AP Calculus AB/BC · Unit 1 · Topic 1.8
Determining Limits Using the Squeeze Theorem
Trap a function between two functions that approach the same value.
1. Topic Focus
Build the language of limits, connect numerical, graphical, and algebraic representations, and use continuity theorems with verified hypotheses.
This topic: Trap a function between two functions that approach the same value.
2. Key Relationship
Read every symbol with its domain, direction, units, and hypotheses before applying the relationship.
3. Visual Connection
4. Worked Example
Since |x²sin(1/x)|≤x², its limit at 0 is 0.
Write the governing relationship first, carry out the algebra cleanly, and finish with a sentence that answers the mathematical question.
5. Concept Development
What the squeeze theorem guarantees
Suppose three functions satisfy
for every sufficiently close input \(x\ne a\). If both outer functions approach the same finite number \(L\), then the middle function must approach that number:
The outer functions may approach \(L\) from different directions. What matters is that the vertical interval between them closes around one common value.
Verify all three hypotheses
| Required condition | What to show | Why it matters |
|---|---|---|
| Ordering | \(g(x)\le f(x)\le h(x)\) near \(a\) | The target is genuinely trapped |
| Common outer limit | Both \(g\) and \(h\) approach the same \(L\) | Different outer limits leave room for many middle behaviors |
| Nearby domain | The inequalities hold on a deleted neighborhood, or on the requested one side | A limit depends on nearby inputs, not only the point \(a\) |
The values \(f(a)\), \(g(a)\), and \(h(a)\) do not need to exist or agree. The inequalities also do not need to hold far away from the target.
The distance form is often faster
To prove that \(f(x)\to L\), it is enough to trap the distance from \(f(x)\) to \(L\):
Because the nonnegative distance is squeezed between 0 and a quantity tending to 0, \(|f(x)-L|\to0\), which forces \(f(x)\to L\). This version prevents sign mistakes when the shrinking factor can be negative.
A reusable bounded-times-vanishing pattern
If \(|b(x)|\le M\) near \(a\), where \(M\) is a fixed positive constant, and \(q(x)\to0\), then
Therefore \(q(x)b(x)\to0\), even when \(b(x)\) oscillates and has no limit. Common bounds include
The vanishing factor controls the amplitude; the oscillation is allowed to continue indefinitely inside an envelope whose height shrinks to zero.
How to construct useful bounds
| Middle expression | Useful fact | Resulting bound |
|---|---|---|
| \(q(x)\sin(u(x))\) | \(|\sin(u(x))|\le1\) | \(|q(x)\sin(u(x))|\le|q(x)|\) |
| \(q(x)\cos(u(x))\) | \(|\cos(u(x))|\le1\) | \(|q(x)\cos(u(x))|\le|q(x)|\) |
| \(L+q(x)b(x)\) | Center the expression at \(L\) | \(|[L+q(x)b(x)]-L|\le M|q(x)|\) |
| A function with supplied inequalities | Use the given lower and upper functions directly | Compute both outer limits and compare them |
If multiplying an inequality by a quantity whose sign is unknown, use absolute values or split into cases. Multiplication by a negative quantity reverses the inequality signs.
One-sided squeezing
The theorem also applies to one-sided limits. For \(x\to a^+\), the ordering and outer limits need only hold for inputs immediately to the right of \(a\). The same principle applies from the left.
Then \(\lim\limits_{x\to a^+}f(x)=L\). A one-sided conclusion alone does not establish a two-sided limit.
Connection to the foundational sine limit
For angles measured in radians and \(0<\theta<\pi/2\), a geometric comparison gives
As \(\theta\to0^+\), both outer functions approach 1, so the middle ratio approaches 1. The ratio \(\sin\theta/\theta\) is even, giving the same conclusion from the left:
This result becomes a standard building block for later trigonometric limits. The radian condition is essential.
A complete justification routine
- Identify the difficult target function and the approach direction.
- State explicit lower and upper bounds, or an absolute-value bound.
- Specify that the inequality holds for all sufficiently close valid inputs.
- Calculate the limits of both outer bounds.
- Confirm that the outer limits are equal.
- Invoke the squeeze theorem and state the target limit.
When squeezing does not justify a conclusion
- The lower and upper limits are different.
- Only one inequality is known, so the function is not trapped on both sides.
- The proposed ordering fails arbitrarily close to the target.
- The bounds are themselves harder to evaluate than the original expression.
- A student says only that a trig factor is bounded without showing a shrinking envelope.
- The conclusion uses a two-sided limit after verifying the inequality on only one side.
6. Detailed Worked Example and Error Check
Example 1: A bounded oscillation with shrinking amplitude.
Since \(|\cos(3/x)|\le1\),
The right side approaches 0, so the distance form of the squeeze theorem gives
The cosine factor itself has no limit, but it cannot escape the shrinking envelope.
Example 2: Squeeze around a nonzero target.
Measure the distance from 5:
Therefore the entire expression approaches \(\boxed{5}\). Squeezing is not limited to targets of 0.
Example 3: Use supplied outer functions. Suppose that near \(x=2\),
Both bounds approach 4:
All three hypotheses are satisfied, so \(\boxed{\lim\limits_{x\to2}f(x)=4}\). No formula or value for \(f(2)\) is required.
Example 4: A one-sided squeeze. Assume \(0\le r(x)\le\sqrt{x-3}\) for \(3<x<3.1\). Then
Thus \(\boxed{\lim\limits_{x\to3^+}r(x)=0}\). The information does not determine the left-hand or two-sided limit because no left-side domain or bounds were supplied.
Example 5: Recognize insufficient bounds. If
the lower limit is 1 and the upper limit is 3. The inequalities constrain \(k\), but the gap does not close. The squeeze theorem gives no unique value for \(\lim\limits_{x\to1}k(x)\).
7. AP Reasoning Routine
Read one-sided behavior first, choose a matching limit procedure, and justify conclusions with definitions or theorem conditions.
- Identify the representation and requested quantity.
- State the rule or theorem and verify its conditions.
- Keep exact values until the final requested approximation.
- Interpret sign, units, interval, and context.
Use the squeeze theorem where justified. State the bounds and verify their limits.
(a) Evaluate \(\lim\limits_{x\to0}x^3\cos(1/x^2)\).
(b) Evaluate \(\lim\limits_{x\to0}[7+x^2\sin(5/x)]\).
(c) Suppose \(4-x^2\le m(x)\le4+x^2\) near 0. Find \(\lim\limits_{x\to0}m(x)\).
(d) Suppose \(0\le n(x)\le(x-2)^2\) only for \(x>2\) sufficiently close to 2. State exactly which limit is guaranteed.
(e) If \(x\le p(x)\le2x+3\) near \(x=1\), explain whether the squeeze theorem determines \(\lim\limits_{x\to1}p(x)\).
(f) Given \(|b(x)|\le6\) near \(x=-1\), prove that \(\lim\limits_{x\to-1}(x+1)^2b(x)=0\).
(g) Use the foundational sine limit to evaluate \(\lim\limits_{x\to0}\frac{\sin(6x)}{4x}\).
(h) Explain why changing the value of \(x^2\cos(3/x)\) at \(x=0\) cannot change its limit there.
Check the solution
For part (a), \(|x^3\cos(1/x^2)|\le|x|^3\to0\), so the limit is 0. For part (b), the distance from 7 satisfies \(|x^2\sin(5/x)|\le x^2\to0\), so the limit is 7. In part (c), both \(4-x^2\) and \(4+x^2\) approach 4, so \(m(x)\to4\). Part (d) guarantees only \(\lim\limits_{x\to2^+}n(x)=0\); no left-side information is given. In part (e), the outer limits are 1 and 5, so the theorem does not determine a unique limit. In part (f), \(|(x+1)^2b(x)|\le6(x+1)^2\to0\), proving the limit is 0. In part (g), \(\sin(6x)/(4x)=(3/2)[\sin(6x)/(6x)]\), so the limit is \(3/2\). In part (h), a limit depends on values at nearby nonzero inputs; changing one value at the target does not alter the squeezing inequalities on a deleted neighborhood.