^{2}+1}\ge1$ for all x, we have $0\le\frac{x^{2}}{\sqrt{x^{2}+1}}\le x^{2}$ . Both bounding expressions tend to 0 as $x\to0$ ; thus, by squeezing, the limit of the given fraction is 0.

Q11. In the one‑sided version of the Squeeze Theorem for $x\to c^{+}$ , which hypothesis must be ensured?

A.The interval becomes

(c,c+\delta)

B.The inequality may fail at c

C.Both g and h need the same right‑hand limit ✅

D.None of the above

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: For a right‑hand limit, the bounding functions must agree on their right‑hand limits at c. The inequalities need only hold on a punctured right‑hand neighborhood, and the interval is naturally $(c,c+\delta)$ . Hence the essential condition is the equality of the right‑hand limits.

Q12. Which pair of functions can be used to squeeze $\displaystyle f(x)=\frac{\sin x}{x}$ as $x\to\infty$ ?

-\tfrac{1}{x}\le f(x)\le \tfrac{1}{x}

✅

-1\le f(x)\le 1

0\le f(x)\le \tfrac{1}{x}

\tfrac{1}{x}\le f(x)\le 1

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Since $|\sin x|\le1$ , dividing by $|x|$ gives $-1/|x|\le\sin x/x\le1/|x|$ . As $x\to\infty$ , both bounds approach 0, so the squeeze theorem shows $\lim_{x\to\infty}\sin x/x=0$ .

Q13. Using the Squeeze Theorem, what is $\displaystyle\lim_{x\to\infty}\frac{\sin x}{x^{2}}$ ?

A.0 ✅

B.does not exist

C.1

D.-1

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Because $-1\le\sin x\le1$ , dividing by $x^{2}>0$ yields $-1/x^{2}\le\sin x/x^{2}\le1/x^{2}$ . Both outer terms tend to 0 as $x\to\infty$ ; thus the squeezed function also tends to 0.

Q14. If $g(x)\le f(x)\le h(x)$ holds for all $x\neq c$ near c, but $\lim_{x\to c^{-}}g(x)=L_{1}$ and $\lim_{x\to c^{+}}h(x)=L_{2}$ with $L_{1}\neq L_{2}$ , what can be said about $\lim_{x\to c}f(x)$ ?

A.The limit does not exist ✅

B.The limit equals the average of

L_{1}

and

L_{2}

C.The limit may exist if extra conditions are met

D.The limit equals

L_{1}

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: When the left‑ and right‑hand limits of the bounding functions differ, the squeeze theorem cannot guarantee a common limit for f at c. Without a single shared limit for both sides, we cannot conclude the existence or value of $\lim_{x\to c}f(x)$ .

Q15. To prove $\displaystyle\lim_{x\to0}\frac{1-\cos x}{x^{2}}=\tfrac12$ via the Squeeze Theorem, which inequality is pivotal?

0\le 1-\cos x\le x^{2}

\tfrac{x^{2}}{2}\le 1-\cos x\le x^{2}

0\le 1-\cos x\le \tfrac{x^{2}}{2}

✅

x^{2}\le 1-\cos x\le 2x^{2}

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Using the identity $1-\cos x=2\sin^{2}(x/2)$ and the fact that $|\sin(u)|\le|u|$ , we obtain $0\le1-\cos x\le x^{2}/2$ . Dividing by $x^{2}$ yields $0\le\frac{1-\cos x}{x^{2}}\le\frac12$ ; the lower bound approaches $\frac12$ as well, forcing the limit to be $\frac12$ .

Q16. Given $x^{3}\le f(x)\le x$ for $x>0$ and both bounding limits as $x\to0^{+}$ equal 0, what is $\displaystyle\lim_{x\to0^{+}}f(x)$ ?

A.0

B.May be non‑zero

C.Does not exist

D.0 ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: Both $x^{3}$ and $x$ approach 0 as $x\to0^{+}$ . Since $f(x)$ is trapped between them on a right‑hand neighborhood, the squeeze theorem forces $\lim_{x\to0^{+}}f(x)=0$ . The statement holds regardless of the specific form of f within the bounds.

Q17. Which statement correctly generalizes the Squeeze Theorem for limits at $+\infty$ ?

A.If

g(x)\le f(x)\le h(x)

for large x and

\lim_{x\to\infty}g=\lim_{x\to\infty}h=L

, then

\lim_{x\to\infty}f=L

. ✅

B.If

g(x)\le f(x)\le h(x)

for large x and

\lim_{x\to\infty}g

exists, then

\lim_{x\to\infty}f

exists.

C.If

g(x)\le f(x)\le h(x)

for all x, then

\lim_{x\to\infty}f

equals the average of the limits of g and h.

D.If

g(x)\le f(x)\le h(x)

and both limits are finite, then f is bounded.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The theorem extends to infinite limits: when two functions bound a third for all sufficiently large x and share the same finite limit L as $x\to\infty$ , the squeezed function must also converge to L. The other statements either omit necessary conditions or assert incorrect conclusions.

Q18. Find $\displaystyle\lim_{x\to\infty}x\ln\!\left(1+\frac{1}{x}\right)$ using the Squeeze Theorem.

A.0

\infty

C.1 ✅

D.does not exist

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: For $x>0$ , the inequality $\frac{1}{x+1}\le\ln\!\left(1+\frac{1}{x}\right)\le\frac{1}{x}$ holds. Multiplying by $x$ gives \(\frac{x}{x+1}\le x\ln(1+1/x)\le1\

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