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4.1.5 · D5Calculus I — Limits & Derivatives

Question bank — Squeeze theorem (sandwich theorem)

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Recall the shape of the theorem before you start: we need three ingredients at once — a lower wall , an upper wall , and the trapped filling with near a point , PLUS both walls sharing the same limit . Miss any one and the conclusion collapses. The traps below all attack one of these ingredients.

True or false — justify

The two walls must tend to the same limit for any conclusion to follow.
True — this is the whole engine. If and , the filling could wander anywhere in between, so nothing is forced.
If holds only at the single point , the theorem still applies.
False — limits ignore the value at and depend on a whole neighbourhood around it, so the inequality must hold on an interval around (except possibly at itself).
The squeeze theorem can prove a limit equals , but never that a limit fails to exist.
True — it is a one-directional tool: matching walls force existence. To show non-existence you need a different argument (e.g. two paths giving different values).
If both walls tend to , the squeeze theorem forces to tend to too.
True — the "same limit" can be (or ); if and , then as well (this is the one-sided "push" version).
You must know an explicit formula for to apply the squeeze theorem.
False — the beauty is that can be a mystery. We only need the two inequalities and the two wall-limits; 's inner workings never appear.
If is squeezed to , then must actually reach the value somewhere near .
False — squeezing controls the limit, not the values. E.g. has limit but takes both positive and negative values arbitrarily close to .
The theorem works identically for sequences with replacing .
True — the Limits of sequences version just reads "for all large enough" instead of "for all near "; the trapping logic is unchanged.
If and only the lower wall converges to , the theorem gives the limit of .
False — you need both walls at . A lower wall alone leaves the ceiling free, so could rise away from .

Spot the error

" because ." — what is wrong?
When the claimed bounds are backwards, since multiplied by a negative flips the order. Correct bounds are .
"Since has no limit at , the product has no limit either." — error?
A limit can exist even when a factor does not, via Bounded times vanishing: bounded () times vanishing () gives . The wild factor is tamed by the shrinking one.
" and but fails at itself, so I cannot conclude." — error?
The value exactly at is irrelevant to a limit; as long as the inequality holds for all other near , the squeeze still delivers .
"I averaged the walls: , , so ." — error?
Averaging is not a theorem. With different wall-limits the squeeze concludes nothing; could approach any value in or oscillate without a limit.
"To bound I multiplied by and flipped the signs to be safe." — error?
No flip is needed. always, and multiplying by a non-negative quantity preserves the inequality direction; flipping would give a false bound.
" holds for all , so the limit is ." — error in the range claim?
The unit-circle derivation assumes ; the two-sided limit follows only after noting is even, extending the bound to negative . See Limits of trigonometric functions.

Why questions

Why does the proof use instead of just one ?
Each wall gives its own window ( for , for ); taking the smaller one guarantees both -bounds hold simultaneously on that single interval.
Why is the squeeze theorem needed at all instead of just substituting ?
For Oscillating functions like there is nothing to substitute — the function is undefined or wildly varying at , so direct evaluation is impossible; bounding sidesteps this.
Why must the same appear on both sides for the – argument to close?
The chain only pins to within of one number ; if the walls used different centres, the two -bands would not overlap into a single .
Why does bounding by (rather than ) work near without case-splitting?
is non-negative for every real , so multiplying preserves order in one shot; changes sign at and would force separate and cases.
Why does "squeeze" connect naturally to Continuity?
Once the squeeze establishes , comparing with is exactly the continuity test; the theorem often supplies the limit that continuity then checks against the function value.
Why can't the squeeze theorem be run "in reverse" to bound the walls from the filling?
The logic flows from known wall-limits to an unknown filling-limit. Knowing tells you nothing forced about or — many different walls can trap the same .

Edge cases

If for all near (walls touch the filling everywhere), does the theorem still apply?
Yes, trivially — equality is a valid case of , and the shared limit of the identical functions is ; the "squeeze" is just a single line collapsing to a point.
What if the walls converge to but is undefined at infinitely many points near ?
The theorem still gives the limit provided is defined and trapped on a punctured neighbourhood; scattered undefined points off that set are fine, but must exist where the inequality is claimed.
Both walls tend to from strictly positive values, i.e. . What sign must have?
The limit is , not "strictly positive" — strict inequalities between functions become non-strict in the limit, so even though throughout.
For a sequence, the bound holds only for . Does the missing matter?
No — sequence limits care only about the tail (all beyond some point), so finitely many early terms, or an undefined , never affect .
If one wall equals the constant and the other genuinely varies toward , is that allowed?
Yes — a constant function is a perfectly valid wall with ; only the shared limit matters, not that both walls move.
Recall One-line self-test before you leave

If you can state, without looking: (1) both walls, same limit, (2) inequality on a neighbourhood not a point, (3) strict inequalities relax in the limit, (4) sign-flip trap when multiplying by something that can be negative — you have covered every trap on this page.

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