Squeeze theorem for sequences: Question bank — Maths deep dive

Spot the error

sin n \to 0

because

\frac{s i n n}{n} \to 0

." — find the flaw.

The flaw is dividing back out:

\frac{s i n n}{n} \to 0

comes from the

\frac{1}{n}

factor collapsing, not from

sin n

; in fact

sin n

has no limit, it keeps oscillating in

[- 1, 1]

0 \leq \frac{n !}{n ^{n}} \leq \frac{1}{n}

, and since the upper wall

\frac{1}{n} \to 0

I'm done — no need for the lower wall." — is the reasoning complete?

Here it happens to work only because the explicit lower wall

0

is also stated and also

\to 0

; without a lower wall an upper bound alone proves nothing, so the "

0 \leq

" is essential, not decorative.

"To bound

\frac{s i n n}{n}

, multiply

- 1 \leq sin n \leq 1

- \frac{1}{n}

to get

- \frac{1}{n} \leq \frac{- s i n n}{n} \leq \frac{1}{n}

." — spot the sign slip.

Multiplying an inequality by the negative number

- \frac{1}{n}

must flip both signs, giving

\frac{1}{n} \geq \frac{- s i n n}{n} \geq - \frac{1}{n}

; forgetting to flip is the classic error, so only divide/multiply by positive quantities like

n > 0

"Walls are

a_{n} = - \frac{1}{n} \to 0

and

c_{n} = 2 - \frac{1}{n} \to 2

; since

b_{n}

is trapped and both walls converge,

b_{n}

converges." — where is the mistake?

Both walls converge, yes, but to different limits (

0

and

2

); the squeeze needs a common limit, so this conclusion is unjustified.

"The bound

a_{n} \leq b_{n} \leq c_{n}

fails at

n = 1, 2, 3

, so the squeeze theorem cannot be used." — correct or not?

Not correct — the theorem only needs the bound for

n \geq N_{0}

; a failure at finitely many early indices is irrelevant, just take

N_{0} = 4

"Since

\frac{l n n}{n} \to 0

, I can conclude

n n \to 0

." — trace the error.

The limit of the log is

0

, and

n n = e^{l n n / n} \to e^{0} = 1

, not

0

; forgetting to undo the log (exponentiate) sends the answer to the wrong place.

Why questions

Why must both outer sequences converge to the same limit?

Because the

ε

-band argument needs a single target

L

: the lower wall gives

L - ε < a_{n}

and the upper gives

c_{n} < L + ε

around the same

L

; two different limits give two different bands that don't pin

b_{n}

Why is it enough for the inequality to hold "eventually" (for

n \geq N_{0}

A limit is determined only by the tail of a sequence; changing or dropping finitely many terms cannot change convergence or the limit value, so early violations are harmless.

Why do we bound

sin n

instead of trying to compute

lim sin n

directly?

Because

sin n

has no limit — it oscillates forever in

[- 1, 1]

; the squeeze lets us sidestep the messy oscillation by wrapping it in tame walls

\pm \frac{1}{n}

that we can evaluate.

Why does dividing

- 1 \leq sin n \leq 1

n

preserve the direction of the inequalities?

Because

n > 0

, and dividing by a positive number keeps order; only division by a negative number would reverse the

\leq

signs.

Why can't a single upper bound ever prove a limit exists?

An upper bound only caps how high the sequence goes; it says nothing about how low it dips, so the sequence can diverge downward while still staying below the wall.

Why is

∣ b_{n} - L ∣ < ε

equivalent to

L - ε < b_{n} < L + ε

Absolute value measures distance, so "

b_{n}

is within

ε

L

" literally means

b_{n}

sits in the open interval

(L - ε, L + ε)

— the two statements are the same fact written two ways.

Why does the proof set

N = max (N_{0}, N_{1}, N_{2})

rather than just one of them?

We need all three facts (the bound, and each wall being inside its band) to be true simultaneously; taking the largest index guarantees every one of them has already "switched on."

Edge cases

b_{n} = L

is constant and

a_{n} \leq L \leq c_{n}

with both walls

\to L

, does the squeeze still apply?

Yes — a constant sequence is a perfectly valid

b_{n}

; it is trapped and trivially converges to

L

, consistent with (though not needing) the theorem.

What if

a_{n} = c_{n}

for all

n

(the walls coincide)?

Then

a_{n} \leq b_{n} \leq a_{n}

forces

b_{n} = a_{n}

exactly, so

b_{n}

inherits the wall's limit directly — the squeeze becomes an equality with no room to wiggle.

Can the walls oscillate (not be monotone) and the squeeze still work?

Yes — the theorem never requires monotone walls, only that each wall converges to

L

; e.g.

a_{n} = - \frac{1 + ( - 1 ) ^{n}}{n}

still

\to 0

and is a valid wall.

What happens if

b_{n}

touches or equals a wall infinitely often?

Nothing breaks — the inequalities are non-strict (

\leq

), so

b_{n}

may equal

a_{n}

c_{n}

as often as it likes and still be squeezed to

L

If both walls diverge to

+ \infty

(

a_{n}, c_{n} \to \infty

) with

a_{n} \leq b_{n} \leq c_{n}

, what can we say about

b_{n}

We conclude

b_{n} \to + \infty

too — this is the "squeeze to infinity" variant, driven by the lower wall

a_{n} \to \infty

pushing

b_{n}

up; the finite-limit squeeze theorem does not directly apply.

Does the squeeze theorem require

b_{n}

to be defined by a formula?

No —

b_{n}

can be any real sequence whatsoever (even one with no closed form); all that matters is the two-sided bound and the common wall limit.

If only

a_{n} \leq b_{n}

holds (no upper wall) but

a_{n} \to L

, can we salvage anything?

Only a one-sided statement: we get

lim inf b_{n} \geq L

, i.e.

b_{n}

cannot ultimately drop below

L

; that is weaker than convergence and needs an upper wall to complete.

Recall One-line survival kit

Two walls, same limit, holding eventually, trapping $b_{n}$ on both sides — miss any one and the theorem is silent. Two walls, same limit, eventual bound, two-sided — all four? ::: Then and only then is $b_{n} \to L$ guaranteed.

Connections

Squeeze theorem for sequences — the parent statement and proof these traps stress-test.
Limit of a sequence (epsilon-N definition) — why "eventually" and the $ε$ -band reasoning work.
Bounded sequences — where the walls $\pm \frac{1}{n}$ come from.
Algebra of limits (sum, product, quotient) — the tame alternative when no squeeze is needed.
Squeeze theorem for functions — the continuous cousin of every trap above.
Standard limits ( n-th roots, n!/n^n, ln n / n ) — the limits assumed in the examples.

Question bank — Squeeze theorem for sequences

True or false — justify

Spot the error

Why questions

Edge cases

Connections