Lesson 10 · Calculus and the squeeze theorem

Learning intentionTo prove inequalities using calculus, and to use inequalities to find limits with the squeeze theorem.

Success criteria · I can…

prove $f(x)\ge g(x)$ by studying $h(x)=f(x)-g(x)$ and its derivative
use a minimum value, concavity or a known inequality in a proof
state the squeeze theorem and use it to find limits
prove $\displaystyle\lim_{x\to0}\frac{\sin x}{x}=1$

ResourcesLesson slides
Practice worksheet
Worked solutions

Lesson flow

LaunchWarm-upSlide 2 Key ideasInequalities using calculusSlide 3 ExamplesExample 2 · Trigonometric inequalitiesSlide 4 HSC-styleHSC practice · From a function inequality to a sequenceSlide 5 ExamplesExample 3 · Concavity and a famous comparisonSlide 6 Exam practiceExam practice · Bounds for the harmonic seriesSlide 7 Key ideasThe squeeze theoremSlide 8 ExamplesExample 5 · A limit at infinitySlide 9 ExamplesExample 6 · The fundamental trigonometric limitSlide 10 ThinkSpot the flaw · Circular reasoningSlide 11 HSC-styleHSC practice · Squeezing a series to eSlide 12 Mixed practiceWhich method? · The whole unitSlide 13 Exit ticketExit ticketSlide 14

Warm-up

Slide 2

(a) Find the minimum value of

f(x)=e^x-1-x

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Answer

The minimum value is

0

, at

x=0

. So

e^x\ge1+x

for all

x

Why

f'(x)=e^x-1=0

x=0

, and

f''(x)=e^x>0

, so this is a minimum.

f(0)=1-1-0=0

(b) If

h(0)=0

and

h'(x)>0

for

x>0

, what can you say about

h(x)

for

x>0

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Answer

h(x)>0

for

x>0

Why

h

is increasing for

x>0

, so

h(x)>h(0)=0

(c) Evaluate

\displaystyle\lim_{x\to\infty}\frac1x

and

\displaystyle\lim_{x\to0}x^2

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Answer

0

and

0

Why

\dfrac1x

becomes arbitrarily small as

x

grows, and

x^2\to0^2=0

(d) Since

-1\le\sin x\le1

, find bounds for

\dfrac{\sin x}x

when

x>0

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Answer

-\dfrac1x\le\dfrac{\sin x}x\le\dfrac1x

Why

Divide the inequality by

x>0

; the direction does not change. This is used in Example 5.

Inequalities using calculus

Slide 3

Key idea

Three strategies for proving $f(x)\ge g(x)$

Start and slope: let $h(x)=f(x)-g(x)$ . Show $h(a)=0$ and $h'(x)\ge0$ for $x\ge a$ . So $h(x)\ge0$ for $x\ge a$ .
Global minimum: show the minimum value of $h(x)$ on the domain is $\ge0$ .
Graphs and concavity: a concave down curve lies above its chords and below its tangents (and the reverse for concave up).

Example 1

Prove that

\ln(1+x)<x

for all

x>0

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Answer

Let

h(x)=x-\ln(1+x)

for

x>0

. Then

h(0)=0

and

h'(x)=1-\frac1{1+x}=\frac x{1+x}>0\quad\text{for }x>0.

h

is increasing, and

h(x)>h(0)=0

for

x>0

. Hence

\ln(1+x)<x

Why

This is strategy 1. State the domain and give a reason for every sign. Warm-up (a) was strategy 2.

Example 2 · Trigonometric inequalities

Slide 4

Example 2

(a) Prove that

\sin x<x

for all

x>0

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Answer

Let

h(x)=x-\sin x

. Then

h(0)=0

and

h'(x)=1-\cos x\ge0

, with equality only at isolated points. So

h

is increasing, and

h(x)>0

for

x>0

Why

Because

h'(x)=0

only at isolated points (

x=2\pi k

h

is still strictly increasing, so the inequality is strict. Caution: these proofs use

\dfrac d{dx}(\sin x)=\cos x

, which is itself proved from

\displaystyle\lim_{x\to0}\frac{\sin x}x=1

. They are fine for proving inequalities, but cannot be used to prove that limit (see Example 6).

(b) Prove that

\cos x\ge1-\dfrac{x^2}2

for all real

x

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Answer

Let

h(x)=\cos x-1+\dfrac{x^2}2

. Then

h(0)=0

and

h'(x)=x-\sin x\ge0

for

x\ge0

by (a). So

h(x)\ge0

for

x\ge0

. Since

h

is an even function, the result also holds for

x<0

Why

Part (a) is used inside part (b): a “previously obtained inequality”. For the even function,

h(-x)=\cos(-x)-1+\dfrac{(-x)^2}2=h(x)

HSC practice · From a function inequality to a sequence

Slide 5

2023 HSC Q16(b)(i) Show that

x>\ln x

for all

x>0

.
(ii) Hence, or otherwise, show that

e^{n^2+n}>(n!)^2

for all positive integers

n

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Answer

(i) Let

h(x)=x-\ln x

for

x>0

. Then

h'(x)=1-\dfrac1x=0

x=1

. Also

h''(x)=\dfrac1{x^2}>0

, so

x=1

is the minimum, with

h(1)=1>0

. So

h(x)>0

for all

x>0

.
(ii) By (i),

\ln k<k

for each

k=1,2,\dots,n

. Then

\ln\left((n!)^2\right)=2\left(\ln1+\ln2+\cdots+\ln n\right)<2(1+2+\cdots+n)=n(n+1)=n^2+n.

Since

\ln

is increasing,

(n!)^2<e^{n^2+n}

Why

Part (i) is strategy 2 (global minimum). In (ii), apply (i) to each term and use

1+2+\cdots+n=\dfrac{n(n+1)}2

. This can also be done by induction. (Wording adapted from the paper.)

Example 3 · Concavity and a famous comparison

Slide 6

Example 3

(a) Use concavity to prove that

\sin x\ge\dfrac{2x}\pi

for

0\le x\le\dfrac\pi2

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Answer

y=\sin x

has

y''=-\sin x\le0

\left[0,\dfrac\pi2\right]

, so it is concave down there. The chord from

(0,0)

\left(\dfrac\pi2,1\right)

is the line

y=\dfrac{2x}\pi

. A concave down curve lies on or above its chord, so

\sin x\ge\dfrac{2x}\pi

Why

Strategy 3. The gradient of the chord is

\dfrac{1-0}{\pi/2-0}=\dfrac2\pi

. Sketch both graphs.

(b) By finding the maximum of

f(x)=\dfrac{\ln x}x

, prove that

\pi^e<e^\pi

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Answer

f'(x)=\dfrac{1-\ln x}{x^2}=0

x=e

. Also

f'>0

for

x<e

and

f'<0

for

x>e

, so the maximum is

f(e)=\dfrac1e

. Since

\pi\ne e

\dfrac{\ln\pi}\pi<\dfrac1e

, so

e\ln\pi<\pi

, that is

\ln\left(\pi^e\right)<\ln\left(e^\pi\right)

. Hence

\pi^e<e^\pi

Why

Because

x=e

is the only maximum,

f(\pi)

is strictly less than

f(e)

. Multiply by

e\pi>0

, then use

e\ln\pi=\ln(\pi^e)

and

\pi=\ln(e^\pi)

, and finally the fact that

\ln

is increasing.

Exam practice · Bounds for the harmonic series

Slide 7

Exam practice(i) Show that

\dfrac x{1+x}<\ln(1+x)<x

for

x>0

.
(ii) By putting

x=\dfrac1k

, show that

\dfrac1{k+1}<\ln\left(\dfrac{k+1}k\right)<\dfrac1k

.
(iii) Hence show that

\ln(n+1)<1+\dfrac12+\dfrac13+\cdots+\dfrac1n<1+\ln n

for all integers

n\ge2

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Answer

(i) The right-hand inequality is Example 1. For the left, let

h(x)=\ln(1+x)-\dfrac x{1+x}

. Then

h(0)=0

and

h'(x)=\frac1{1+x}-\frac1{(1+x)^2}=\frac x{(1+x)^2}>0\quad(x>0),

h(x)>0

for

x>0

.
(ii)

\dfrac{1/k}{1+1/k}=\dfrac1{k+1}

and

\ln\left(1+\dfrac1k\right)=\ln\dfrac{k+1}k

.
(iii) Write

H_n=1+\dfrac12+\cdots+\dfrac1n

. Sum

\ln\dfrac{k+1}k<\dfrac1k

for

k=1,\dots,n

: the logs telescope to

\ln(n+1)-\ln1

, so

\ln(n+1)<H_n

. Sum

\dfrac1{k+1}<\ln\dfrac{k+1}k

for

k=1,\dots,n-1

H_n-1<\ln n

, so

H_n<1+\ln n

Why

Telescoping:

\sum_{k=1}^{n}\ln\dfrac{k+1}k=\ln\dfrac21+\ln\dfrac32+\cdots+\ln\dfrac{n+1}n=\ln(n+1)

. For the second sum the left side is

\dfrac12+\cdots+\dfrac1n=H_n-1

. At

n=1

the right-hand bound is an equality (

H_1=1=1+\ln1

), so the strict inequality fails; this is why

n\ge2

. Discussion:

H_n\to\infty

, but very slowly, like

\ln n

The squeeze theorem

Slide 8

Theorem

Squeeze theorem (limit version)

If $f(x)\le g(x)\le h(x)$ for all $x$ near $k$ (or for all large $x$ ), and $\displaystyle\lim_{x\to k}f(x)=\lim_{x\to k}h(x)=L$ , then $\displaystyle\lim_{x\to k}g(x)=L$ .

Typical sources of the bounds: $-1\le\sin\theta\le1$ , $-1\le\cos\theta\le1$ , or an inequality proved earlier. The theorem is also called the sandwich or pinching theorem.

Example 4

Find

\displaystyle\lim_{x\to0}x^2\sin\left(\frac1x\right)

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Answer

For

x\ne0

-1\le\sin\left(\dfrac1x\right)\le1

. Multiply by

x^2\ge0

-x^2\le x^2\sin\left(\frac1x\right)\le x^2.

Both

-x^2

and

x^2

tend to

0

x\to0

, so by the squeeze theorem the limit is

0

Why

You cannot just substitute

x=0

, as

\sin\left(\dfrac1x\right)

is not defined there and oscillates wildly. Sketch

y=x^2

y=-x^2

and the curve squeezed between them.

Example 5 · A limit at infinity

Slide 9

Example 5

Use the squeeze theorem to find

\displaystyle\lim_{x\to\infty}\frac{\sin x}x

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Answer

For

x>0

-\dfrac1x\le\dfrac{\sin x}x\le\dfrac1x

(warm-up (d)). Both bounds tend to

0

x\to\infty

, so

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

Why

The curve oscillates between

y=\pm\dfrac1x

and is squeezed to the

x

-axis. Contrast this with

\displaystyle\lim_{x\to0}\frac{\sin x}x

, the next example.

Example 6 · The fundamental trigonometric limit

Slide 10

Why not use Example 2? Those proofs use $\dfrac d{dx}(\sin x)=\cos x$ , which is proved from this limit. Using them here would be circular, so we use areas.

Example 6

(i) In the unit circle,

A=(1,0)

B=(\cos x,\sin x)

and

C

lies on

OB

extended, on the tangent at

A

. Compare the areas of

\triangle OAB

, sector

OAB

and

\triangle OAC

to prove

\sin x<x<\tan x

for

0<x<\dfrac\pi2

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Answer

With radius

1

\triangle OAB

has area

\tfrac12\times1\times1\times\sin x=\tfrac12\sin x

; sector

OAB

has area

\tfrac12\times1^2\times x=\tfrac12x

;

\triangle OAC

has area

\tfrac12\times OA\times AC=\tfrac12\tan x

. The triangle

OAB

lies inside the sector, which lies inside

\triangle OAC

, so

\tfrac12\sin x<\tfrac12x<\tfrac12\tan x,\quad\text{that is}\quad \sin x<x<\tan x.

Why

AC=\tan x

because

\triangle OAC

is right-angled at

A

with

OA=1

and angle

x

O

. The sector area

\tfrac12r^2\theta

comes from geometry, not calculus, so there is no circularity.

(ii) Hence show that

\cos x<\dfrac{\sin x}x<1

for

0<x<\dfrac\pi2

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Answer

From

\sin x<x<\tan x

, divide by

\sin x>0

1<\dfrac x{\sin x}<\dfrac1{\cos x}

. Take reciprocals (all positive, so the inequalities reverse):

\cos x<\dfrac{\sin x}x<1

Why

Dividing

\tan x=\dfrac{\sin x}{\cos x}

\sin x

gives

\dfrac1{\cos x}

. Taking reciprocals of positive numbers reverses the inequality.

(iii) Hence prove that

\displaystyle\lim_{x\to0}\frac{\sin x}x=1

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Answer

x\to0^+

\cos x\to1

and the upper bound is the constant

1

, so by the squeeze theorem

\displaystyle\lim_{x\to0^+}\frac{\sin x}x=1

. The function

\dfrac{\sin x}x

is even, so the limit from the left is also

1

. Hence

\displaystyle\lim_{x\to0}\frac{\sin x}x=1

Why

This is the limit and the small-angle result used in Extension 1; here is its proof. For the left limit,

\dfrac{\sin(-x)}{-x}=\dfrac{\sin x}x

. Discussion: what is a proof allowed to assume?

Spot the flaw · Circular reasoning

Slide 11

A student wrote this. Find the flaw, explain why it matters, then write a correct version.

Student’s claim

$\displaystyle\lim_{x\to0}\frac{\sin x}x=1$ .

$\displaystyle\lim_{x\to0}\frac{\sin x}x=\lim_{x\to0}\frac{\sin x-\sin0}{x-0}$ .
This is the derivative of $\sin x$ at $x=0$ , which is $\cos0=1$ . □

Spot the flawFind the flaw and write a correct version.

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Answer

Flaw: the argument is circular. The result

\dfrac d{dx}(\sin x)=\cos x

is derived from first principles using

\displaystyle\lim_{h\to0}\frac{\sin h}h=1

, so line 2 assumes what is being proved.
Correct version: Example 6. Compare areas to get

\sin x<x<\tan x

, then squeeze.

Why

A proof may only assume results that do not depend on the statement being proved. In Extension 1 the derivative of

\sin x

was taken on trust; Example 6 is the missing foundation.

HSC practice · Squeezing a series to e

Slide 12

Adapted from 2022 HSC Q14(b)(v)It can be shown that for all integers

n\ge0

0\le1-\frac1e\sum_{r=0}^{n}\frac1{r!}\le\frac1{(n+1)!}.

Use this to show that

\displaystyle\lim_{n\to\infty}\sum_{r=0}^{n}\frac1{r!}=e

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Answer

n\to\infty

\dfrac1{(n+1)!}\to0

. The lower bound is

0

, so by the squeeze theorem

1-\dfrac1e\displaystyle\sum_{r=0}^{n}\frac1{r!}\to0

. Therefore

\dfrac1e\displaystyle\sum_{r=0}^{n}\frac1{r!}\to1

, that is

\displaystyle\sum_{r=0}^{n}\frac1{r!}\to e

Why

The squeeze theorem applies to sequences as well as functions. The earlier parts of the original question use integration by parts on

\displaystyle\int_0^1x^ne^{-x}\,dx

, which we revisit in the Integration unit. (Earlier parts omitted; wording adapted from the paper.)

Which method? · The whole unit

Slide 13

For each statement, name the method you would use and write the first line of the proof. Then prove any two.

(a)

1+3+5+\cdots+(2n-1)=n^2

for

n\ge1

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Answer

Method: induction on a series (or the AP sum formula). First line:

P(1)

1=1^2

Why

Assume

\sum_{r=1}^{k}(2r-1)=k^2

, then add the next term:

k^2+(2k+1)=(k+1)^2

(b)

e^x\ge1+x

for all real

x

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Answer

Method: calculus, global minimum. First line: let

h(x)=e^x-1-x

; then

h'(x)=e^x-1=0

x=0

Why

h''(x)=e^x>0

, so

h(0)=0

is the minimum, hence

h(x)\ge0

(warm-up (a)).

(c)

3^n>n^3

for all integers

n\ge4

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Answer

Method: induction with an inequality. First line:

P(4)

3^4=81>64=4^3

Why

Assume

3^k>k^3

with

k\ge4

. Then

3^{k+1}>3k^3\ge(k+1)^3

, because

\left(1+\dfrac1k\right)^3\le\left(\dfrac54\right)^3<3

for

k\ge4

. (For

n=3

the two sides are equal, so the start is

n=4

(d)

\log_3 7

is irrational.

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Answer

Method: proof by contradiction. First line: suppose

\log_37=\dfrac pq

with

p,q

positive integers.

Why

Then

3^{p/q}=7

, so

3^p=7^q

. The left side is divisible by

3

but the right side is not, a contradiction. (A parity argument does not work here, as both sides are odd.)

(e)

\displaystyle\lim_{x\to\infty}\frac{\cos x}{x^2}=0

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Answer

Method: squeeze theorem. First line: for

x>0

-\dfrac1{x^2}\le\dfrac{\cos x}{x^2}\le\dfrac1{x^2}

Why

Both bounds tend to

0

x\to\infty

, so the limit is

0

(f) For integers

n

, if

n^3

is even, then

n

is even.

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Answer

Method: contrapositive. First line: suppose

n

is odd, so

n=2k+1

Why

Then

n^3=8k^3+12k^2+6k+1=2\left(4k^3+6k^2+3k\right)+1

is odd. So if

n

is odd then

n^3

is odd, which is the contrapositive of the statement.

Exit ticket

Slide 14

Using

\cos x\ge1-\dfrac{x^2}2

(Example 2), prove that

\sin x>x-\dfrac{x^3}6

for all

x>0

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Answer

Let

h(x)=\sin x-x+\dfrac{x^3}6

. Then

h(0)=0

and

h'(x)=\cos x-1+\dfrac{x^2}2\ge0

by Example 2(b), with equality only at isolated points. So

h

is increasing, and

h(x)>0

for

x>0

, that is

\sin x>x-\dfrac{x^3}6

Why

Same pattern each time: build

h

, check

h(0)=0

, and sign

h'

using the previous inequality. Together with Example 2(a):

x-\dfrac{x^3}6<\sin x<x

for

x>0

, another squeeze for

\dfrac{\sin x}x