lesson 4: calculating limits (slides)

..

Sec on 1.4Calcula ng Limits

V63.0121.001: Calculus IProfessor Ma hew Leingang

New York University

February 2, 2011

Announcements

I First wri en HW due todayI Get-to-know-you survey and photo deadline is February 11

Announcements

I First wri en HW duetoday

I Get-to-know-you surveyand photo deadline isFebruary 11

ObjectivesI Know basic limits like lim

x→ax = a

and limx→a

c = c.I Use the limit laws to computeelementary limits.

I Use algebra to simplify limits.I Understand and state theSqueeze Theorem.

I Use the Squeeze Theorem todemonstrate a limit.

..

Limit

Yoda on teaching course concepts

You must unlearnwhat you havelearned.

In other words, we arebuilding up concepts andallowing ourselves only tospeak in terms of what wepersonally have produced.

OutlineRecall: The concept of limit

Basic Limits

Limit LawsThe direct subs tu on property

Limits with AlgebraTwo more limit theorems

Two important trigonometric limits

Heuristic Definition of a LimitDefini onWe write

limx→a

f(x) = L

and say

“the limit of f(x), as x approaches a, equals L”

if we can make the values of f(x) arbitrarily close to L (as close to Las we like) by taking x to be sufficiently close to a (on either side ofa) but not equal to a.

The error-tolerance gameA game between two players (Dana and Emerson) to decide if a limitlimx→a

f(x) exists.

Step 1 Dana proposes L to be the limit.Step 2 Emerson challenges with an “error” level around L.Step 3 Dana chooses a “tolerance” level around a so that points x

within that tolerance of a (not coun ng a itself) are taken tovalues y within the error level of L. If Dana cannot, Emersonwins and the limit cannot be L.

Step 4 If Dana’s move is a good one, Emerson can challenge againor give up. If Emerson gives up, Dana wins and the limit is L.

The error-tolerance game

.

.

This tolerance is too big

.

S ll too big

.

This looks good

.

So does this

.a

.

L

I To be legit, the part of the graph inside the blue (ver cal) stripmust also be inside the green (horizontal) strip.

I Even if Emerson shrinks the error, Dana can s ll move.

Limit FAIL: Jump

.. x.

y

..

−1

..

1

...

Part of graphinside blueis not insidegreen

.

Part of graphinside blueis not insidegreen

I So limx→0

|x|x

does notexist.

I But limx→0+

f(x) = 1

andlimx→0−

f(x) = −1.

Limit FAIL: Jump

.. x.

y

..

−1

..

1

..

.

Part of graphinside blueis not insidegreen

.

Part of graphinside blueis not insidegreen

I So limx→0

|x|x

does notexist.

I But limx→0+

f(x) = 1

andlimx→0−

f(x) = −1.

Limit FAIL: Jump

.. x.

y

..

−1

..

1

..

.

Part of graphinside blueis not insidegreen

.

Part of graphinside blueis not insidegreen

I So limx→0

|x|x

does notexist.

I But limx→0+

f(x) = 1

andlimx→0−

f(x) = −1.

Limit FAIL: unboundedness

.. x.

y

.0..

L?

.

The graph escapesthe green, so nogood

.

Even worse!

.

limx→0+

1xdoes not exist be-

cause the func on is un-bounded near 0

Limit EPIC FAILHere is a graph of the func on f(x) = sin

(πx

):

.. x.

y

..

−1

..

1

For every y in [−1, 1], there are infinitely many points x arbitrarilyclose to zero where f(x) = y. So lim

x→0f(x) cannot exist.

OutlineRecall: The concept of limit

Basic Limits

Limit LawsThe direct subs tu on property

Limits with AlgebraTwo more limit theorems

Two important trigonometric limits

Really basic limits

FactLet c be a constant and a a real number.(i) lim

x→ax = a

(ii) limx→a

c = c

Proof.The first is tautological, the second is trivial.

Really basic limits

FactLet c be a constant and a a real number.(i) lim

x→ax = a

(ii) limx→a

c = c

Proof.The first is tautological, the second is trivial.

ET game for f(x) = x

.. x.

y

..a

..

aI Se ng error equal totolerance works!

ET game for f(x) = x

.. x.

y

..a

..

aI Se ng error equal totolerance works!

ET game for f(x) = x

.. x.

y

..a

..

a

I Se ng error equal totolerance works!

ET game for f(x) = x

.. x.

y

..a

..

a

I Se ng error equal totolerance works!

ET game for f(x) = x

.. x.

y

..a

..

a

I Se ng error equal totolerance works!

ET game for f(x) = x

.. x.

y

..a

..

a

I Se ng error equal totolerance works!

ET game for f(x) = x

.. x.

y

..a

..

aI Se ng error equal totolerance works!

ET game for f(x) = c

.

. x.

y

..a

..

cI any tolerance works!

ET game for f(x) = c

.. x.

y

..a

..

cI any tolerance works!

ET game for f(x) = c

.. x.

y

..a

..

cI any tolerance works!

ET game for f(x) = c

.. x.

y

..a

..

c

I any tolerance works!

ET game for f(x) = c

.. x.

y

..a

..

c

I any tolerance works!

ET game for f(x) = c

.. x.

y

..a

..

c

I any tolerance works!

ET game for f(x) = c

.. x.

y

..a

..

cI any tolerance works!

Really basic limits

FactLet c be a constant and a a real number.(i) lim

x→ax = a

(ii) limx→a

c = c

Proof.The first is tautological, the second is trivial.

OutlineRecall: The concept of limit

Basic Limits

Limit LawsThe direct subs tu on property

Limits with AlgebraTwo more limit theorems

Two important trigonometric limits

Limits and arithmetic

FactSuppose lim

x→af(x) = L and lim

x→ag(x) = M and c is a constant. Then

1. limx→a

[f(x) + g(x)] = L+M

(errors add)

2. limx→a

[f(x)− g(x)] = L−M

(combina on of adding and scaling)

3. limx→a

[cf(x)] = cL

(error scales)

4. limx→a

[f(x)g(x)] = L ·M

(more complicated, but doable)

Limits and arithmetic

FactSuppose lim

x→af(x) = L and lim

x→ag(x) = M and c is a constant. Then

1. limx→a

[f(x) + g(x)] = L+M (errors add)

2. limx→a

[f(x)− g(x)] = L−M

(combina on of adding and scaling)

3. limx→a

[cf(x)] = cL

(error scales)

4. limx→a

[f(x)g(x)] = L ·M

(more complicated, but doable)

Limits and arithmetic

FactSuppose lim

x→af(x) = L and lim

x→ag(x) = M and c is a constant. Then

1. limx→a

[f(x) + g(x)] = L+M (errors add)

2. limx→a

[f(x)− g(x)] = L−M

(combina on of adding and scaling)

3. limx→a

[cf(x)] = cL

(error scales)

4. limx→a

[f(x)g(x)] = L ·M

(more complicated, but doable)

Limits and arithmetic

FactSuppose lim

x→af(x) = L and lim

x→ag(x) = M and c is a constant. Then

1. limx→a

[f(x) + g(x)] = L+M (errors add)

2. limx→a

[f(x)− g(x)] = L−M

(combina on of adding and scaling)

3. limx→a

[cf(x)] = cL

(error scales)

4. limx→a

[f(x)g(x)] = L ·M

(more complicated, but doable)

Limits and arithmetic

FactSuppose lim

x→af(x) = L and lim

x→ag(x) = M and c is a constant. Then

1. limx→a

[f(x) + g(x)] = L+M (errors add)

2. limx→a

[f(x)− g(x)] = L−M

(combina on of adding and scaling)

3. limx→a

[cf(x)] = cL (error scales)

4. limx→a

[f(x)g(x)] = L ·M

(more complicated, but doable)

Justification of the scaling lawI errors scale: If f(x) is e away from L, then

(c · f(x)− c · L) = c · (f(x)− L) = c · e

That is, (c · f)(x) is c · e away from cL,I So if Emerson gives us an error of 1 (for instance), Dana can usethe fact that lim

x→af(x) = L to find a tolerance for f and g

corresponding to the error 1/c.I Dana wins the round.

Limits and arithmetic

FactSuppose lim

x→af(x) = L and lim

x→ag(x) = M and c is a constant. Then

1. limx→a

[f(x) + g(x)] = L+M (errors add)

2. limx→a

[f(x)− g(x)] = L−M (combina on of adding and scaling)

3. limx→a

[cf(x)] = cL (error scales)

4. limx→a

[f(x)g(x)] = L ·M

(more complicated, but doable)

Limits and arithmetic

FactSuppose lim

x→af(x) = L and lim

x→ag(x) = M and c is a constant. Then

1. limx→a

[f(x) + g(x)] = L+M (errors add)

2. limx→a

[f(x)− g(x)] = L−M (combina on of adding and scaling)

3. limx→a

[cf(x)] = cL (error scales)

4. limx→a

[f(x)g(x)] = L ·M

(more complicated, but doable)

Limits and arithmetic

FactSuppose lim

x→af(x) = L and lim

x→ag(x) = M and c is a constant. Then

1. limx→a

[f(x) + g(x)] = L+M (errors add)

2. limx→a

[f(x)− g(x)] = L−M (combina on of adding and scaling)

3. limx→a

[cf(x)] = cL (error scales)

4. limx→a

[f(x)g(x)] = L ·M (more complicated, but doable)

Limits and arithmetic IIFact (Con nued)

5. limx→a

f(x)g(x)

=LM

, if M ̸= 0.

6. limx→a

[f(x)]n =[limx→a

f(x)]n

(follows from 4 repeatedly)

7. limx→a

xn = an

(follows from 6)

8. limx→a

n√x = n

√a

9. limx→a

n√

f(x) = n

√limx→a

f(x) (If n is even, we must addi onallyassume that lim

x→af(x) > 0)

Caution!I The quo ent rule for limits says that if lim

x→ag(x) ̸= 0, then

limx→a

f(x)g(x)

=limx→a f(x)limx→a g(x)

I It does NOT say that if limx→a

g(x) = 0, then

limx→a

f(x)g(x)

does not exist

I In fact, limits of quo ents where numerator and denominatorboth tend to 0 are exactly where the magic happens.

Limits and arithmetic IIFact (Con nued)

5. limx→a

f(x)g(x)

=LM

, if M ̸= 0.

6. limx→a

[f(x)]n =[limx→a

f(x)]n

(follows from 4 repeatedly)

7. limx→a

xn = an

(follows from 6)

8. limx→a

n√x = n

√a

9. limx→a

n√

f(x) = n

√limx→a

f(x) (If n is even, we must addi onallyassume that lim

x→af(x) > 0)

Limits and arithmetic IIFact (Con nued)

5. limx→a

f(x)g(x)

=LM

, if M ̸= 0.

6. limx→a

[f(x)]n =[limx→a

f(x)]n

(follows from 4 repeatedly)

7. limx→a

xn = an

(follows from 6)

8. limx→a

n√x = n

√a

9. limx→a

n√

f(x) = n

√limx→a

f(x) (If n is even, we must addi onallyassume that lim

x→af(x) > 0)

Limits and arithmetic IIFact (Con nued)

5. limx→a

f(x)g(x)

=LM

, if M ̸= 0.

6. limx→a

[f(x)]n =[limx→a

f(x)]n

(follows from 4 repeatedly)

7. limx→a

xn = an

(follows from 6)

8. limx→a

n√x = n

√a

9. limx→a

n√

f(x) = n

√limx→a

f(x) (If n is even, we must addi onallyassume that lim

x→af(x) > 0)

Limits and arithmetic IIFact (Con nued)

5. limx→a

f(x)g(x)

=LM

, if M ̸= 0.

6. limx→a

[f(x)]n =[limx→a

f(x)]n

(follows from 4 repeatedly)

7. limx→a

xn = an

(follows from 6)

8. limx→a

n√x = n

√a

9. limx→a

n√

f(x) = n

√limx→a

f(x) (If n is even, we must addi onallyassume that lim

x→af(x) > 0)

Limits and arithmetic IIFact (Con nued)

5. limx→a

f(x)g(x)

=LM

, if M ̸= 0.

6. limx→a

[f(x)]n =[limx→a

f(x)]n

(follows from 4 repeatedly)

7. limx→a

xn = an (follows from 6)

8. limx→a

n√x = n

√a

9. limx→a

n√

f(x) = n

√limx→a

f(x) (If n is even, we must addi onallyassume that lim

x→af(x) > 0)

Limits and arithmetic IIFact (Con nued)

5. limx→a

f(x)g(x)

=LM

, if M ̸= 0.

6. limx→a

[f(x)]n =[limx→a

f(x)]n

(follows from 4 repeatedly)

7. limx→a

xn = an (follows from 6)

8. limx→a

n√x = n

√a

9. limx→a

n√

f(x) = n

√limx→a

f(x) (If n is even, we must addi onallyassume that lim

x→af(x) > 0)

Applying the limit lawsExample

Find limx→3

(x2 + 2x+ 4

).

Solu onBy applying the limit laws repeatedly:

limx→3

(x2 + 2x+ 4

)

= limx→3

(x2)+ lim

x→3(2x) + lim

x→3(4)

=(limx→3

x)2

+ 2 · limx→3

(x) + 4

= (3)2 + 2 · 3+ 4 = 9+ 6+ 4 = 19.

Applying the limit lawsExample

Find limx→3

(x2 + 2x+ 4

).

Solu onBy applying the limit laws repeatedly:

limx→3

(x2 + 2x+ 4

)

= limx→3

(x2)+ lim

x→3(2x) + lim

x→3(4)

=(limx→3

x)2

+ 2 · limx→3

(x) + 4

= (3)2 + 2 · 3+ 4 = 9+ 6+ 4 = 19.

Applying the limit lawsExample

Find limx→3

(x2 + 2x+ 4

).

Solu onBy applying the limit laws repeatedly:

limx→3

(x2 + 2x+ 4

)= lim

x→3

(x2)+ lim

x→3(2x) + lim

x→3(4)

=(limx→3

x)2

+ 2 · limx→3

(x) + 4

= (3)2 + 2 · 3+ 4 = 9+ 6+ 4 = 19.

Applying the limit lawsExample

Find limx→3

(x2 + 2x+ 4

).

Solu onBy applying the limit laws repeatedly:

limx→3

(x2 + 2x+ 4

)= lim

x→3

(x2)+ lim

x→3(2x) + lim

x→3(4)

=(limx→3

x)2

+ 2 · limx→3

(x) + 4

= (3)2 + 2 · 3+ 4 = 9+ 6+ 4 = 19.

Applying the limit lawsExample

Find limx→3

(x2 + 2x+ 4

).

Solu onBy applying the limit laws repeatedly:

limx→3

(x2 + 2x+ 4

)= lim

x→3

(x2)+ lim

x→3(2x) + lim

x→3(4)

=(limx→3

x)2

+ 2 · limx→3

(x) + 4

= (3)2 + 2 · 3+ 4

= 9+ 6+ 4 = 19.

Applying the limit lawsExample

Find limx→3

(x2 + 2x+ 4

).

Solu onBy applying the limit laws repeatedly:

limx→3

(x2 + 2x+ 4

)= lim

x→3

(x2)+ lim

x→3(2x) + lim

x→3(4)

=(limx→3

x)2

+ 2 · limx→3

(x) + 4

= (3)2 + 2 · 3+ 4 = 9+ 6+ 4 = 19.

Your turnExample

Find limx→3

x2 + 2x+ 4x3 + 11

Solu on

The answer is1938

=12.

Your turnExample

Find limx→3

x2 + 2x+ 4x3 + 11

Solu on

The answer is1938

=12.

Direct Substitution Property

As a direct consequence of the limit laws and the really basic limitswe have:Theorem (The Direct Subs tu on Property)

If f is a polynomial or a ra onal func on and a is in the domain of f,then

limx→a

f(x) = f(a)

OutlineRecall: The concept of limit

Basic Limits

Limit LawsThe direct subs tu on property

Limits with AlgebraTwo more limit theorems

Two important trigonometric limits

Limits do not see the point!

(in a good way)

TheoremIf f(x) = g(x) when x ̸= a, and lim

x→ag(x) = L, then lim

x→af(x) = L.

Example of the MTP principleExample

Find limx→−1

x2 + 2x+ 1x+ 1

, if it exists.

Solu on

Sincex2 + 2x+ 1

x+ 1= x+ 1 whenever x ̸= −1, and since

limx→−1

x+ 1 = 0, we have limx→−1

x2 + 2x+ 1x+ 1

= 0.

Example of the MTP principleExample

Find limx→−1

x2 + 2x+ 1x+ 1

, if it exists.

Solu on

Sincex2 + 2x+ 1

x+ 1= x+ 1 whenever x ̸= −1, and since

limx→−1

x+ 1 = 0, we have limx→−1

x2 + 2x+ 1x+ 1

= 0.

ET game for f(x) =x2 + 2x + 1

x + 1

.. x.

y

...−1

I Even if f(−1) were something else, it would not effect the limit.

ET game for f(x) =x2 + 2x + 1

x + 1

.. x.

y

...−1

I Even if f(−1) were something else, it would not effect the limit.

Limit of a piecewise functionExample

Let f(x) =

{x2 x ≥ 0−x x < 0

. Does limx→0

f(x) exist?

Solu on

I We have limx→0+

f(x) MTP= lim

x→0+x2 DSP

= 02 = 0

I Likewise: limx→0−

f(x) = limx→0−

−x = −0 = 0

I So limx→0

f(x) = 0.

.

.

Limit of a piecewise functionExample

Let f(x) =

{x2 x ≥ 0−x x < 0

. Does limx→0

f(x) exist?

Solu on

I We have limx→0+

f(x) MTP= lim

x→0+x2 DSP

= 02 = 0

I Likewise: limx→0−

f(x) = limx→0−

−x = −0 = 0

I So limx→0

f(x) = 0.

.

.

Limit of a piecewise functionExample

Let f(x) =

{x2 x ≥ 0−x x < 0

. Does limx→0

f(x) exist?

Solu on

I We have limx→0+

f(x) MTP= lim

x→0+x2 DSP

= 02 = 0

I Likewise: limx→0−

f(x) = limx→0−

−x = −0 = 0

I So limx→0

f(x) = 0.

.

.

Limit of a piecewise functionExample

Let f(x) =

{x2 x ≥ 0−x x < 0

. Does limx→0

f(x) exist?

Solu on

I We have limx→0+

f(x) MTP= lim

x→0+x2 DSP

= 02 = 0

I Likewise: limx→0−

f(x) = limx→0−

−x = −0 = 0

I So limx→0

f(x) = 0.

..

Limit of a piecewise functionExample

Let f(x) =

{x2 x ≥ 0−x x < 0

. Does limx→0

f(x) exist?

Solu on

I We have limx→0+

f(x) MTP= lim

x→0+x2 DSP

= 02 = 0

I Likewise: limx→0−

f(x) = limx→0−

−x = −0 = 0

I So limx→0

f(x) = 0.

..

Limit of a piecewise functionExample

Let f(x) =

{x2 x ≥ 0−x x < 0

. Does limx→0

f(x) exist?

Solu on

I We have limx→0+

f(x) MTP= lim

x→0+x2 DSP

= 02 = 0

I Likewise: limx→0−

f(x) = limx→0−

−x = −0 = 0

I So limx→0

f(x) = 0.

.

.

Limit of a piecewise functionExample

Let f(x) =

{x2 x ≥ 0−x x < 0

. Does limx→0

f(x) exist?

Solu on

I We have limx→0+

f(x) MTP= lim

x→0+x2 DSP

= 02 = 0

I Likewise: limx→0−

f(x) = limx→0−

−x = −0 = 0

I So limx→0

f(x) = 0.

.

.

Finding limits by algebraExample

Find limx→4

√x− 2x− 4

.

Solu onWrite the denominator as x− 4 =

√x2 − 4 = (

√x− 2)(

√x+ 2). So

limx→4

√x− 2x− 4

= limx→4

√x− 2

(√x− 2)(

√x+ 2)

= limx→4

1√x+ 2

=14

Finding limits by algebraExample

Find limx→4

√x− 2x− 4

.

Solu onWrite the denominator as x− 4 =

√x2 − 4 = (

√x− 2)(

√x+ 2).

So

limx→4

√x− 2x− 4

= limx→4

√x− 2

(√x− 2)(

√x+ 2)

= limx→4

1√x+ 2

=14

Finding limits by algebraExample

Find limx→4

√x− 2x− 4

.

Solu onWrite the denominator as x− 4 =

√x2 − 4 = (

√x− 2)(

√x+ 2). So

limx→4

√x− 2x− 4

= limx→4

√x− 2

(√x− 2)(

√x+ 2)

= limx→4

1√x+ 2

=14

Your turnExample

Let f(x) =

{1− x2 x ≥ 12x x < 1

. Find limx→1

f(x) if it exists.

Solu on

I limx→1+

f(x) = limx→1+

(1− x2

) DSP= 0

I limx→1−

f(x) = limx→1−

(2x) DSP= 2

I The le - and right-hand limits disagree, so thelimit does not exist.

.

..1

..

Your turnExample

Let f(x) =

{1− x2 x ≥ 12x x < 1

. Find limx→1

f(x) if it exists.

Solu on

I limx→1+

f(x) = limx→1+

(1− x2

) DSP= 0

I limx→1−

f(x) = limx→1−

(2x) DSP= 2

I The le - and right-hand limits disagree, so thelimit does not exist.

.

..1

..

Your turnExample

Let f(x) =

{1− x2 x ≥ 12x x < 1

. Find limx→1

f(x) if it exists.

Solu on

I limx→1+

f(x) = limx→1+

(1− x2

) DSP= 0

I limx→1−

f(x) = limx→1−

(2x) DSP= 2

I The le - and right-hand limits disagree, so thelimit does not exist.

...1

.

.

Your turnExample

Let f(x) =

{1− x2 x ≥ 12x x < 1

. Find limx→1

f(x) if it exists.

Solu on

I limx→1+

f(x) = limx→1+

(1− x2

) DSP= 0

I limx→1−

f(x) = limx→1−

(2x) DSP= 2

I The le - and right-hand limits disagree, so thelimit does not exist.

...1

.

.

Your turnExample

Let f(x) =

{1− x2 x ≥ 12x x < 1

. Find limx→1

f(x) if it exists.

Solu on

I limx→1+

f(x) = limx→1+

(1− x2

) DSP= 0

I limx→1−

f(x) = limx→1−

(2x) DSP= 2

I The le - and right-hand limits disagree, so thelimit does not exist.

...1

..

Your turnExample

Let f(x) =

{1− x2 x ≥ 12x x < 1

. Find limx→1

f(x) if it exists.

Solu on

I limx→1+

f(x) = limx→1+

(1− x2

) DSP= 0

I limx→1−

f(x) = limx→1−

(2x) DSP= 2

I The le - and right-hand limits disagree, so thelimit does not exist.

...1

..

A message fromthe Mathematical Grammar Police

Please do not say “limx→a

f(x) = DNE.” Does not compute.

I Too many verbsI Leads to FALSE limit laws like “If lim

x→af(x) DNE and lim

x→ag(x) DNE,

then limx→a

(f(x) + g(x)) DNE.”

A message fromthe Mathematical Grammar Police

Please do not say “limx→a

f(x) = DNE.” Does not compute.I Too many verbs

I Leads to FALSE limit laws like “If limx→a

f(x) DNE and limx→a

g(x) DNE,then lim

x→a(f(x) + g(x)) DNE.”

A message fromthe Mathematical Grammar Police

Please do not say “limx→a

f(x) = DNE.” Does not compute.I Too many verbsI Leads to FALSE limit laws like “If lim

x→af(x) DNE and lim

x→ag(x) DNE,

then limx→a

(f(x) + g(x)) DNE.”

Two Important Limit Theorems

TheoremIf f(x) ≤ g(x) when x is near a(except possibly at a), then

limx→a

f(x) ≤ limx→a

g(x)

(as usual, provided these limitsexist).

Theorem (The Squeeze/Sandwich/ Pinching Theorem)

If f(x) ≤ g(x) ≤ h(x) when x isnear a (as usual, exceptpossibly at a), and

limx→a

f(x) = limx→a

h(x) = L,

then limx→a

g(x) = L.

Using the Squeeze TheoremWe can use the Squeeze Theorem to replace complicatedexpressions with simple ones when taking the limit.

Example

Show that limx→0

x2 sin(πx

)= 0.

Solu onWe have for all x,

−1 ≤ sin(πx

)≤ 1 =⇒ −x2 ≤ x2 sin

(πx

)≤ x2

The le and right sides go to zero as x → 0.

Using the Squeeze TheoremWe can use the Squeeze Theorem to replace complicatedexpressions with simple ones when taking the limit.Example

Show that limx→0

x2 sin(πx

)= 0.

Solu onWe have for all x,

−1 ≤ sin(πx

)≤ 1 =⇒ −x2 ≤ x2 sin

(πx

)≤ x2

The le and right sides go to zero as x → 0.

Using the Squeeze TheoremWe can use the Squeeze Theorem to replace complicatedexpressions with simple ones when taking the limit.Example

Show that limx→0

x2 sin(πx

)= 0.

Solu onWe have for all x,

−1 ≤ sin(πx

)≤ 1 =⇒ −x2 ≤ x2 sin

(πx

)≤ x2

The le and right sides go to zero as x → 0.

Illustrating the Squeeze Theorem

.. x.

y

.

h(x) = x2

.

f(x) = −x2

.

g(x) = x2 sin(πx

)

Illustrating the Squeeze Theorem

.. x.

y

.

h(x) = x2

.

f(x) = −x2

.

g(x) = x2 sin(πx

)

Illustrating the Squeeze Theorem

.. x.

y

.

h(x) = x2

.

f(x) = −x2

.

g(x) = x2 sin(πx

)

OutlineRecall: The concept of limit

Basic Limits

Limit LawsThe direct subs tu on property

Limits with AlgebraTwo more limit theorems

Two important trigonometric limits

Two trigonometric limits

TheoremThe following two limits hold:

I limθ→0

sin θθ

= 1

I limθ→0

cos θ − 1θ

= 0

Proof of the Sine LimitProof.

.. θ

.sin θ

.cos θ

.θ

.tan θ

.−1

.1

I No ce

sin θ ≤

θ

≤ 2 tanθ

2≤ tan θ

I Divide by sin θ: 1 ≤ θ

sin θ≤ 1

cos θI Take reciprocals: 1 ≥ sin θ

θ≥ cos θ

As θ → 0, the le and right sides tend to 1. So, then, must themiddle expression.

Proof of the Sine LimitProof.

.. θ.sin θ

.cos θ

.θ

.tan θ

.−1

.1

I No ce sin θ ≤ θ

≤ 2 tanθ

2≤ tan θ

I Divide by sin θ: 1 ≤ θ

sin θ≤ 1

cos θI Take reciprocals: 1 ≥ sin θ

θ≥ cos θ

As θ → 0, the le and right sides tend to 1. So, then, must themiddle expression.

Proof of the Sine LimitProof.

.. θ.sin θ

.cos θ

.θ

.tan θ

.−1

.1

I No ce sin θ ≤ θ

≤ 2 tanθ

2≤

tan θ

I Divide by sin θ: 1 ≤ θ

sin θ≤ 1

cos θI Take reciprocals: 1 ≥ sin θ

θ≥ cos θ

As θ → 0, the le and right sides tend to 1. So, then, must themiddle expression.

Proof of the Sine LimitProof.

.. θ.sin θ

.cos θ

.θ

.tan θ

.−1

.1

I No ce sin θ ≤ θ ≤ 2 tanθ

2≤ tan θ

I Divide by sin θ: 1 ≤ θ

sin θ≤ 1

cos θI Take reciprocals: 1 ≥ sin θ

θ≥ cos θ

As θ → 0, the le and right sides tend to 1. So, then, must themiddle expression.

Proof of the Sine LimitProof.

.. θ.sin θ

.cos θ

.θ

.tan θ

.−1

.1

I No ce sin θ ≤ θ ≤ 2 tanθ

2≤ tan θ

I Divide by sin θ: 1 ≤ θ

sin θ≤ 1

cos θ

I Take reciprocals: 1 ≥ sin θθ

≥ cos θ

As θ → 0, the le and right sides tend to 1. So, then, must themiddle expression.

Proof of the Sine LimitProof.

.. θ.sin θ

.cos θ

.θ

.tan θ

.−1

.1

I No ce sin θ ≤ θ ≤ 2 tanθ

2≤ tan θ

I Divide by sin θ: 1 ≤ θ

sin θ≤ 1

cos θI Take reciprocals: 1 ≥ sin θ

θ≥ cos θ

As θ → 0, the le and right sides tend to 1. So, then, must themiddle expression.

Proof of the Sine LimitProof.

.. θ.sin θ

.cos θ

.θ

.tan θ

.−1

.1

I No ce sin θ ≤ θ ≤ 2 tanθ

2≤ tan θ

I Divide by sin θ: 1 ≤ θ

sin θ≤ 1

cos θI Take reciprocals: 1 ≥ sin θ

θ≥ cos θ

As θ → 0, the le and right sides tend to 1. So, then, must themiddle expression.

Proof of the Cosine LimitProof.

1− cos θθ

=1− cos θ

θ· 1+ cos θ1+ cos θ

=1− cos2 θθ(1+ cos θ)

=sin2 θ

θ(1+ cos θ)=

sin θθ

· sin θ1+ cos θ

So

limθ→0

1− cos θθ

=

(limθ→0

sin θθ

)·(limθ→0

sin θ1+ cos θ

)= 1 · 0

2= 0.

Proof of the Cosine LimitProof.

1− cos θθ

=1− cos θ

θ· 1+ cos θ1+ cos θ

=1− cos2 θθ(1+ cos θ)

=sin2 θ

θ(1+ cos θ)=

sin θθ

· sin θ1+ cos θ

So

limθ→0

1− cos θθ

=

(limθ→0

sin θθ

)·(limθ→0

sin θ1+ cos θ

)= 1 · 0

2= 0.

Proof of the Cosine LimitProof.

1− cos θθ

=1− cos θ

θ· 1+ cos θ1+ cos θ

=1− cos2 θθ(1+ cos θ)

=sin2 θ

θ(1+ cos θ)

=sin θθ

· sin θ1+ cos θ

So

limθ→0

1− cos θθ

=

(limθ→0

sin θθ

)·(limθ→0

sin θ1+ cos θ

)= 1 · 0

2= 0.

Proof of the Cosine LimitProof.

1− cos θθ

=1− cos θ

θ· 1+ cos θ1+ cos θ

=1− cos2 θθ(1+ cos θ)

=sin2 θ

θ(1+ cos θ)=

sin θθ

· sin θ1+ cos θ

So

limθ→0

1− cos θθ

=

(limθ→0

sin θθ

)·(limθ→0

sin θ1+ cos θ

)= 1 · 0

2= 0.

Proof of the Cosine LimitProof.

1− cos θθ

=1− cos θ

θ· 1+ cos θ1+ cos θ

=1− cos2 θθ(1+ cos θ)

=sin2 θ

θ(1+ cos θ)=

sin θθ

· sin θ1+ cos θ

So

limθ→0

1− cos θθ

=

(limθ→0

sin θθ

)·(limθ→0

sin θ1+ cos θ

)

= 1 · 02= 0.

Proof of the Cosine LimitProof.

1− cos θθ

=1− cos θ

θ· 1+ cos θ1+ cos θ

=1− cos2 θθ(1+ cos θ)

=sin2 θ

θ(1+ cos θ)=

sin θθ

· sin θ1+ cos θ

So

limθ→0

1− cos θθ

=

(limθ→0

sin θθ

)·(limθ→0

sin θ1+ cos θ

)= 1 · 0

2= 0.

Try theseExample

1. limθ→0

tan θθ

2. limθ→0

sin 2θθ

Answer

1. 12. 2

Try theseExample

1. limθ→0

tan θθ

2. limθ→0

sin 2θθ

Answer

1. 12. 2

Solutions1. Use the basic trigonometric limit and the defini on of tangent.

limθ→0

tan θθ

= limθ→0

sin θθ cos θ

= limθ→0

sin θθ

· limθ→0

1cos θ

= 1 · 11= 1.

2. Change the variable:

limθ→0

sin 2θθ

= lim2θ→0

sin 2θ2θ · 1

2= 2 · lim

2θ→0

sin 2θ2θ

= 2 · 1 = 2

OR use a trigonometric iden ty:

limθ→0

sin 2θθ

= limθ→0

2 sin θ cos θθ

= 2·limθ→0

sin θθ

·limθ→0

cos θ = 2·1·1 = 2

Solutions1. Use the basic trigonometric limit and the defini on of tangent.

limθ→0

tan θθ

= limθ→0

sin θθ cos θ

= limθ→0

sin θθ

· limθ→0

1cos θ

= 1 · 11= 1.

2. Change the variable:

limθ→0

sin 2θθ

= lim2θ→0

sin 2θ2θ · 1

2= 2 · lim

2θ→0

sin 2θ2θ

= 2 · 1 = 2

OR use a trigonometric iden ty:

limθ→0

sin 2θθ

= limθ→0

2 sin θ cos θθ

= 2·limθ→0

sin θθ

·limθ→0

cos θ = 2·1·1 = 2

Solutions1. Use the basic trigonometric limit and the defini on of tangent.

limθ→0

tan θθ

= limθ→0

sin θθ cos θ

= limθ→0

sin θθ

· limθ→0

1cos θ

= 1 · 11= 1.

2. Change the variable:

limθ→0

sin 2θθ

= lim2θ→0

sin 2θ2θ · 1

2= 2 · lim

2θ→0

sin 2θ2θ

= 2 · 1 = 2

OR use a trigonometric iden ty:

limθ→0

sin 2θθ

= limθ→0

2 sin θ cos θθ

= 2·limθ→0

sin θθ

·limθ→0

cos θ = 2·1·1 = 2

Summary

I The limit laws allow us tocompute limitsreasonably.

I BUT we cannot make upextra laws otherwise weget into trouble.

.. x.

y