Notes 2.1 - Limits

I. Limit Notation

A.) Two-sided Notation:

Read as “the limit of f(x) as x approaches a is L.”

lim ( )x a

f x L

B.) One-sided Notation:

Read as “the limit of f(x) as x approaches a from the right is L.”

Read as “the limit of f(x) as x approaches a from the left is L.”

lim ( )x a

f x L

lim ( )x a

f x L

II. Limit Definition

A.) Def: The function f has a limit L as x approaches c iff:

B.) Visual Representations:

lim ( ) lim ( ) lim ( )x c x c x c

f x L f x f x

2 11.) ( )

1

xf x

x

2 1, 1

2.) ( ) 11, 1

xx

g x xx

B.) Visual Reps(cont.)-

3.) ( ) 1h x x

III. Non-existent Limits

A.) fails to exist when:

1.) The right-side limit and left-side limit equal different real numbers.

2.) The are infinite oscillations .

3.) The limit(s) approach

lim ( )x a

f x

.

Ex. – Evaluate 0

1lim .x x

0

1lim x x

0

1lim x x

0

1lim Does Not Exist.x x

Although limits approaching infinity do not exist, we must still describe the behavior from both/each side(s)!!!

IV. Computational Techniques for Limits

A.) Properties of Limits: p. 61-62 of your text. KNOW THEM!!!

SUM, DIFFERENCE, PRODUCT, CONSTANT MULTIPLE, QUOTIENT, and POWER RULES

B.) Theorem – Polynomial and Rational Functions-

11 01.) If ( ) ... is any

polynomial fn and is any real number, then

n nn nf x a x a x a

c

11 0lim ( ) ... n n

n nx c

f x a c a c a

B.) Theorem – Polynomial and Rational Functions-

2.) If ( ) and ( ) are polynomial fns

and is any real number, then

f x g x

c

( ) ( )lim , if ( ) 0

( ) ( )x c

f x f cg c

g x g c

C.) Limit Properties and Theorems valid for one-sided limits as well.

V. Techniques for Evaluation

A.) Apply the theorems/properties (Substitute)

B.) Modify the expression:

1.) Simplify

2.) Factor

3.) Expand

4.) Multiply by conjugates

5.) Apply thm./prop. again

C.) Evaluate the following limits:

1.)

2.)

2

2

4lim

1x

x

x

22 4 00

2 1 3

2

2

4lim

2x

x

x

22 4 0???

2 2 0

2 2

2 2lim lim 2

2x x

x xx

x

2 2 4

3.)

4.)

2

2

4lim

2x

x

x

22 4 00

2 2 4

22

2lim

4x

x

x

2

2 2 4DNE

2 2 0

22

2 4lim

4 0x

x

x

22

2 4lim

4 0x

x

x

5.) 22

3lim

2 ( 5)x

x

x x

1DNE

0

22

22

3lim

2 ( 5)

3 1lim

02 ( 5)

x

x

x

x x

x

x x

6.) 4

4lim

2x

x

x

0???

0

4

4

4

2 24lim lim

lim 2 4 2 4

2 2

x

x x

x xx

x x

x

7.) 0

2 2lim x

x

x

0???

0

0

0 0

2 2 2 2lim

2 2

1lim lim

2 22 2

x

x x

x x

x x

x

xx x

1

2 2

8.) 2

4

3 49lim

4x

x

x

0???

0

4

4 4

3 7 3 7lim

44 10

lim lim 104

x

x x

x x

xx x

xx

14

9.) 3

4

64lim

4x

x

x

0???

0

2

4

2

4

4 4 16lim

4

lim 4 16

x

x

x x x

x

x x

48

VI. You Try!

1.)

2.)

3.)

4.)

3

2

2 16lim

2x

x

x

1

1lim

1x

x

x

0

9 3lim x

x

x

24

2

21

2lim

1x

x x

x

1

2

1

6

DNE

VII. Sandwich Theorem

GRAPHICALLY

A.) If ( ) ( ) ( ) for all in an open interval

containing the point (with the possible exception

at ) and lim ( ) lim ( ), then lim ( )x c x c x c

g x f x h x x

x c

x c g x L h x f x L

( )f x

( )h x

( )g x

B.) Example -

What do you know about the sin function?

2 2

0

1lim sinx

xx

11 sin 1

x

2 10 sin 1

x

2 2 2 210 sin 1x x x

x

2 2

0

1lim sin 0x

xx

2 2 210 sinx x

x 2 2 2

0 0 0

1lim 0 lim sin limx x x

x xx

2 2

0

10 lim sin 0

xx

x

C.) Example - 20

1lim cosx

xx

2

11 cos 1

x

2

11 cos 1x x x

x

2

1cosx x x

x

20 0 0

1lim lim cos limx x x

x x xx

20

10 lim cos 0

xx

x

20

1lim cos 0x

xx

VIII. Limit Theorems

0A.) lim cos 1

0B.) lim sin 0

0

sinC.) lim 1

0D.) lim 1

sin

IX. Proof of

Given the point P in the first quadrant on the unit circle:

The area of sector OAP = 21

2 2

0

sinlim 1

P (x, y)

A (1, 0)

O

(0, 0)

𝜃

1

Dropping a perp. From P to Q on the x-axis gives us triangle OPQ

The area of triangle OPQ = 1 1( ) sin cos

2 2x y

P (x, y)

A (1, 0)O

(0, 0)

𝜃1

Q (x,0)

Now, we can all agree that

Area Area Sect. OPQ OPA

1sin cos

2 2

P (x, y)

A (1, 0)O

(0, 0)

𝜃1

Q (x,0)

Extending OP to T and dropping a perp. from T to A on the x-axis gives us triangle OTA

The area of triangle OTA = 1 1

(1) tan tan2 2

P

A O

(0, 0)

1

Q

T(1, tan 𝜃)

𝜃

Now, we can all agree that

Area Area Sect. Area OPQ OPA OTA

1 1sin cos tan

2 2 2

P

A O

(0, 0)

1

Q

T(1, tan 𝜃)

𝜃

1 1sin cos tan

2 2 2

2 1 1sin cos tan

sin 2 2 2

1cos

sin cos

1 sincos

cos

sin 1cos

cos

0 0 0

sin 1lim cos lim lim

cos

0

sin1 lim 1

0

sinlim 1

X. Patching

In order to make our trigonometric limits look like A-D of II, we may need to “PATCH” the trig expression. After, we apply our limit properties and verify on our calculator.

A) Examples -

0

sin 31.) lim

x

x

x

0

sin 3lim x

x

x

0 0 0

sin 3 sin 3lim lim .lim

3

3 3

3x x x

x x

x xx x

x x

0 0

sinlim .lim 3

0 0

sin 3lim .lim 3

3x x

x

x 0LET 3 ; lim 0

xx

1 3 3

0

sin 32.) lim

5x

x

x

0

sin 3lim

5x

x

x

0 0 0

sin 3 sin 3 3 3 sin 3lim lim . lim

5 5 5

3

3 3 3x x x

x

x

x x x

x xx

3 31

5 5

0

sin 33.) lim

sin 2x

x

x

0

sin 3lim

sin 2x

x

x

0 0

sin 3 3 sin 3 2lim lim

sin

2

2 2 3 sin 2

3

3 2x x

x x

x x

x x x

x x x

3 31 1

2 2

3

0

sin4.) lim

x

x

x 2

0

sinlim sinx

xx

x 0 1 0

0

1 sin 3lim

cos3 sin 2x

x

x x

0

tan 35.) lim

sin 2x

x

x

0

sin 3lim

sin 2 cos3x

x

x x

0

1 sin 3 2 3lim

cos3 sin 23 2x

x x x

x xx x 3

2

0

1 sin 2 1lim

cos3 cos 2 7x

x

x x x

0

sec3 tan 26.) lim

7x

x x

x

0

1lim sec3 tan 2

7xx x

x

0

1 sin 2lim

cos3 cos 2 7x

x

x x x

0

1 sin 2 2 2lim

cos3 cos 2 7 72x

x x

x x xx