Learning Objectives
State the hypotheses of the Squeeze Theorem.
Construct useful bounds for a product involving a bounded function.
Distinguish a successful squeeze from bounds that give no conclusion.
Prerequisites: two-sided limits, inequalities, absolute value, and basic trigonometric bounds. All trigonometric inputs in this lesson are measured in radians.
1. Trap an Output Between Two Known Behaviors
Suppose g(x) ≤ f(x) ≤ h(x) for all x sufficiently close to a, except possibly at a itself. If lim(x → a) g(x) = L and lim(x → a) h(x) = L, then lim(x → a) f(x) = L. This is the Squeeze Theorem.
Both outer functions must approach the same finite number. The inequalities must also hold throughout a nearby interval, not merely at a few values in a table. The middle function's value at a does not matter for this limit argument.
Part of the argument | What to establish |
|---|---|
Lower bound | g(x) ≤ f(x) near the target |
Upper bound | f(x) ≤ h(x) near the target |
Matching limits | Both g and h approach the same L |
Conclusion | f also approaches L |
2. Combine Oscillation with a Shrinking Factor
Worked Example: x² sin(1/x)
Evaluate lim(x → 0) x² sin(1/x). The expression is defined for x ≠ 0, and sin(1/x) oscillates as x approaches zero.
Since −1 ≤ sin(1/x) ≤ 1 and x² ≥ 0, multiplication preserves the inequalities: −x² ≤ x² sin(1/x) ≤ x².
Both −x² and x² approach 0. The Squeeze Theorem therefore gives lim(x → 0) x² sin(1/x) = 0.
The factor sin(1/x) does not need its own limit. Its bounded size, together with the shrinking x² factor, is enough.
This explains why multiplying something that tends to zero by a bounded oscillating quantity can still produce a limit of zero. You cannot justify this example using the ordinary product law, because that law assumes both individual finite limits exist.
3. Absolute Values Handle a Changing Sign
Worked Example: x cos(1/x)
Find lim(x → 0) x cos(1/x). Directly multiplying −1 ≤ cos(1/x) ≤ 1 by x would require separate cases for x > 0 and x < 0.
Instead, use |x cos(1/x)| ≤ |x|. This gives −|x| ≤ x cos(1/x) ≤ |x| for every nonzero x.
Both bounds tend to 0, so the limit is 0. Absolute values provide a single argument valid from both sides.
4. Bounds Need to Meet
The inequality −1 ≤ sin(1/x) ≤ 1 does not prove that sin(1/x) has a limit at zero. Its two constant bounds approach −1 and 1, which differ. The Squeeze Theorem gives no conclusion from those bounds.
In fact, inputs xₙ = 1/(π/2 + 2πn) approach zero through positive values while sin(1/xₙ) = 1. Inputs yₙ = 1/(3π/2 + 2πn) also approach zero while sin(1/yₙ) = −1. These incompatible nearby outputs show why there is no single limit.
Common Mistakes
Boundedness alone does not imply a limit. The shrinking gap between matching bounds is what matters.
When multiplying an inequality by an expression that may be negative, track its sign or use an absolute-value bound.
Do not use a product-of-limits argument when one factor has no limit.
5. Practice
1. Evaluate lim(x → 0) x⁴ cos(3/x).
2. Suppose 2 − x² ≤ f(x) ≤ 2 + |x| near zero. Find lim(x → 0) f(x).
3. Suppose −2 ≤ g(x) ≤ 3 near zero. Do these bounds determine lim(x → 0) g(x)? Explain.
4. If |h(x) − 5| ≤ 3|x − 2| near x = 2, find lim(x → 2) h(x).
Worked Solutions
1. −x⁴ ≤ x⁴ cos(3/x) ≤ x⁴, and both bounds tend to 0. The limit is 0.
2. Both bounds approach 2, so the Squeeze Theorem gives a limit of 2.
3. No. The bounds have different limits. A constant function and an oscillating function could both satisfy them, so these inequalities alone are insufficient.
4. Rewrite the bound as 5 − 3|x − 2| ≤ h(x) ≤ 5 + 3|x − 2|. Both outer expressions approach 5, so the limit is 5.
6. Summary
- Find inequalities valid near the target input.
- Check that both outer limits agree before invoking the theorem.
- A bounded factor multiplied by a shrinking factor often suggests a useful squeeze.
- Absolute values can avoid unnecessary sign cases.