Step by Step
1
The core idea
If a function f(x) is always trapped between two other functions g(x) and h(x) — meaning g(x) ≤ f(x) ≤ h(x) — and both g and h approach the same limit L, then f is squeezed into approaching that same limit L too.
2
When this technique is needed
The squeeze theorem is most useful when f(x) is too complicated or oscillates too wildly to evaluate its limit directly, but you can find simpler bounding functions above and below it.
3
The classic example
For lim(x→0) x²sin(1/x), the sin(1/x) part oscillates unpredictably and has no limit on its own. But since −1 ≤ sin(1/x) ≤ 1 always, multiplying through by x² gives −x² ≤ x²sin(1/x) ≤ x².
4
Finishing the proof
Since both bounding functions (−x² and x²) approach 0 as x approaches 0, the squeeze theorem guarantees that x²sin(1/x) is trapped and must also approach 0 — even though sin(1/x) itself never settles down.
Applied Walkthrough
1
Evaluate lim(x→0) x²sin(1/x) using the squeeze theorem, since sin(1/x) oscillates infinitely fast near 0 and has no limit by itself.
2
Use the fact that sine is always between −1 and 1: −1 ≤ sin(1/x) ≤ 1 for all x ≠ 0.
3
Multiply every part of this inequality by x² (which is always non-negative, so the inequality direction doesn't flip): −x² ≤ x²sin(1/x) ≤ x².
4
Since both −x² and x² approach 0 as x approaches 0, the squeezed function x²sin(1/x) is trapped between them and must also approach 0.
Exam Application
Exams test whether you can identify appropriate bounding functions for a difficult limit, and whether you correctly conclude that the squeezed function shares the same limit as both bounds.
⚠ Common Trap
The most common trap is choosing bounding functions that don't actually approach the same limit as each other — the squeeze theorem only works if both the upper and lower bounding functions converge to the identical value.
✓ Quick Self-Check
1. What does the squeeze theorem state?
If g(x) ≤ f(x) ≤ h(x) and both g and h approach the same limit L, then f also approaches L.
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2. Why is the squeeze theorem needed for lim(x→0) x²sin(1/x)?
Because sin(1/x) oscillates infinitely fast near 0 and has no limit on its own, so the limit can't be evaluated directly.
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3. What are the bounding functions used for x²sin(1/x)?
−x² and x², since sin(1/x) is always between −1 and 1.
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4. What is lim(x→0) x²sin(1/x)?
0.
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5. What condition must both bounding functions satisfy for the squeeze theorem to apply?
They must both approach the same limit value as x approaches the point in question.
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