Limit continuity handout

THE LIMIT AND CONTINUITY OF A FUNCTION Electronic version of lecture THE LIMIT AND CONTINUITY OF A FUNCTION ELECTRONIC VERSION OF LECTURE Dr Lê Xuân Đại HoChiMinh City University of Technology Faculty[.]

THE LIMIT AND CONTINUITY OF A FUNCTION

Dr Lê Xuân Đại

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

Email: ytkadai@hcmut.edu.vn

HCM — 2016.

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

c 2 = L 0

s

1 − c 2

c 2 = 0

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) = Lmeans that the values off (x)can be made

C ALCULATING L IMITS USING THE LIMIT LAWS

THEOREM 1.1

Suppose that lim

x→a f (x) = A ∈ Randlim

THEOREM 1.2

The squeeze theorem: If

1 f (x) É g(x) É h(x)wherexis neara(except possibly

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

ifa < x < a + δthen |f (x) − L| < ε

THEOREM 1.3

lim

x→a f (x) = L if and only if

( lim

x→a+ f (x) = L

lim

x→a− f (x) = L

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

lim

x→3 (x 3 − 5x 2 + 7x − 10) = 3 3 − 5.3 2 + 7.3 − 10 = −7

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

lim f (x) = f (a); lim f (x) = f (b)

Suppose that g is continuous at a and f is continuous

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

lim (f ◦g)(x) = lim f (g(x)) = f ³ lim g(x) ´ = f (g(a)) = (f ◦g)(a).

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 bothf andg are continuous for all

DEFINITION 3.1

Letf (x) be a function defined on some open interval that contains the numbera, except possibly ataitself Then

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

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

arbitrarily small (smaller than any negative number N ) by

x→+∞ f (x) = Lmeans that the values off (x)can be

x→−∞ f (x) = Lmeans that the values off (x)can be

EXAMPLE 3.3

Evaluate lim

x→∞

3x 2 − x − 2 5x 2 + 4x + 1

SOLUTION Devide both the numerator and

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

The Limit Laws can not be applied to infinite limits

SOLUTION We divide the numerator and

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

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

lim

x→a+ f (x) = lim

x→a− f (x) 6= f (a). (16)

MATLAB: LIMITS

M AT L AB : F UNCTIONS

⇒ ans = exp(2 ∗ x).

ans = log(x).

M AT L AB : M ATHEMATICAL EXPRESSION

simplify(f ) ⇒ ans = 1.

ans = xˆ3 − x.

MATLAB: INPUT - OUTPUT

THANK YOU FOR YOUR ATTENTION