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Squeeze theorem

Learning goals

  • Identifies lower and upper bounding functions with a common limit.
  • Verifies the target function is trapped between the bounds near the point.
  • Uses the common bounding limit to determine the target limit.
  • Explains why the theorem applies locally.

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Squeeze theorem 1
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

\(0\).
Squeeze theorem 2
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

\(0\).
Squeeze theorem 3
For \(x\ne0\), \(s(x)=\frac{\sin x}{x}\). The curves \(l\) and \(u\) give lower and upper bounds near \(x=0\). What value do the two outer curves approach as \(x\to0\)? What does this suggest for the limit of \(s\)? Explain using the Squeeze Theorem.
Figure for problem 562419

Hints

- Follow the upper and lower curves toward the y-axis. - Do their heights approach the same number? - Think about what that means for any curve trapped between them.

Solution

1. The upper curve stays at height \(1\). The lower curve also approaches height \(1\) as \(x\to0\). 2. The middle curve stays between those bounds for nonzero inputs near \(0\). 3. The Squeeze Theorem therefore gives \(\lim_{x\to0}\frac{\sin x}{x}=1\). The ratio does not have to be defined at zero.

Answer

Both bounds approach \(1\). The Squeeze Theorem forces the middle curve to have limit \(1\).
Squeeze theorem 4
Near \(x=0\), the middle curve \(f\) lies between \(l\) and \(u\). Use the graph to predict \(\lim_{x\to0}f(x)\), and explain why the outer curves justify the answer.
Figure for problem 562447

Hints

- Follow the two outer curves toward the target input from each side. - Does the space between the bounds close up near zero? - What does that tell you about the values trapped between them?

Solution

1. The lower curve \(l\) approaches height \(3\) from both sides of zero. 2. The upper curve \(u\) also approaches \(3\). 3. Since the middle curve stays between them, the Squeeze Theorem gives \(\lim_{x\to0}f(x)=3\).

Answer

\(3\); both outer curves approach \(3\), trapping the middle curve at that height.
Squeeze theorem 5
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

\(4\).
Squeeze theorem 6
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

\(7\).
Squeeze theorem 7
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

\(-3\).
Squeeze theorem 8
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 \(q\).
Squeeze theorem 9
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 \(x\to\infty\).
Squeeze theorem 10
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 \(|x|\).
Squeeze theorem 11
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

\(5\).
Squeeze theorem 12
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

\(1\).
Squeeze theorem 13
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.
Squeeze theorem 14
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 \(x\to0\).
Squeeze theorem 15
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

\(\lim_{x\to0}\frac{\sin x}{x}=1\).
Squeeze theorem 16
In each panel, \(f\) stays between a lower curve \(l\) and an upper curve \(u\) near \(x=0\). Which panel gives a valid Squeeze Theorem argument for the limit of \(f\) at \(0\)? Give that limit and explain why the other panel's bounds do not establish one.
Figure for problem 555926

Hints

- Being trapped between two curves is only part of the theorem's condition. - Compare the limits of the outer curves in each panel. - Look for the panel where the gap shrinks toward the same height on both sides.

Solution

1. In panel a), the lower and upper curves both approach \(0\) as \(x\to0\). 2. Because \(l(x)\le f(x)\le u(x)\), the Squeeze Theorem gives \(\lim_{x\to0}f(x)=0\). 3. In panel b), the lower bound approaches \(-1\) and the upper bound approaches \(1\). The bounds have different limits, so they cannot force a single limiting value for the middle function.

Answer

Panel a) gives \(\lim_{x\to0}f(x)=0\). Panel b)'s bounds approach different values and cannot establish the limit by squeezing.
Squeeze theorem 17
For every \(x\) near \(0\), \(0\le h(x)\le2\). Priya says the Squeeze Theorem shows \(\lim_{x\to0}h(x)=1\), the midpoint of the bounds. Explain why this is wrong, and give two functions that fit the bounds but have different limits.

Hints

- What must be true about the limits of both outer functions? - Do the bounds give an average or a fixed approaching value? - Could different functions fit the same bounds while approaching different heights?

Solution

1. The lower bound approaches \(0\), while the upper bound approaches \(2\). The Squeeze Theorem does not force a limit when the bounds differ. 2. The constant functions \(h_1(x)=0\) and \(h_2(x)=2\) both satisfy the inequality. 3. Their limits are \(0\) and \(2\), showing the bounds do not determine one value, much less the midpoint.

Answer

The bounds approach different values, so squeezing does not apply. For example, \(h_1(x)=0\) and \(h_2(x)=2\) both fit but approach different limits.
Squeeze theorem 18
For \(x\ne0\), let \(f(x)=x^2\cos\left(\frac{1}{x}\right)\). Find two simple functions that bound \(f(x)\) from below and above, then use them to find \(\lim_{x\to0}f(x)\).

Hints

- What range of values can cosine take? - Does multiplying an inequality by a square reverse its direction? - Compare the limits of the bounds you choose.

Solution

1. Since \(-1\le\cos(1/x)\le1\), multiplying by \(x^2\ge0\) gives \(-x^2\le f(x)\le x^2\). 2. Both bounds approach \(0\) as \(x\to0\). 3. The Squeeze Theorem gives \(\lim_{x\to0}f(x)=0\).

Answer

Bounds: \(-x^2\le f(x)\le x^2\); limit \(0\).
Squeeze theorem 19
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) \(x_0=500\)
Squeeze theorem 20
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) \(x_0=20\)
Squeeze theorem 21
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 \(0\).
Squeeze theorem 22
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

\(\lim_{x\to\infty}g(x)=3\)
Squeeze theorem 23
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 \(y=2\).
Squeeze theorem 24
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.
Squeeze theorem 25
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

\(\lim_{x\to a}u(x)v(x)=0\).
Squeeze theorem 26
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

\(6\).
Squeeze theorem 27
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 \(0\).
Squeeze theorem 28
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 \(0\).
Squeeze theorem 29
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 \(\lim_{x\to0}F(x)=0\).
Squeeze theorem 30
For \(0<x<1\), a function \(m\) satisfies \(0\le m(x)\le x^2\). a) Find \(\lim_{x\to0^+}m(x)\) using the Squeeze Theorem. b) Must \(\lim_{x\to0}m(x)\) exist? Give one way to define \(m(x)\) for \(x<0\) that shows your answer.

Hints

- On which side of zero is the bound known to hold? - What do both bounds approach from that side? - What freedom remains when choosing the function on the other side?

Solution

1. Both \(0\) and \(x^2\) approach \(0\) as \(x\to0^+\), and \(m\) lies between them. The Squeeze Theorem gives the right-hand limit \(0\). 2. The bound says nothing about negative inputs. Set \(m(x)=1\) for every \(x<0\), while keeping any permitted values on the right. 3. The resulting left-hand limit is \(1\) and the right-hand limit is \(0\), so the two-sided limit does not exist.

Answer

a) \(\lim_{x\to0^+}m(x)=0\). b) No. For example, set \(m(x)=1\) for \(x<0\); then the one-sided limits disagree.
Squeeze theorem 31
For a small angle \(u\) in radians with \(0<u<\frac{\pi}{2}\), geometry gives \(\sin u\le u\le\tan u\). a) Rearrange these inequalities to place \(\frac{\sin u}{u}\) between \(\cos u\) and \(1\). b) Explain why the same limiting value holds when \(u\to0\) through negative angles.

Hints

- The sign of the angle matters when dividing inequalities. - Which operations keep the inequality directions unchanged for positive angles? - What happens to the sine ratio when the angle changes sign?

Solution

1. Dividing \(\sin u\le u\) by positive \(u\) gives \(\frac{\sin u}{u}\le1\). 2. Since \(u\le\tan u=\frac{\sin u}{\cos u}\), and \(u\) and \(\cos u\) are positive, \(\cos u\le\frac{\sin u}{u}\). 3. Thus \(\cos u\le\frac{\sin u}{u}\le1\), and both bounds approach \(1\) as \(u\to0^+\). 4. The ratio \(\frac{\sin u}{u}\) is even: replacing \(u\) by \(-u\) changes both signs. Hence the left-hand limit also equals \(1\).

Answer

a) \(\cos u\le\frac{\sin u}{u}\le1\). b) \(\frac{\sin(-u)}{-u}=\frac{\sin u}{u}\), so both sides have limit \(1\).
Squeeze theorem 32
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 she right? Explain what is missing.

Hints

- At which inputs is the bound given? - Does a function limit consider only one particular list of nearby inputs? - Try imagining the function at nearby points that are not of the form \(1/n\).

Solution

1. The inequality controls \(F\) only at the inputs \(x=1/n\), not at every input close to \(0\). 2. The Squeeze Theorem for \(x\to0\) needs the bounds to hold at all nearby inputs other than possibly \(0\) itself. 3. For example, \(F(1/n)\) could equal \(0\) for every positive integer \(n\) while \(F(x)=1\) at every other nearby input. Then the claimed limit does not exist.

Answer

No. The inequality holds only at the listed sequence of inputs; it is not known to hold at every input sufficiently close to \(0\).

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