§ 2.1 The Tangent and Velocity Problems 切線與速度問題

§ 2.1 The Tangent and Velocity Problems 切線與速度問題

[Ex1] about the tangent line:

Secant line，

( ) ( )

1 1

PQ

f x f x

m x x

− −

= =

− −

Tangent line，slope

( ) ( )

1 1

lim

1 1

lim lim

1 1

Q P PQ

x x

m

f x f x

x x

→

→ →

=

− −

= =

− −

[Ex3] about the velocity:

Slope of the secant line = average velocity

( ) ( ) ( )

h

a h

a h

a S h a

S + − =4.9 + ²−4.9 ²

=

Slope of the tangent line = instantaneous velocity

( ) ( )

( )

h

a h

a h

a S h a S

h h

2 2 0

0

9 . 4 9

. lim 4 lim

−

= +

−

= +

→

→

（平均速度）

（瞬時加速速度）

（切線）

（割線）

§ 2.2 The Limit of a Function 函數的極限

x<2 x>2

1.5 2.75 2.5 5.75

4.64 4.31 4.03 4.003

1.8 3.44 2.2

1.9 3.71 2.1

1.99 3.97 2.01

3.997 2.001 1.99

9

( ) f x

↓

||

↓ ↓

( ) ^lim ( ² ) ⁴

lim

2

2

= − + =

→

→

f x x x

x x

( ) f x

2

(

(The limit of as x approaches 2 is equal to 4)

2 4

2

)

lim

x

−

f x

→

||

↓

4

2

( )

lim

x

+

f x

→

( )

f

( )

f

（充分接近）

Def ( informal )

We write if we can make the value of arbitrarily

^{（任意地）}

close to L ( as close to L as we like ) by taking x to be sufficiently close to a ( on either side of a ) but not equal to a

（不等於

）

( ) ^{x L}

g

a x

→ =

( ) ^{x L} lim f

a x

→

= lim

Notice : In finding limit of as x approaches a , we never consider x= a The only thing that matters is how defined near a .

( ) ^{x L}

h

a x

→ = lim lim ( )

x a f x L

→

=

( )

( )

f

One – sided limit 單邊極限

the limit of ( ) as approaches lim ( )

the limit of ( ) as approaches from the left

x a

left hand f x x a

f x L

f x x a

→ −

⎛ − ⎞

= ⎜ ⎟

⎝ ⎠

Def ( informal )

We write

if we can make the value of arbitrarily close to L by taking x to be sufficiently close to a and less than a

Def ( informal ) We write

if we can make the value of arbitrarily close to L by taking x to be sufficiently close to a and greater than a

( )

f

( )

f

the limit of ( ) as approaches lim ( )

the limit of ( ) as approaches from the right

x a

right hand f x x a

f x L

f x x a

→ +

⎛ − ⎞

= ⎜ ⎟

⎝ ⎠

[Ex1] Guess the value of ₂

( )

1

1 1

lim let

1 1

x

x x

f x

x x

→

− − ⎛ ⎜ ⎝ = − − ⎞ ⎟ ⎠

x

0.9 0.99 0.999 0.9999 → 1 ← 1.0001 1.001 1.01 1.1

f (x)

0.526 0.5025 0.50025 0.500025 0.499975 0.49975 0.4975 0.476

0.5

1

( )

li mx f x

→ (Notice: does not exist)

( )

( ) ⎥ ⎥

⎦

⎤

⎢ ⎢

⎣

⎡

=

•

=

•

+

−

→

→

5 . 0 lim

5 . 0 lim

1 1

x f

x f

x x

( )

[Ex]

⎪⎩

⎪⎨

⎧

=

− ≠

−

=

1

, 2

1

, 1 1

2

x x x

x x

g

Since and share the same value on either side of a except only when x=a We have

li m

( ) 0 .5

x

g x

→

=

( )

Notice:

1 2 exists

⎛ ⎞

⎜ = ⎟

⎝ ⎠

( )

( )

⎦

⎤

⎢⎢

⎣

⎡

=

•

=

•

+

−

→

→

5 . 0 lim

5 . 0 lim

1 1

x g

x g

x x

( )

f

( )

f ^g

( )

[Ex2] Estimate（估算）

the

X -1.0 -0.5 -0.1 -0.05 -0.001 → 0 ← 0.001 0.05 0.1 0.5 1

0.16228 0.16553 0.16662 0.16666 0.16667 0.16667 0.1666 0.16662 0.16553 0.16558 2

2 0

3 lim 9

t t

t

− +

→

2 2 9 3

t

t + − ^0.1666….

( )

( ) ( )

t 0

t 0

Notice : (1) 0 does not exist (2) lim 1

6 (3) lim 1

6 f

f t f t

−

+

→

→

⎡ ⎤

⎢ ⎥

⎢ ⎥

⎢ = ⎥

⎢ ⎥

⎢ ⎥

⎢ = ⎥

⎣ ⎦

6 1 3 lim ₂9

2

0 + − =

→ t

t

t

But if we take even smaller values of t and get the results from a calculator You can see something strange is happening

X -0.0005 -0.0001 -0.00005 -0.00001 0 0.00001 0.00005 0.0001 0.0005

0.16800 0.20000 0.00000 0.00000 0.00000 0.00000 0.20000 0.16800

⇓

2 2 9 3

t t + −

The calculator gave false value !!

(代入更小的數值）

[Ex4] Guess the value of

x

x

sin π lim

x ¹

0 0 0 0 0 0

( )

f π

= sin

1 2 1 3 1 4

We might be tempted to guess that But our guess is wrong !!

0 sin

lim0 =

→ x

x

π

( )

( ) ( )

1 sin 0

2 (4 1) sin 2 2 1

f n n

f n n

π

π π

= =

+ = + =

⎥⎥

⎥⎥

⎦

⎤

⎢⎢

⎢⎢

⎣

⎡

•

•

+

−

→

→

sin lim

sin lim

0 x

0 x

x x π π

The value of is actually oscillating

（振盪）

^（無限）

In fact, for any integer

（整數）

s i n x π

Hence , does not exist

x

x

sin π lim

does not exist does not exist

3 x 0

cos 5

lim 10000

x x

→

⎛ + ⎞

⎜ ⎟

⎝ ⎠

1 0.5 0.1 0.05 0.01

0.000028 0.124920 0.001088 0.00022 0.000101

x

( )

f x

x

= π

0.005 0.001

0.00010009 0.00010000

3 3

x 0 x 0

cos5 cos5

You might intend to guess lim 0. In fact, lim 0.0001

10000 10000

x x

x x

→ →

⎛ + ⎞= ⎛ + ⎞=

⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

[Ex6] The Heaviside function H is defined by

( ) 0 , if t 0 1 , if t 0

H t

⎧ <

= ⎨ ⎩ ≥

Obviously

^{（明顯地）}

( )

( )

0 0

=

≠

= ₊

− →

→ H t H t

t t

There is not any number approaches as approaches 0 Hence dose not exist.^H

( )

t 0

lim→

( ) ( ) ( )

lim lim lim

t a

t a t a

f t f t L f t L

− +

→

→ = → = ⇔ =

( )

H t

[Ex5] Find

[Ex7] The graph of a function g is shown in the figure below.

Use it to state the value (if they exist) of the following:

( ) ( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )

2

2

2

5

5

5

a lim b lim c lim

d lim e lim f lim

x

x

x

x

x

x

g x g x g x

g x g x g x

−

+

−

+

→

→

→

→

→

→

[Sol] :

( ) ( )

( ) ( )

( ) ( ) ( ) ( )

( ) ( )

( ) ( )

( ) ( ) ( ) ( )

2

2

2 2 2

x 5

x 5

x 5

x 5 x 5

a lim 3 b lim 1

c lim lim lim does not exist.

d lim 2 e lim 2

f lim lim 2 lim 2

x

x

x x x

g x g x

g x g x g x

g x g x

g x g x g x

−

+

− +

−

+

− +

→

→

→ → →

→

→

→ → →

=

=

≠ ∴

=

=

= = ∴ =

∵

∵

Notice:

lim ( ) ( ) 5

5

g x g

x

≠

→

Exercise:

The graph of f is given below, find the following.

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

( ) ( )

2 2 -2

1 1 1

3

i (a) lim (b) lim (c) lim 2 (d) 2 ii (a) lim (b) lim (c) lim 1 (d) 1 iii (a) lim

x x x

x x x

x

f x f x f f -

f x f x f f

f x

+ −

+ −

+

→− →− →

→ → →

→

−

( ) ( ) ( )

3 3

(b) lim (c) lim (d) 3

x x

f x f x f

→ − →

[Sol] :

( ) ( ) ( )

i (a) -1 (b) 1 (c) D.N.E (d) 1 ii (a) 3 (b) 1 (c) D.N.E (d) 2 iii (a) 0.5 (b) 0.5 (c) 0.5 (d) 1

§ 2.3 Calulating Limits Using the Limit Laws

（利用極限法則計算極限）

(

)

5

2 2 2

1

[Ex] (1) lim 2 3 4 2 5 3 5 4 61 ( 2 3 4 )

4 1 1 4 1 1 4 2 4 1

(2) lim ( )

5 1 6 6 3 5

No

x

x

x x x x -

x x x x

x x

→

→

+ − = ⋅ + ⋅ − = +

− + = − ⋅ + = − = − − +

+ + +

∵

∵

2 2

1

4 3 3 1

0

4 3 0 4 3

tice: lim

1 0 1

(3) lim 4

lim 1 1

Actually , lim 0

x

x

x -

x

x x x x

x x

x x

+ x

→

→

→

→

⎛ − + ≠ − + ⎞

⎜ − − ⎟

⎝ ⎠

=

= − = −

=

0

, lim

x ₋ x

→

Direct substitution Property（直接代入性質）

If f is a polynomial（多項式）, a rational function（有理函數）or a root

function（根式函數） ⁿ

x

( )

( )

a

x =

lim→

is a polynomial is a rational function because 1 is not in the domain of

doesn’t exist since is undefined when x < 0

x

Limit Laws (極限法則）

( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

2 2

4 4 4

2 4

Suppose lim and lim exist. Then

lim lim lim lim 5 lim lim 5 2 11 13

lim lim lim lim 5 lim

x a x a

x a x a x a x x x

x a x a x a x

f x g x

f x g x f x g x x x x x

f x g x f x g x x x

→ →

→ → → → → →

→ → → →

⎡ ⎤

+ = + ⇒ + − = + − = + =

⎡ ⎤

⎣ ⎦ ⎣ ⎦

⎡ ⎤

− = − ⇒ − − =

⎡ ⎤

⎣ ⎦ ⎣ ⎦

( )

( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

2

4 4

4 4

2 2

4 4 4

lim 5 2 11 9

lim lim lim3 3lim 3 2 6

lim lim lim lim 5 lim lim 5 2 11 22

li

x x

x a x a x x

x a x a x a x x x

x x

cf x c f x x x

f x g x f x g x x x x x

→ →

→ → → →

→ → → → → →

− − = − = −

= ⇒ = = ⋅ =

= ⋅ ⇒ − = ⋅ − = ⋅ =

( ) ( )

( )

( ) ( ) ( ) ( ( ) )

( ) ( ) ^{( ) ( )}

4 2

2 2

x a 4 4

4 6 6

2 2 6

4 4

lim lim 2

m if lim 0 . lim lim 5 0

lim 5 lim 5 11

lim ( ) lim ( ) , n:positive integer. lim 5 lim 5 11 lim

x a x

x a x x

x a x

n n

x a x a x x

x a

f x x

f x x

g x x

g x g x x x

f x f x x x

→ →

→ → → →

→ →

→ → → →

→

= ≠ ⇒ = = − ≠

− −

⎡ ⎤

= ⇒ − = ⎣ − ⎦ =

∵

( )

( )

( )

(

)

n n n

x a x a x x a

f x f x f x x x

→ → → →

= ⇒ − = − =

[Ex1] Use the Limit Laws and the graph of f and g below to evaluate the following limits , if they exist

( ) ( )

( ) ( ) ( ) ( )

2

1

2

(a ) lim 5

(b ) lim (c ) lim

x

x

x

f x g x

f x g x

f x g x

→ −

→

→

+

⎡ ⎤

⎣ ⎦

( )

( )

( ) ^{( ) ( )}

10

3 2

2

10

3 2

2 2 2

10

3 2 2 3

2 2

2

[Ex] lim 4 1 1

5 lim 4 lim 1 lim 1

5

lim 1 1

lim 4 lim 1 2 4 5

lim 5 10

x

x x x

x

x x

x

x x x

x

x x x

x x

x x

x

→

→ → →

→

→ →

→

⎛ − ⎞

⎜ + + − ⎟

⎜ ⎟

⎝ ⎠

= + + − −

= + + − − = + −

(b)

( )

lim

2

x

f x

→

= ( )

lim

x

g x

→

( ) ( )

x 1 x 1

lim

g x 1 lim

g x 2

→

=− ≠

=−

So we can

not

x 1 x 1 x 1

lim f x g x ( ) ( ) lim f x ( ) lim g x ( )

→

=

However we can do by considering one-sided limits

(單邊極限）

( ) ( ) ( ) ( ) ( )

1 1 1

lim lim lim 2 1 2

x x x

f x g x f x g x

+ + +

→

=

⋅

= ⋅ − = −

and

( ) ( ) ( ) ( ) ( )

1 1 1

lim lim lim 2 2 4

x x x

f x g x f x g x

− − −

→

=

= ⋅ − = −

Since

( ) ( ) ( ) ( )

x 1 x 1

lim

f x g x lim

f x g x

→

≠

does not exist

( ) ( )

x 1

lim f x g x

→

[Sol]

(a) From the graphs we have

( ) ( )

2 2

lim 1 an d lim 1

x

f x

g x

→ −

=

= −

Therefore

( ) ( ) ( ) ( ) ( )

x -2 x -2 x -2

lim f x 5 g x lim f x 5lim g x 1 5 1 4

→

⎡ ⎣ + ⎤ ⎦ =

+

= + − = −

(c) To find we intend to use the law But we can not because .

Actually , the limit does not exist because the denominator

^（分母）

approach 0 while the numerator

（分子）

（非零的）

( ) ( )

lim

f x

→

g x

( ) ( )

( )

2

( )

2

2

lim

lim lim

x x

x

f x f x

g x g x

→

→

→

=

2

( )

lim 0

x

g x

→

=

( ) ( )

x 2

lim f x

→

g x

[Ex3] Find

2 1

l i m 1

1

x

x

→

x

−

−

[Sol] ²

( )( ) ( )

1 1 1

1 1

lim 1 lim lim 1 2

1 1

x x x

x x

x x

x x

→ → →

− +

− = = + =

− −

When we're taking the limit as approaches 1 , we have

1 , so 1 0

x x

⎡ ⎤

⎢ ⎥

⎢ ⎥

⎢ ≠ − ≠ ⎥

⎣ ⎦

Direct substitution

（直接代入）

[Ex4] find

( ) ( )

1

1 , if 1 lim where g

, 1

x

x x

g x x

π x

→

+ ≠

= ⎨⎧⎩ =

[Sol 1]

Let

( )

¹

1 f x x

x

= −

−

Since

^f ( ) ^{x = g} ( ) ^x

[Sol 2]

( ) ( )

1 1

lim lim 1 2

x

g x

x

→

=

+ =

[Ex5] Evaluate [Sol]

( )

(

) _{( )}

0 0 0

9 6 9

3 9

lim lim lim 6 6

h h h

h h

h h

h h

→ → →

+ + −

+ −

= = + =

(

) direct substitution

( ) ( )

1 1

li m li m 2

x

f x

g x

→

=

=

( )

0

3 9

lim

h

→

h

+ −

[Ex] Find

2 2 0

3 lim 9

t t

t

− +

→

[Sol]

( )

( ) ⁽ ( ⁾ )

( )

2

2 2 2

2 2 2

2 2 2

0 0 0 2 2 0 2 2

2

0 2 2 0 2

9 3 9 9

9 3 9 3 9 3

lim lim lim lim

9 3 9 3 9 3

1 1 1

lim lim

9 3 6 9 3 9 3

t t t t

t t

t t

t t t

t t t t t t t

t t t t

→ → → →

→ →

+ − + −

+ − = + − ⋅ + + = =

+ + + + + +

= = = =

+ + + + +

[Ex] Show that

l i m

0

x

x

→

=

[Sol 1]

0 0 0 0

( )

0 0 0

,if 0 ,if 0

lim lim 0 an d lim lim 0

lim lim 0 lim 0

x x x x

x x x

x x

x x x

x x x x

x x x

+ + − −

+ −

→ → → →

→ → →

⎧ ≥

= ⎨ ⎩ − <

= = = − =

= = ∴ =

∵

[Sol 2] ²

0 0

l i m l i m 0

x x x x

→

=

=

[Ex8] Prove

^（證明）

l i m

x

→ x does not exist [Sol]

0 0

0 0

lim lim 1

lim lim 1

x x

x x

x x

x x

x x

x x

+ +

− −

→ →

→ →

= =

= − = −

[Ex9] If

( ) ^{4 4}

8 2 4 x - , x f x - x , x

⎧ >

= ⎨ ⎪

⎪⎩ <

^（判定）

^f ( ) ^x

x 4

lim

[Sol]

( )

( ) ( )

( ) ( )

( )

4 4

4 4

4 4

4

lim lim 4 4 4 0

lim lim 8 2 8 2 4 0

lim lim 0

lim 0

x x

x x

x x

x

f x x

f x x

f x f x

f x

+ +

− −

+ −

→ →

→ →

→ →

→

= − = − =

= − = − ⋅ =

= =

∴ =

∵

0 0 0

lim lim lim does not exist

x x x

x x x

x x x

+ − →

→

≠

∴

∵

[Ex10] The greatest integer function（最大整數函數） is defined by [x] = the largest that is less than or equal to x . Show that

[ ]

x 3

lim x

→ does not exist [Sol]

3

[ ]

li m li m 3 3

x ₊

x

→

=

=

3

[ ]

lim lim 2 2

x x

−

x

→

=

=

[ ] [ ] [ ]

3 3 3

lim lim lim does not exist

x x x

x x x

+ − →

→

≠

∴

∵

Actually, does not exist for any integer n

^lim [ ]

x n

x

→

Exercise 1. Find

2.

( ) ( )

2 2

1 2

4

4 2

3 1

(a) lim (c) lim where 2

3 3 2

(b) lim 4

2

x x

x

x - ,x x - x -

f x f x x -

x , x

x - x -

→ →

→

⎧ ≠

= ⎨⎪

+ ⎪⎩ =

( ) ( )

2

1 2 2 2

4 6 2

3 2

,x x ,x

f x x , g x

x ,x

x ,x

⎧ ≥ ⎧ >

= ⎪⎨⎪ −⎩ < = ⎨⎩ − ≤

(1) Determine whether and exist

( )

lim

x

f x

→

( )

lim2

x g x

→

(2) Does exist ?

^f ( ) ( ) ^{x g x}

x 2

lim

[Sol]

( )

4 4 4 4

2

2 2 2

2 2 2 2

1.(a) 1 2

4 ( 4)( 2) ( 4)( 2)

(b) lim lim lim lim( 2) 4

2 ( 2)( 2) 4

( 4)

(c) lim lim lim( 2) 4 ( 2)

1 1

2.(1) lim ( ) lim lim ( ) lim ( 2

x x x x

x x x

x x x x

x - x - x x - x

x x

x - x - x

f x xx -

x -

f x f x x

+ + x − −

→ → → →

→ → →

→ → → →

+ +

= = = + =

+ −

= = + =

= = ≠ =

∵

( )( )

( )( )

2

2

2 2 2 2 2

2 2 2

2 2 2

3) 1 lim ( ) D.N.E lim ( ) lim 2 4 lim ( ) lim (4 6) 2 lim ( ) D.N.E (2) lim ( ) ( ) lim ( ) lim ( ) 1 4 2

2 lim ( ) ( ) lim ( ) lim ( ) 1

x

x x x x x

x x x

x x x

f x

g x x g x x g x

f x g x f x g x

f x g x f x g x

+ + − −

+ + +

− − −

→

→ → → → →

→ → →

→ → →

− = ∴

= = ≠ = − = ∴

= = ⋅ =

= =

∵

2 2 2

2 2

lim ( ) ( ) lim ( ) ( ) lim ( ) ( ) exists.

x x x

f x g x f x g x f x g x

+ − →

→ →

⋅ =

= ∴

∵

Theorem （定理）

If when x is near a (except possibly at a) and the limits of f and g both exist as x approaches a, then

( )

( )

f ≤

( ) ( )

lim lim

x a f x x a g x

→

≤

[Ex] If

^{3 x ≤ f} ( ) ^{x ≤ x}

^{+ 2}

^{0 ≤ ≤ x 2}

( )

lim

x

f x

→

[Sol]

Since and

x

+ 2

( ) ( )

( ) ( )

3

1 1 1

1

1

lim3 lim lim 2

3 lim 3

lim 3

x x x

x

x

x f x x

f x f x

→ → →

→

→

≤ ≤ +

⇒ ≤ ≤

⇒ =

lim3 1

x x

→ ^lim_{x→ 1}

(

)

By the theorem, we have

3x

The Squeeze Theorem （夾擊定理）

If when x is near a (except

possibly at a ) and ,

then

( ) ( ) ( )

f x ≤ g x ≤ h x

( ) ( )

lim lim

x a f x x ah x L

→ = → =

( )

lim

→

=

[E11] Show that

2 0

sin 1

lim 0

x

x

x

→

=

Notice: 0 ²

( )

1 1 1

lim sin lim limsin limsin does not exist

x x x x x x

x x x

→ → → →

⎛ ⎞⎛ ⎞

≠ ⎜⎝ ⎟⎜⎠⎝∵ ⎟⎠

[Sol]:

2

2 2

2

1 sin 1 1

0

x x x

x

− ≤

≤

> ∴− ≤ ≤

∵

Since , by the Squeeze Theorem, we have

( )

0 0

lim lim

0

x x x x

→

− =

=

2 0

lim sin 1 0

x

x

→

x =

The Squeeze Theorem (right-hand limit version) If when x is near a and x > a and , then

The Squeeze Theorem (left-hand limit version)

If when x is near a and x < a and , then

( ) ( ) ( )

f x ≤ g x ≤ h x

( ) ( )

lim lim

x a x a

x h x L

f →

→ = + =

+

( ) ( ) ( )

f x ≤ g x ≤ h x

( ) ( )

lim lim

x a x a

h

f x x L

→ − = → − =

( )

lim

x a

x L

→ +

g =

( )

lim

x a

x L

→ −

g =

[Ex] Prove that

0

lim 1 1

x

x

x

→ +

⎡ ⎤ = ⎢ ⎥ ⎣ ⎦

[Sol]:

for any real number

As x approaches 0 from the right, x>0

[ ]

1

t − < t < t

∵

1 1 1 1

x x x

∴ − < ⎡ ⎤ ⎢ ⎥ ⎣ ⎦ ≤

( ) ^{1 1 1 1}

x x x

x x x

⇒ − < ⎡ ⎤ ⎢ ⎥ ⎣ ⎦ ≤ ⋅

i.e.

Since , By the Squeeze Theorem, we get

1 x x 1 1

x

⎡ ⎤

− < ⎢ ⎣ ⎥ ⎦ ≤

( )

0 0

lim 1 lim 1 1

x x

+

x

→

− =

=

0

lim 1 1

x

x x

→ +

⎡ ⎤ = ⎢ ⎥ ⎣ ⎦

t

Exercise:

1. If for all

x

( ) ^{2 2}

1 ≤ f x ≤ x

+ x + ^f ( ) ^x

x

lim

−

→ 5

0 3

lim 1 0

x

x x

→ +

⎡ ⎤ = ⎢ ⎥ ⎣ ⎦

[Sol]

[ ]

2

1 1 1

2

1 1 1

3 3 3

5 5 5

3 3 3

1. lim 1 lim ( ) lim ( 2 2)

lim (1) lim ( 2 2) 1 lim ( ) 1

1 1 1

2. 1 1

1 1 1

w hen > 0 , 1 .

.

x x x

x x x

f x x x

x x f x

t t t

x x x

x x x x

x x x

i

→ − → − → − +

→ − → − → −

≤ ≤ + +

= + + = ∴ =

⎡ ⎤

− < ≤ ⇒ − < ⎢⎣ ⎥⎦ ≤

⎛ ⎞ ⎡ ⎤

⎜⎝ − ⎟⎠ < ⎢⎣ ⎥⎦ ≤

∵ ∵

( )

( )

2 5 5 2

3

2 5 2 5

0 0 0 3

. 1

lim lim 0 lim 1 0

x x x

e x x x x

x

x x x x

+ + + x

→ → →

⎡ ⎤

− < ⎢⎣ ⎥⎦ ≤

⎡ ⎤

− = = ∴ ⎢⎣ ⎥⎦ =

∵

Def

A function is continuous at a (在 a 點連續） if

^lim ( ) ( )

x a

f x f a

→

=

This def. implicitly requires three things for to be continuous at a, (1) is defined

(2) exists (3)

( ) ^x

a

f lim

( ) ( ) ^{x f a}

a

f

x

=

lim

If is not continuous at a , we say that is discontinuous at a (在 a 點不連續）

or f has a discontinuity at a ( f 在 a 有一個不連續點）

f f

( )

f

§ 2.5 Continuity 連續性

f

[Ex1] At which numbers is f discontinuous ? Why ?

[Sol]: (1) At x=1, is not defined . is discontinuous at x=1 (2) At x=3. is defined.

But does not exist because . Therefore , f is discontinuous at x=3.

(3) At x=5 . is defined and exists

But .Hence f is discontinuous at x=5.

( )

∵

3

( )

lim 3

x f

→

( ) ( )

3 3

lim lim

x ₊ f x x ₋ f x

→ ≠ →

( ) ( )

lim5 5

x f x f

→ ≠

∴

( )

f

( )

f

( )

lim5

x f x

→

Suppose has a discontinuity at x=a

(1)If exist, the discontinuity is called removable

（可移除的）

because we can remove the discontinuity by redefining

(2)If does not exist , the discontinuity is called nonremovable

（不可移除的）

( )

f

x→a

lim

( )

a f

lim

f

( )

In the example , determine whether each discontinuity is removable or not [Sol]:

(1) f is discontinuous at x= 1 because is undefined .

Since exists, the discontinuity x = 1 is removable.

f has a removable discontinuity at x = 1.

because we can remove the discontinuity by defining as the value of .

1

( )

lim

x f x

→

( ) ¹

f

( ) ¹

f ( )

lim1

x f x

→

(2) f has a discontinuous at x=3.

Since does not exist. f has a nonremovable discontinuity at x=3 because no matter what value we define to be, we can’t make f

become continuous at x=1.So the discontinuity is not removable.

3

( )

lim

f x

→

(3) f has a discontinuity at x=5 and exists. So f has a removable discontinuity at x=5 (Because again by redefining to be the value

of , we can make f become continuous at x=5. Hence the discontinuity x=5 is removable)

5

( )

lim

f x

→

( )

f

( )

f

5

( )

lim

f x

→

[Ex2] Where is each of the following functions discontinuous?

( )

²

(a)

2

x x

f x x

= − − −

[Sol]: (a) is not defined.

is discontinuous at x=2.

Besides,

The limit exists.

Therefore , has a removable discontinuity at x=2

( ) ²

∵ f

( )

∴ f x

( )

( )( )

2 2 2

2 1

lim lim 2 lim

2 2

x x x

x x

x x

f x x x

→ → →

− +

= − − =

− −

( ) ^{1 ,}

⁰

(b)

1 , 0

x x f x

x

⎧ ≠

= ⎨ ⎪

⎪ =

⎩

( )

2

2 , 2

(c) 2

1, 2

x x x

f x x

x

⎧ − − ≠

⎪ −

= ⎨ ⎪ ⎩ = ^{(d) f x} ( ) ⁼ [ ] ^x

( )

lim

1 3

x

x

=

+ =

( )

f x

(b) f is defined at x= 0.

But The limit does not exist.

So f has a discontinuity at x=0. and it’s nonremovable.

(It’s called infinite discontinuity

^{無窮不連續}

( )

0 0

lim lim 1

x f x x

→

=

= ∞

(c) is defined and

The limit exists.

But ,so f is discontinuous at x=2.

In addition , since exists , the discontinuity is removable.

( ) ( )

lim

2

x

f x f

→

≠

2

( )

lim

f x

→

( )

f =

( )

( )( )

2 2 2

2 1

lim lim 2 lim

2 2

x x x

x x

x x

f x x x

→ → →

− +

= − − − = −

( )

lim

1 3

x

x

=

+ =

(d) f has discontinuities at all of the integers

because does not exist for any integer n.

Therefore,

the discontinuities x= n are nonremovable.

(They are called jump discontinuities

跳躍不連續

[ ]

limx n x

→

Def

A function f is continuous from the right at x=a if and f is continuous from the left at x=a if

( ) ( )

lim

x a

f x f a

→ +

=

( ) ( )

lim

x a₋

f x f a

→

=

[Ex3] the function is continuous from the right but discontinuous from the left at x=n ( any integer) because

but

( ) [ ]

f x = x

( ) ( )

lim

x n

f x n f n

→ +

= = ^lim ( ) ¹ ( )

x n

f x n f n

→ −

= − ≠