The squeeze theorem: statement for functions and sequences, building the two bounding inequalities, limits of sin n over n and x squared sin of 1 over x, the limit of sin x over x, sums and integrals trapped between bounds (new in the 2024 Extension 2 syllabus)

Syllabus dot point

“Use inequalities and the squeeze theorem to prove results about limits of sequences and functions”

HSCMaths Extension 2Proof (MEX-P1)12 min readUpdated 7 Oct 2026

Quick answer

If $f(x) \leq g(x) \leq h(x)$ near $a$ and $f$ and $h$ both tend to $L$ , then $g \to L$ (and the same for sequences). Build an inequality that traps the awkward quantity, show both bounds tend to the same limit and name the theorem: used for $\frac{\sin n}{n}$ , $x^2 \sin\frac{1}{x}$ , $\frac{n!}{n^n}$ , sums of $n$ terms, integral bounds and $\frac{\sin x}{x}$ . New to Extension 2 in the 2024 syllabus.

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What this dot point is asking
The answer
Practice questions

What this dot point is asking

The squeeze theorem is new to the Mathematics Extension 2 11-12 Syllabus (2024), first examined in the 2027 HSC. NESA lists it in Year 12 The nature of proof, under "Proof of inequalities", alongside the triangle inequality, the arithmetic and geometric mean inequality, and graphical and calculus-based proofs. You are expected to use inequalities to prove that a limit exists and has a certain value, so the theorem sits where two ideas meet: building a correct inequality, and drawing a conclusion from it.

Note

Suppose a runner is always ahead of a slow walker and always behind a fast walker, and the slow walker and the fast walker are heading to the same finish line. The runner has no choice but to finish at that line as well. The squeeze theorem says exactly this about numbers: a quantity that is trapped between two others cannot escape the place where they meet.

The answer

Key fact

Squeeze theorem for functions. Suppose $f(x) \leq g(x) \leq h(x)$ for all $x$ near $a$ (except possibly at $a$ itself) and $\displaystyle\lim_{x \to a} f(x) = \lim_{x \to a} h(x) = L$ . Then $\displaystyle\lim_{x \to a} g(x) = L$ .

Squeeze theorem for sequences. If $a_n \leq b_n \leq c_n$ for all large $n$ and $a_n \to L$ and $c_n \to L$ , then $b_n \to L$ .

The same statement holds as $x \to \infty$ and for one-sided limits.

How to write a squeeze proof

Find a lower bound and an upper bound for the awkward quantity. Use a bound that you can prove, such as $-1 \leq \sin t \leq 1$ , $\frac{k}{n} \leq 1$ , or a comparison of integrands.
State the inequality chain in one line, making sure each direction is justified (watch out for multiplying by negative numbers).
Find the limits of the two bounding quantities and show that they are equal.
Name the theorem and conclude with the common limit.

You do not need the theorem when the quantity has an ordinary limit by the limit laws, and you may not use it when the two bounds have different limits, because then the argument proves nothing.

Typical sources of bounds

Bounded trigonometric functions: $|\sin t| \leq 1$ and $|\cos t| \leq 1$ trap oscillating factors, giving $\frac{\sin n}{n} \to 0$ and $x^2 \sin\frac{1}{x} \to 0$ .
Largest and smallest term in a sum: a sum of $n$ positive terms lies between $n$ times its smallest term and $n$ times its largest term.
Products with factors at most one: $\frac{n!}{n^n} \leq \frac{1}{n}$ .
Dominant term: $3^n \leq 2^n + 3^n \leq 2 \cdot 3^n$ .
Integral comparison: if $f(x) \leq g(x)$ on $[a, b]$ then $\int_a^b f \leq \int_a^b g$ .

The classic limit $\frac{\sin x}{x} \to 1$

For $0 < x < \frac{\pi}{2}$ , comparing the areas of a triangle, a circular sector and a larger triangle in a unit circle gives $\sin x < x < \tan x$ . Dividing by $\sin x > 0$ and taking reciprocals,

\cos x < \frac{\sin x}{x} < 1.

Because $\cos x \to 1$ as $x \to 0^+$ , the squeeze theorem gives $\frac{\sin x}{x} \to 1$ from the right, and since $\frac{\sin x}{x}$ is an even function the left limit is the same.

How exam questions use this

"Show that $\lim \dots = 0$ " for an oscillating factor times a decaying one.
"Prove that the sequence converges to $L$ " for a sum, a product or a root of a sum.
"Show that $0 \leq I_n \leq \dots$ and hence find $\lim I_n$ " for a reduction-formula integral.
"Hence" parts that use an inequality proved in an earlier part as one of the bounds.

Worked examples

A rational expression with an oscillating term

Find $\displaystyle\lim_{x \to \infty} \frac{2x + \cos x}{x + 3}$ .

Step 1: bound the oscillating term

For all $x$ , $-1 \leq \cos x \leq 1$ , so for $x > 0$

\frac{2x - 1}{x + 3} \leq \frac{2x + \cos x}{x + 3} \leq \frac{2x + 1}{x + 3}.

Step 2: limits of the bounds

Dividing the top and bottom by $x$ , $\frac{2x - 1}{x + 3} = \frac{2 - \frac{1}{x}}{1 + \frac{3}{x}} \to 2$ and $\frac{2x + 1}{x + 3} = \frac{2 + \frac{1}{x}}{1 + \frac{3}{x}} \to 2$ .

Step 3: conclude

By the squeeze theorem the limit is $2$ .

Marker's note: one mark for the double inequality, one for the limits of the two bounds, one for the conclusion.

A factor that oscillates faster and faster

Show that $\displaystyle\lim_{x \to 0} x\cos\left( \frac{1}{x^2} \right) = 0$ .

Step 1: bound the absolute value

For $x \neq 0$ , $\left| \cos\left( \frac{1}{x^2} \right) \right| \leq 1$ , so $\left| x\cos\left( \frac{1}{x^2} \right) \right| \leq |x|$ , which means

-|x| \leq x\cos\left( \frac{1}{x^2} \right) \leq |x|.

Step 2: conclude

Both $-|x|$ and $|x|$ tend to $0$ as $x \to 0$ , so the squeeze theorem gives the limit $0$ .

Marker's note: working with the absolute value avoids splitting into the cases $x > 0$ and $x < 0$ .

Common traps

Two bounds with different limits: The squeeze theorem needs the same limit on both sides. A loose upper bound that tends to $2$ proves nothing about a quantity that tends to $1$ .
Splitting a product when one factor has no limit: In $x^2 \sin\left( \frac{1}{x} \right)$ the factor $\sin\left( \frac{1}{x} \right)$ has no limit as $x \to 0$ , so the product-of-limits law cannot be used.
Reversing inequalities incorrectly: Multiplying or dividing by a negative number flips the signs, as in the one-sided limits of $\frac{1 - \cos x}{x}$ .
Arguing "bounded over infinity": Statements such as $\frac{\sin n}{n} = \frac{\text{bounded}}{\infty} = 0$ are not proofs.
A bound that holds only for some values: The inequality chain must be true for every $n$ (or every $x$ near $a$ ) in the range that matters, not just for a few test values. Prove each bound, then test it with a small value as a check.

Exam technique

Write the double inequality as the very first line of your answer, then the limits of both ends on the next line, then the conclusion in one sentence that names the squeeze theorem. If a "hence" asks you to use an earlier result, that result is nearly always one of the bounds. Test your bound with a small value such as $n = 1$ or $n = 4$ before you rely on it; a wrong inequality direction is the commonest way to lose all the marks.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marks

Use the squeeze theorem to show that

\displaystyle\lim_{n \to \infty} \frac{\sin n}{n} = 0

Show worked solution →

Bound the numerator. For every integer $n \geq 1$ , $-1 \leq \sin n \leq 1$ . Dividing by $n > 0$ keeps the inequalities:

-\frac{1}{n} \leq \frac{\sin n}{n} \leq \frac{1}{n}.

Limits of the bounds. $\displaystyle\lim_{n \to \infty}\left( -\frac{1}{n} \right) = 0$ and $\displaystyle\lim_{n \to \infty}\frac{1}{n} = 0$ .

By the squeeze theorem, $\displaystyle\lim_{n \to \infty}\frac{\sin n}{n} = 0$ .

Marker's note: one mark for the two-sided bound, one for the limits of the bounds and the conclusion. Writing " $\frac{\sin n}{n} \to \frac{\text{something}}{\infty}$ " earns nothing, because $\sin n$ has no limit.

core3 marks

Show that

\displaystyle\lim_{x \to 0} x^2 \sin\left( \frac{1}{x} \right) = 0

Show worked solution →

Bound the oscillating factor. For $x \neq 0$ , $-1 \leq \sin\left( \frac{1}{x} \right) \leq 1$ . Multiply by $x^2 > 0$ :

-x^2 \leq x^2 \sin\left( \frac{1}{x} \right) \leq x^2.

Limits of the bounds. As $x \to 0$ , both $-x^2 \to 0$ and $x^2 \to 0$ .

By the squeeze theorem the middle expression also tends to $0$ .

Marker's note: one mark for the bound on the sine, one for multiplying through (the multiplier is positive, so the inequality directions are unchanged), one for the conclusion. The limit of $\sin\left( \frac{1}{x} \right)$ alone does not exist, so you may not split the product into separate limits.

core3 marks

Prove that

\displaystyle\lim_{n \to \infty} \frac{n!}{n^n} = 0

Show worked solution →

Write the ratio as a product of fractions.

\frac{n!}{n^n} = \frac{1}{n} \times \frac{2}{n} \times \frac{3}{n} \times \cdots \times \frac{n}{n}.

Bound it. Each factor from $\frac{2}{n}$ to $\frac{n}{n}$ is at most $1$ and positive, so removing them can only increase the product:

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

Conclude. The bounds $0$ and $\frac{1}{n}$ both tend to $0$ , so by the squeeze theorem $\frac{n!}{n^n} \to 0$ .

Marker's note: one mark for the product form, one for $\frac{n!}{n^n} \leq \frac{1}{n}$ with a reason, one for the conclusion. A spot check: at $n = 4$ the ratio is $\frac{24}{256} = 0.09375 \leq 0.25$ .

core3 marks

Show that

\displaystyle\lim_{n \to \infty} \left( 2^n + 3^n \right)^{1/n} = 3

Show worked solution →

Lower bound. Since $2^n > 0$ , we have $2^n + 3^n > 3^n$ , so $\left( 2^n + 3^n \right)^{1/n} > 3$ .

Upper bound. Since $2^n < 3^n$ , we have $2^n + 3^n < 2 \cdot 3^n$ , so

\left( 2^n + 3^n \right)^{1/n} < \left( 2 \cdot 3^n \right)^{1/n} = 3 \cdot 2^{1/n}.

Combine. $3 < \left( 2^n + 3^n \right)^{1/n} < 3 \cdot 2^{1/n}$ . As $n \to \infty$ , $2^{1/n} \to 2^0 = 1$ , so the upper bound tends to $3$ and the lower bound is the constant $3$ . By the squeeze theorem the limit is $3$ .

Marker's note: one mark for each bound, one for the limit of $2^{1/n}$ and the conclusion.

exam4 marks

Let

S_n = \dfrac{1}{\sqrt{n^2 + 1}} + \dfrac{1}{\sqrt{n^2 + 2}} + \cdots + \dfrac{1}{\sqrt{n^2 + n}}

. Show that

\displaystyle\lim_{n \to \infty} S_n = 1

Show worked solution →

The sum has $n$ terms. Each term is $\frac{1}{\sqrt{n^2 + k}}$ for $k = 1, 2, \dots, n$ . The terms decrease as $k$ increases, so the largest is $\frac{1}{\sqrt{n^2 + 1}}$ and the smallest is $\frac{1}{\sqrt{n^2 + n}}$ .

Bound the sum by $n$ times the smallest and $n$ times the largest term:

\frac{n}{\sqrt{n^2 + n}} \leq S_n \leq \frac{n}{\sqrt{n^2 + 1}}.

Limits of the bounds. Divide the numerator and denominator by $n$ :

\frac{n}{\sqrt{n^2 + n}} = \frac{1}{\sqrt{1 + \frac{1}{n}}} \to 1 \quad \text{and} \quad \frac{n}{\sqrt{n^2 + 1}} = \frac{1}{\sqrt{1 + \frac{1}{n^2}}} \to 1.

By the squeeze theorem $S_n \to 1$ .

Marker's note: one mark for noting there are $n$ terms, one for the two bounds, one for each limit of a bound (or one for both) and one for the conclusion. Numerically, $S_{100} \approx 0.9975$ , between the bounds $0.9950$ and $0.99995$ .

exam4 marks

For

n \geq 1

let

I_n = \displaystyle\int_0^1 \frac{x^n}{1 + x}\,dx

. Show that

0 \leq I_n \leq \dfrac{1}{n + 1}

and hence find

\displaystyle\lim_{n \to \infty} I_n

Show worked solution →

Lower bound. For $0 \leq x \leq 1$ the integrand $\frac{x^n}{1 + x}$ is non-negative, so $I_n \geq 0$ .

Upper bound. For $x \geq 0$ , $1 + x \geq 1$ , so $\frac{1}{1 + x} \leq 1$ and therefore $\frac{x^n}{1 + x} \leq x^n$ . Integrating over $[0, 1]$ ,

I_n \leq \int_0^1 x^n\,dx = \left[ \frac{x^{n+1}}{n + 1} \right]_0^1 = \frac{1}{n + 1}.

Conclude. $0 \leq I_n \leq \frac{1}{n + 1}$ , and $\frac{1}{n + 1} \to 0$ , so by the squeeze theorem $\displaystyle\lim_{n \to \infty} I_n = 0$ .

Marker's note: one mark for the lower bound, one for the pointwise upper bound, one for integrating it, one for the conclusion. A check: $I_1 \approx 0.307 \leq 0.5$ and $I_5 \approx 0.090 \leq 0.167$ .

exam4 marks

Use the inequality

0 \leq 1 - \cos x \leq \dfrac{x^2}{2}

(which holds for all real

x

) to show that

\displaystyle\lim_{x \to 0}\frac{1 - \cos x}{x} = 0

Show worked solution →

Case $x > 0$ . Divide the inequality by $x > 0$ :

0 \leq \frac{1 - \cos x}{x} \leq \frac{x}{2}.

As $x \to 0^+$ the upper bound tends to $0$ , so $\frac{1 - \cos x}{x} \to 0$ from the right.

Case $x < 0$ . Dividing by $x < 0$ reverses every inequality:

0 \geq \frac{1 - \cos x}{x} \geq \frac{x}{2}.

As $x \to 0^-$ the lower bound $\frac{x}{2} \to 0$ , so $\frac{1 - \cos x}{x} \to 0$ from the left.

Conclusion. Both one-sided limits equal $0$ , so $\displaystyle\lim_{x \to 0}\frac{1 - \cos x}{x} = 0$ .

Marker's note: one mark for each one-sided squeeze and one for noting the reversal of the inequality for negative $x$ ; one for the combined conclusion. Dividing by a negative number without reversing the signs is the classic error.

Practise this

Flashcards Maths Extension 2 deck

Sources & how we know this

Mathematics Extension 2 11-12 Syllabus (2024): Year 12 The nature of proof — NESA (2024)
Mathematics Extension 2 11-12 Syllabus (2024): course overview and implementation — NESA (2024)
Mathematics Extension 2 Stage 6 Syllabus (2017) — NESA
Mathematics Extension 2 HSC exam papers — NESA

What this dot point is asking

The answer

How to write a squeeze proof

Typical sources of bounds

The classic limit sin⁡xx→1\frac{\sin x}{x} \to 1xsinx​→1

How exam questions use this

A rational expression with an oscillating term

Step 1: bound the oscillating term

Step 2: limits of the bounds

Step 3: conclude

A factor that oscillates faster and faster

Step 1: bound the absolute value

Step 2: conclude

Practice questions

Practise this

Sources & how we know this

The classic limit $\frac{\sin x}{x} \to 1$