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Lê Xuân Đại HoChiMinh City University of Technology Faculty of Applied Science, Department of Applied Mathematics... Lê Xuân Đại HCMUT-OISP THE LIMIT AND CONTINUITY OF A FUNCTION HCM — 2

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THE LIMIT AND CONTINUITY OF A FUNCTION

E LECTRONIC VERSION OF LECTURE

Dr Lê Xuân Đại

HoChiMinh City University of Technology Faculty of Applied Science, Department of Applied Mathematics

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P HYSICS

According to the special theory of relativity developed by Albert Einstein, the length of a moving object, as measured by an observer at rest, shrinks as

s

1 − v

2

c 2 , where c is the speed of light.

Question: If the space shuttle were able to approach

Dr Lê Xuân Đại (HCMUT-OISP) THE LIMIT AND CONTINUITY OF A FUNCTION HCM — 2016 3 / 68

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c 2 = L 0

s

1 − c 2

c 2 = 0

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Let f (x) be defined on some open interval that

DEFINITION 1.1

The number L ∈ R is called the limit of f (x) as x

approaches a, and we write

x→a f (x) = L means that the values of f (x) can be made

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C ALCULATING L IMITS USING THE LIMIT LAWS

THEOREM 1.1

Suppose that lim

x→a f (x) = A ∈ R and lim

B if B 6= 0.

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THEOREM 1.2

The squeeze theorem: If

1 f (x) É g(x) É h(x) where x is near a (except possibly

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DEFINITION 1.2

The number L ∈ R is called the limit of f (x) as x

approaches a from the left if for every number ε > 0

there is a number δ > 0 such that

if a − δ < x < a then |f (x) − L| < ε

DEFINITION 1.3

The number L ∈ R is called the limit of f (x) as x

approaches a from the right if for every number ε > 0

there is a number δ > 0 such that

if a < x < a + δ then |f (x) − L| < ε

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THEOREM 1.3

lim

x→a f (x) = L if and only if

( lim

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x→0 f (x) does not exist.

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The elementary functions such as polynomials,

x α , sin x, cos x, a x , log a x(x > 0) are continuous at every number in their domains.

EXAMPLE 2.1

Evaluate lim

x→3 (x 3 − 5x 2 + 7x − 10)

SOLUTION Since f (x) = x 3 − 5x 2 + 7x − 10 is a polynomial function, it is continuous at every

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1 If f (x) is continuous at every number on an open

interval (a, b), then f (x) is continuous on (a, b)

2 f (x) is continuous on the closed interval [a, b], if

f (x) is continuous on the open interval (a, b) and

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Suppose that g is continuous at a and f is continuous

at g(a) Then, the composition f ◦ g is continuous at a

lim

x→a (f ◦g)(x) = lim

x→a f (g(x)) = f ³ lim

x→a g(x) ´ = f (g(a)) = (f ◦g)(a).

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Determine where h(x) = cos(x 2 − 5x + 2) is continuous.

SOLUTION h(x) = f (g(x)), where g(x) = x 2 − 5x + 2 and

f (x) = cosx. Since both f and g are continuous for all

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DEFINITION 3.1

Let f (x) be a function defined on some open interval that contains the number a, except possibly at a itself Then

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DEFINITION 3.2

Let f (x) be a function defined on some open interval that contains the number a, except possibly at a itself Then

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x→+∞ f (x) = L means that the values of f (x) can be

ε ).

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x→−∞ f (x) = L means that the values of f (x) can be

on ε ).

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EXAMPLE 3.3

Evaluate lim

x→∞

3x 2 − x − 2 5x 2 + 4x + 1

SOLUTION Devide both the numerator and

the denominator

lim

x→∞

3x 2 − x − 2 5x 2 + 4x + 1 = lim x→∞

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p 2

3 , y = −

p 2 3

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x→+∞ f (x) = +∞ means that the values of f (x) can be

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The Limit Laws can not be applied to infinite limits

lim

x→+∞ (x 2 − x) = lim

x→+∞ x(x − 1) = ∞ × ∞ = ∞

so their product does too.

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SOLUTION We divide the numerator and

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This means that:

x→a+ f (x) and

lim

x→a− f (x) does not exist or is equal ∞

x→a+ f (x) and lim

x→a− f (x) exist but at least one of the above equalities is not true.

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DEFINITION 4.2

A function f has a removable discontinuity at a if

lim

x→a+ f (x) and lim

x→a− f (x) exist and either f (a) is undefined or

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M AT L AB : L IMITS

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M AT L AB : F UNCTIONS

⇒ ans = exp(2 ∗ x).

ans = log(x).

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M AT L AB : M ATHEMATICAL EXPRESSION

simplify(f ) ⇒ ans = 1.

ans = xˆ3 − x.

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M AT L AB : I NPUT - O UTPUT

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THANK YOU FOR YOUR ATTENTION

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