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Calculus learning

Squeeze Theorem: Bounding an Oscillating Function Near a Point

The squeeze theorem proves a limit when direct substitution cannot settle an oscillating factor. Rather than trying to evaluate every oscillation, bound the whole expression by functions whose limits are already known.

Direct answer

If g(x)≤f(x)≤h(x) near a and both outer functions tend to L, then f(x) tends to L.

The inequalities must hold throughout a punctured neighborhood, not merely at selected sample points.

Squeeze Theorem worked method

For x² sin(1/x) with x≠0, use −1≤sin(1/x)≤1. Since x² is nonnegative, multiplying preserves the inequality.

This gives −x²≤x² sin(1/x)≤x². Both bounds tend to zero as x→0, so the middle limit is zero.

For x sin(1/x), use |x sin(1/x)|≤|x|. Multiplying −1≤sin≤1 by negative x would reverse the inequalities; absolute values avoid that error.

For sin(1/x) without a shrinking multiplier, the simple bounds −1 and 1 approach different values and prove nothing. In fact sequences approaching zero can make the sine equal 1 or −1.

A plotted or sampled envelope is a clue, not a proof of the neighborhood inequality. Write the bound and its limit explicitly.

1
−x²≤x²sin(1/x)≤x²
2
lim x→0 x²sin(1/x)=0
3
|xsin(1/x)|≤|x|→0

Worked examples

lim x→0 x² sin(1/x)
00

Bound by ±x².

lim x→0 x sin(1/x)
00

Use the absolute-value bound.

lim x→0 sin(1/x)
Does not exist

Different subsequences have different limits.

Common mistakes

  • • Using unequal outer limits.
  • • Checking only a few numeric points.
  • • Multiplying an inequality by a negative quantity without reversing it.

Try the supported calculator preset

The preset opens a supported expression with its variable and conditions. Teaching examples elsewhere in this guide are reference explanations, not claims that the engine supports every method.

The button evaluates the bound x²→0, not the oscillatory target. The rule engine does not currently certify arbitrary squeeze proofs.

極限:x2x^{2}
Open the calculator without a preset: 極限

References

Frequently asked questions

Does a failed bound mean the limit does not exist?+

No. The bound may simply be too weak.

Must the function be defined at a?+

No. The inequality and limit concern nearby points.

Does the calculator certify every squeeze argument?+

No. The preset checks a simple bounding function. The oscillatory proof here is a human-readable reference, not an engine claim.

Conclusion

Squeeze Theorem starts with the stated domain and conditions. Recheck those before carrying a worked example into a different problem.

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