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Squeeze theorem

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54255912
A function \(f\) satisfies \(-x^2\le f(x)\le x^2\) for all \(x\) sufficiently close to \(0\). Use the Squeeze Theorem to find \(\lim_{x\to0}f(x)\).

Hints

- Evaluate the limits of the lower and upper bounds first. - Check whether those two limits agree. - Use the ordering only in a neighborhood of the target input.

Solution

1. As \(x\to0\), both bounding functions satisfy \(-x^2\to0\) and \(x^2\to0\). 2. The function \(f(x)\) remains between these two bounds near \(0\). 3. By the Squeeze Theorem, \(\lim_{x\to0}f(x)=0\).

Answer

\(0\).
54256312
A function \(h\) is defined for \(x\ge0\) and satisfies \(0\le h(x)\le\sqrt{x}\) for \(0<x<1\). Use the Squeeze Theorem to evaluate \(\lim_{x\to0^+}h(x)\).

Hints

- Match the direction of approach to the stated domain. - Find the limits of the lower and upper bounds. - Apply the theorem when those limits agree.

Solution

1. As \(x\to0^+\), the lower bound remains \(0\). 2. The upper bound \(\sqrt{x}\) approaches \(0\). 3. Since \(h(x)\) lies between two functions that both approach \(0\), \(\lim_{x\to0^+}h(x)=0\).

Answer

\(0\).
54256012
For \(x\) near \(1\), a function \(g\) satisfies \(4-2|x-1|\le g(x)\le4+3|x-1|\). Use the Squeeze Theorem to determine \(\lim_{x\to1}g(x)\).

Hints

- Start with the behavior of \(|x-1|\) near the target. - Evaluate each bound independently. - The bounds do not need to be symmetric to squeeze toward the same value.

Solution

1. As \(x\to1\), \(|x-1|\to0\). 2. The lower bound \(4-2|x-1|\) and upper bound \(4+3|x-1|\) both approach \(4\). 3. Therefore the Squeeze Theorem gives \(\lim_{x\to1}g(x)=4\).

Answer

\(4\).
54256112
For all \(x\) sufficiently close to \(2\), a function \(f\) satisfies \(|f(x)-7|\le5(x-2)^2\). Use the Squeeze Theorem to determine \(\lim_{x\to2}f(x)\).

Hints

- Convert the absolute-value inequality into a two-sided bound. - First determine the limit of the deviation from \(7\). - Translate that deviation limit back to the function itself.

Solution

1. The inequality is equivalent to \(-5(x-2)^2\le f(x)-7\le5(x-2)^2\). 2. Both outer expressions approach \(0\) as \(x\to2\). 3. Thus \(f(x)-7\to0\), so \(f(x)\to7\).

Answer

\(7\).
54256212
For every \(x>1\), a function \(g\) satisfies \(|g(x)+3|\le\frac{2}{x}\). Use the Squeeze Theorem to find \(\lim_{x\to\infty}g(x)\).

Hints

- Treat the absolute-value inequality as a bound on the distance from \(-3\). - Examine what happens to the right side as \(x\) grows. - Once the error term is squeezed to zero, recover the function's limit.

Solution

1. Rewrite the bound as \(-\frac{2}{x}\le g(x)+3\le\frac{2}{x}\). 2. Both outer bounds approach \(0\) as \(x\to\infty\). 3. Therefore \(g(x)+3\to0\), which gives \(g(x)\to-3\).

Answer

\(-3\).
54256512
Suppose that for \(x\) near \(1\), \(x^2\le q(x)\le2x+1\). Oskar says the Squeeze Theorem proves that \(\lim_{x\to1}q(x)=2\), because \(2\) lies between the limiting values of the bounds. Evaluate Oskar's reasoning.

Hints

- Compute the two bounding limits before looking at possible middle values. - Recall the exact hypothesis required by the Squeeze Theorem. - Being numerically between two different limiting values is not enough.

Solution

1. The lower bound approaches \(1^2=1\). 2. The upper bound approaches \(2\cdot1+1=3\). 3. The Squeeze Theorem requires the lower and upper bounds to approach the same value. Since they approach \(1\) and \(3\), the theorem does not determine \(\lim_{x\to1}q(x)\).

Answer

The reasoning is invalid. The bounds approach different values, \(1\) and \(3\), so the Squeeze Theorem does not determine the limit of \(q\).
54256612
For all sufficiently large positive \(x\), a function \(g\) satisfies \(x^2-1\le g(x)\le x^2+1\). Use the Squeeze Theorem for infinite limits to describe \(\lim_{x\to\infty}g(x)\).

Hints

- Check the long-run behavior of both bounding functions. - The common behavior can be unbounded rather than a finite number. - Use the ordering only for sufficiently large inputs.

Solution

1. As \(x\to\infty\), both \(x^2-1\) and \(x^2+1\) grow without bound. 2. The function \(g(x)\) remains between these two quantities for all sufficiently large \(x\). 3. Therefore \(g(x)\to+\infty\) as \(x\to\infty\).

Answer

\(g(x)\to+\infty\) as \(x\to\infty\).
54257012
Near \(x=0\), a nonnegative function \(F\) satisfies both \(0\le F(x)\le|x|\) and \(0\le F(x)\le\sqrt{|x|}\). Use the Squeeze Theorem to find \(\lim_{x\to0}F(x)\). Which upper bound is tighter for \(0<|x|<1\)?

Hints

- First check whether each upper bound approaches the same value as the lower bound. - To compare tightness, compare the two positive upper bounds for numbers between \(0\) and \(1\). - A tighter bound is useful but not required if either one already squeezes to the target value.

Solution

1. Both \(|x|\) and \(\sqrt{|x|}\) approach \(0\) as \(x\to0\), so either inequality is enough to squeeze \(F(x)\) to \(0\). 2. For \(0<|x|<1\), \(|x|<\sqrt{|x|}\). 3. Therefore \(|x|\) is the tighter upper bound, and \(\lim_{x\to0}F(x)=0\).

Answer

\(\lim_{x\to0}F(x)=0\). The tighter upper bound near \(0\) is \(|x|\).
54257112
For all sufficiently large positive \(x\), a function \(g\) satisfies \(-\frac{1}{x^2}\le g(x)-5\le\frac{3}{x}\). Use the Squeeze Theorem to find \(\lim_{x\to\infty}g(x)\).

Hints

- Focus first on the quantity \(g(x)-5\). - Evaluate both error bounds at large positive inputs. - Add the center value back after squeezing the error to zero.

Solution

1. As \(x\to\infty\), both \(-\frac{1}{x^2}\) and \(\frac{3}{x}\) approach \(0\). 2. Therefore \(g(x)-5\) is squeezed to \(0\). 3. It follows that \(g(x)\to5\).

Answer

\(5\).
54257312
For every \(x>0\), a function \(H\) satisfies \(\frac{x}{x+1}\le H(x)\le\frac{x+1}{x+2}\). Use the Squeeze Theorem to determine \(\lim_{x\to\infty}H(x)\).

Hints

- Find the end behavior of each rational bound. - Compare the leading terms in numerator and denominator. - Apply the theorem after the two outer limits agree.

Solution

1. Dividing numerator and denominator by \(x\) shows \(\frac{x}{x+1}\to1\). 2. Likewise, \(\frac{x+1}{x+2}\to1\). 3. Since \(H(x)\) stays between two functions approaching \(1\), the Squeeze Theorem gives \(\lim_{x\to\infty}H(x)=1\).

Answer

\(1\).
54257412
A function \(p\) satisfies \(5-|x-2|\le p(x)\le5+|x-2|\) for every \(x\ne2\) sufficiently close to \(2\). In addition, \(p(2)=100\). Use the Squeeze Theorem to find \(\lim_{x\to2}p(x)\), and explain why the value \(p(2)=100\) does not change the conclusion.

Hints

- Evaluate the bounds as \(x\) approaches \(2\). - Keep the function value at exactly \(2\) separate from nearby behavior. - The theorem only needs the ordering in a punctured neighborhood of the target.

Solution

1. As \(x\to2\), both \(5-|x-2|\) and \(5+|x-2|\) approach \(5\). 2. The Squeeze Theorem gives \(\lim_{x\to2}p(x)=5\). 3. The limit depends on values for inputs near \(2\), while the inequality already controls all nearby \(x\ne2\); the single value \(p(2)=100\) is irrelevant to the limit.

Answer

\(\lim_{x\to2}p(x)=5\). The isolated value \(p(2)=100\) does not affect the nearby limiting behavior.
54257512
Near \(x=0\), you already know that \(-2|x|\le F(x)\). Which one of the following upper bounds, if it were also known to hold near \(0\), would be sufficient by the Squeeze Theorem to prove \(\lim_{x\to0}F(x)=0\)? Explain. a) \(F(x)\le|x|\) b) \(F(x)\le1\) c) \(F(x)\le\frac{1}{|x|}\)

Hints

- Start with the limit of the known lower bound. - Test the limiting value of each proposed upper bound. - The Squeeze Theorem needs both outer bounds to approach the same target value.

Solution

1. The known lower bound \(-2|x|\) approaches \(0\) as \(x\to0\). 2. Candidate a) has upper bound \(|x|\to0\), so the lower and upper bounds have the same limit and squeeze \(F(x)\) to \(0\). 3. The upper bounds in b) and c) do not approach \(0\), so those pairs do not satisfy the matching-limit condition.

Answer

a) \(F(x)\le|x|\). This works because both the lower bound \(-2|x|\) and the upper bound \(|x|\) approach \(0\) as \(x\to0\).
55592612
The two panels show the same middle function \(F\) between a lower graph \(L\) and an upper graph \(U\) near \(x=0\). Which panel provides a valid Squeeze Theorem argument for \(\lim_{x\to0}F(x)\)? State the limit and explain why the other panel does not establish it.
Figure for problem 555926

Hints

- A middle graph staying between two others is not enough by itself. - Compare the limiting heights of the two outer graphs in each panel. - Identify the panel in which the vertical gap between the outer bounds collapses to one y-value.

Solution

1. In panel a), the lower and upper graphs both approach y-value \(0\) as \(x\to0\). 2. Because \(F\) stays between those two graphs near \(0\), the Squeeze Theorem gives \(\lim_{x\to0}F(x)=0\). 3. In panel b), the lower graph approaches \(-1\) and the upper graph approaches \(1\). Since the outer limits are different, those bounds alone cannot squeeze \(F\) to one value.

Answer

Panel a) gives the valid squeeze, and \(\lim_{x\to0}F(x)=0\). Panel b) does not establish the limit because its two outer bounds approach different values.
52181112
Let \(f(x)=\frac{5\sin(x)}{x}\), with its maximal real domain. a) State the domain. b) Find all zeros. c) Find \(\lim_{x\to\infty}f(x)\). d) Find a value \(x_0>0\) such that \(|f(x)|<0.01\) for every \(x>x_0\).

Hints

- Exclude inputs that make the denominator zero. - Use the zeros of sine. - Bound the absolute value using \(|\sin(x)|\le1\). - Solve the simpler inequality involving \(\frac{5}{x}\).

Solution

1. The denominator cannot be zero, so the domain is \(\mathbb{R}\setminus\{0\}\). 2. The factor \(\frac{5}{x}\) is never zero. Therefore, zeros occur where \(\sin(x)=0\), at \(x=k\pi\) for \(k\in\mathbb{Z}\setminus\{0\}\). 3. Since \(|\sin(x)|\le1\), \(\left|\frac{5\sin(x)}{x}\right|\le\frac{5}{x}\) for \(x>0\). Because \(\frac{5}{x}\to0\), the squeeze theorem gives \(f(x)\to0\). 4. To guarantee \(|f(x)|<0.01\), it is enough to require \(\frac{5}{x}<0.01\). This gives \(x>500\), so \(x_0=500\) works.

Answer

a) \(\mathbb{R}\setminus\{0\}\) b) \(x=k\pi\), where \(k\in\mathbb{Z}\setminus\{0\}\) c) \(0\) d) \(x_0=500\)
52181212
Let \(g(x)=2+\frac{\cos(x)}{x^2}\), with its maximal real domain. a) State the domain. b) Show that \(g\) has no zeros on \([1, \infty)\). c) Find \(L=\lim_{x\to-\infty}g(x)\). d) Find a positive value \(x_0\) such that \(|g(x)-L|<0.0025\) for every \(x>x_0\).

Hints

- Use \(|\cos(x)|\le1\). - Bound the variable fraction above and below. - Replace the oscillating numerator with its maximum possible absolute value to find a sufficient threshold.

Solution

1. The denominator is zero at \(x=0\), so the domain is \(\mathbb{R}\setminus\{0\}\). 2. For \(x\ge1\), \(\left|\frac{\cos(x)}{x^2}\right|\le\frac{1}{x^2}\le1\). Therefore, \(g(x)\ge2-1=1>0\), so there are no zeros on \([1, \infty)\). 3. Since \(|\cos(x)|\le1\) and \(x^2\to\infty\), the squeeze theorem gives \(\frac{\cos(x)}{x^2}\to0\) as \(x\to-\infty\). Thus, \(L=2\). 4. To guarantee \(|g(x)-2|<0.0025\), it is enough to require \(\frac{1}{x^2}<0.0025\). For positive \(x\), this gives \(x>20\), so \(x_0=20\) works.

Answer

a) \(\mathbb{R}\setminus\{0\}\) b) \(g(x)\ge1>0\) for \(x\ge1\). c) \(L=2\) d) \(x_0=20\)
52190112
Let \(f(x)=(x-2)\sin\left(\frac{1}{x-2}\right)\), with domain \(\mathbb{R}\setminus\{2\}\). Determine whether \(\lim_{x\to2}f(x)\) exists, and find it if it does.

Hints

- Bound the sine factor by its maximum absolute value. - Multiply the bound by \(|x-2|\). - Apply the squeeze theorem.

Solution

1. Since \(\left|\sin\left(\frac{1}{x-2}\right)\right|\le1\), \(|f(x)|\le|x-2|\). 2. Equivalently, \(-|x-2|\le f(x)\le|x-2|\). 3. Both bounding functions approach \(0\) as \(x\to2\). 4. By the squeeze theorem, \(\lim_{x\to2}f(x)=0\).

Answer

The limit exists and equals \(0\).
52190212
Let \(g(x)=\frac{3x+\cos(x^2)}{x}\) for \(x>0\). Determine the behavior of \(g(x)\) as \(x\to\infty\).

Hints

- Split the fraction into two terms. - Use the bounds on cosine. - Apply the squeeze theorem to the oscillating term.

Solution

1. Split the fraction: \(g(x)=3+\frac{\cos(x^2)}{x}\). 2. Since \(-1\le\cos(x^2)\le1\), \(-\frac{1}{x}\le\frac{\cos(x^2)}{x}\le\frac{1}{x}\) for \(x>0\). 3. Both bounds approach \(0\) as \(x\to\infty\), so the squeeze theorem gives \(\frac{\cos(x^2)}{x}\to0\). 4. Therefore, \(g(x)\to3\).

Answer

\(\lim_{x\to\infty}g(x)=3\)
53455112
The graph shows a function \(f\) for positive \(x\). a) From the graph, describe how \(f(x)\) behaves as \(x\) becomes large. Identify the horizontal line around which the graph oscillates, state whether the graph crosses that line, and describe what happens to the height of the oscillations. b) For \(x>0\), you are also told that \(2-\frac{8}{x}\le f(x)\le2+\frac{8}{x}\). Use the Squeeze Theorem to determine \(\lim_{x\to\infty}f(x)\) and the corresponding horizontal asymptote.
Figure for problem 534551

Hints

- In part a), use only visible features of the graph: crossings, center line, and changing vertical spread. - In part b), compare the limits of the supplied lower and upper bounds. - A finite end limit determines a horizontal asymptote.

Solution

1. The graph oscillates above and below \(y=2\), crosses \(y=2\) repeatedly, and the vertical size of the oscillations decreases as \(x\) increases. 2. As \(x\to\infty\), both \(2-\frac{8}{x}\) and \(2+\frac{8}{x}\) approach \(2\). 3. Since \(f(x)\) stays between those two functions for positive \(x\), the Squeeze Theorem gives \(\lim_{x\to\infty}f(x)=2\). 4. Therefore the horizontal asymptote is \(y=2\).

Answer

a) The graph oscillates above and below \(y=2\), crosses \(y=2\) repeatedly, and its oscillation height shrinks as \(x\) increases. b) Since \(2-\frac{8}{x}\to2\) and \(2+\frac{8}{x}\to2\), the Squeeze Theorem gives \(\lim_{x\to\infty}f(x)=2\), so the horizontal asymptote is \(y=2\).
54256412
Near \(x=1\), a function \(p\) satisfies \(|p(x)|\le3|x-1|^2\). Find a positive radius \(r\) such that every \(x\) with \(0<|x-1|<r\) is guaranteed to satisfy \(|p(x)|<0.03\). Use the bounding inequality rather than a formula for \(p\).

Hints

- Make the known upper bound smaller than the required output tolerance. - Solve the resulting inequality for the distance \(|x-1|\). - Any smaller positive radius would also work.

Solution

1. It is enough to make the upper bound satisfy \(3|x-1|^2<0.03\). 2. Dividing by \(3\) gives \(|x-1|^2<0.01\), so \(|x-1|<0.1\) is sufficient. 3. Therefore \(r=0.1\) guarantees \(|p(x)|<0.03\).

Answer

\(r=0.1\) works.
54256712
Suppose \(\lim_{x\to a}u(x)=0\) and \(|v(x)|\le4\) for all \(x\) sufficiently close to \(a\). Use the Squeeze Theorem to show that \(\lim_{x\to a}u(x)v(x)=0\).

Hints

- Convert the bound on \(v\) into a bound on the absolute value of the product. - Use the factor whose limit is known to be zero as the shrinking envelope. - Turn the absolute-value estimate into a two-sided squeeze.

Solution

1. The bound on \(v\) gives \(|u(x)v(x)|=|u(x)||v(x)|\le4|u(x)|\). 2. Since \(u(x)\to0\), we have \(4|u(x)|\to0\). 3. Thus \(-4|u(x)|\le u(x)v(x)\le4|u(x)|\), and the Squeeze Theorem gives \(u(x)v(x)\to0\).

Answer

\(\lim_{x\to a}u(x)v(x)=0\).
54256812
Near \(x=3\), a function \(f\) satisfies \(|f(x)-2x|\le(x-3)^2\). Use the Squeeze Theorem to determine \(\lim_{x\to3}f(x)\).

Hints

- Convert the error bound around \(2x\) into lower and upper functions. - Evaluate both bounding limits at the target input. - The center of the squeeze can itself vary with \(x\).

Solution

1. The inequality gives \(2x-(x-3)^2\le f(x)\le2x+(x-3)^2\). 2. As \(x\to3\), both bounds approach \(6\). 3. Therefore the Squeeze Theorem gives \(\lim_{x\to3}f(x)=6\).

Answer

\(6\).
54256912
To evaluate \(\lim_{x\to0^-}x\sin\left(\frac{1}{x}\right)\), Amara starts with \(-1\le\sin\left(\frac{1}{x}\right)\le1\) and writes \(-x\le x\sin\left(\frac{1}{x}\right)\le x\). Explain the inequality error, write the correct squeeze for \(x<0\), and find the limit.

Hints

- Check the sign of the quantity used to multiply the original inequality. - Reorder the resulting expressions from least to greatest. - Evaluate the two outer limits from the specified one-sided direction.

Solution

1. For \(x<0\), multiplying an inequality by \(x\) reverses both inequality signs. 2. The correct ordering is \(x\le x\sin\left(\frac{1}{x}\right)\le -x\). 3. As \(x\to0^-\), both \(x\) and \(-x\) approach \(0\). The Squeeze Theorem therefore gives the limit \(0\).

Answer

Amara failed to reverse the inequality signs when multiplying by negative \(x\). The correct squeeze is \(x\le x\sin\left(\frac{1}{x}\right)\le -x\), so the limit is \(0\).
54266212
For \(0<|x|<\frac{\pi}{2}\), suppose \(\cos x\le\frac{\sin x}{x}\le1\). Use the Squeeze Theorem to establish \(\lim_{x\to0}\frac{\sin x}{x}\). Angles are measured in radians.

Hints

- Identify the lower and upper bounding functions. - Evaluate each bound as \(x\to0\). - Use the common limiting value in the Squeeze Theorem.

Solution

1. The inequality places \(\frac{\sin x}{x}\) between \(\cos x\) and \(1\) for all sufficiently small nonzero \(x\). 2. As \(x\to0\), \(\cos x\to1\). 3. The upper bound is the constant \(1\), whose limit is also \(1\). 4. Because both bounds tend to \(1\), the Squeeze Theorem gives \(\lim_{x\to0}\frac{\sin x}{x}=1\).

Answer

\(\lim_{x\to0}\frac{\sin x}{x}=1\).
55592712
Evaluate \(\lim_{x\to0^+}\sqrt{x}\cos\left(\frac{1}{x}\right)\) by constructing bounds that are strong enough to apply the Squeeze Theorem. The bounds are not supplied.

Hints

- Start with a bound that is always true for cosine, regardless of its input. - Consider how multiplying an inequality by a nonnegative quantity affects its direction. - Choose outer expressions whose right-hand limits at \(0\) are easy to determine.

Solution

1. For all real inputs where it is defined, \(-1\le\cos(1/x)\le1\). 2. For \(x>0\), \(\sqrt{x}\ge0\), so multiplying preserves the inequalities: \(-\sqrt{x}\le\sqrt{x}\cos(1/x)\le\sqrt{x}\). 3. As \(x\to0^+\), both \(-\sqrt{x}\) and \(\sqrt{x}\) approach \(0\). 4. By the Squeeze Theorem, \(\lim_{x\to0^+}\sqrt{x}\cos(1/x)=0\).

Answer

A sufficient pair of bounds is \(-\sqrt{x}\le\sqrt{x}\cos(1/x)\le\sqrt{x}\), so the limit is \(0\).
55593012
For \(x\ne0\), let \(F(x)=x\sin\left(\frac{1}{x}\right)\). A numerical table samples \(F\) at positive inputs that approach \(0\): <table><tr><th>\(x\)</th><td>\(\frac{2}{\pi}\)</td><td>\(\frac{2}{3\pi}\)</td><td>\(\frac{2}{5\pi}\)</td><td>\(\frac{2}{7\pi}\)</td></tr><tr><th>\(F(x)\)</th><td>\(\frac{2}{\pi}\)</td><td>\(-\frac{2}{3\pi}\)</td><td>\(\frac{2}{5\pi}\)</td><td>\(-\frac{2}{7\pi}\)</td></tr></table> Sasha says the alternating signs show that the limit does not exist. Jordan says a finite table cannot establish the limit by itself. Decide whose reasoning is valid. Then use the most appropriate method to determine \(\lim_{x\to0}F(x)\).

Hints

- Separate what a finite set of sampled values can suggest from what is needed to establish a limit. - Focus on the largest possible magnitude of the sine factor rather than its rapidly changing sign. - Look for two simple functions that trap \(F(x)\) and have the same limit.

Solution

1. Alternating signs do not by themselves prevent convergence; the sampled magnitudes are shrinking. 2. A finite table gives evidence but cannot establish behavior for all sufficiently small nonzero \(x\). 3. Since \(-1\le\sin(1/x)\le1\), multiplying by \(|x|\) gives \(|F(x)|\le|x|\), equivalently \(-|x|\le F(x)\le|x|\). 4. Both bounding functions approach \(0\) as \(x\to0\). 5. By the Squeeze Theorem, \(\lim_{x\to0}F(x)=0\). Jordan's caution about the table is valid, and Sasha's conclusion is invalid.

Answer

Jordan's reasoning about the table is valid. The finite table does not prove nonexistence, and alternating signs can still occur while values approach \(0\). Since \(-|x|\le F(x)\le|x|\) and both bounds approach \(0\), the Squeeze Theorem gives \(\lim_{x\to0}F(x)=0\).
54257212
For every positive integer \(n\), a function \(F\) satisfies \(-\frac{1}{n}\le F\left(\frac{1}{n}\right)\le\frac{1}{n}\). Elif claims the Squeeze Theorem proves \(\lim_{x\to0}F(x)=0\). Is that conclusion justified? Explain what hypothesis is missing.

Hints

- Check where the stated inequality is guaranteed to hold. - A limit concerns all sufficiently nearby inputs, not only one sequence of inputs. - Consider how the function could behave at nearby points not listed by the condition.

Solution

1. The inequality controls \(F\) only at the discrete inputs \(x=1/n\), not at all inputs sufficiently close to \(0\). 2. The Squeeze Theorem requires a neighborhood-wise inequality around the approach point, apart from the point itself if necessary. 3. For example, a function could satisfy \(F(1/n)=0\) for every \(n\) but equal \(1\) at other nearby inputs, so the given information does not force \(\lim_{x\to0}F(x)=0\).

Answer

No. The bounds hold only on the sequence \(x=1/n\), not throughout a punctured neighborhood of \(0\), so the Squeeze Theorem does not apply to the full limit.

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