Continuity (Section 1.8)

Alex Karassev

Definition

A function f is continuous at a number a if

Thus, we can use direct substitution to compute the limit of function that is continuous at a

)()(lim afxfax

Some remarks

Definition of continuity requires three things:

f(a) is defined (i.e. a is in the domain of f)

exists

Limit is equal to the value of the function

The graph of a continuous functions does not have any "gaps" or "jumps"

)(lim xfax

Continuous functions and limits

TheoremSuppose that f is continuous at band Then

Example

bxgax

)(lim

))(lim())((lim xgfxgfaxax

2422)2(lim

2

)2)(2(lim

2

4lim

2

4lim

2

2

2

2

2

2

x

x

xx

x

x

x

x

x

xxx

Properties of continuous functions

Suppose f and g are both continuous at a Then f + g, f – g, fg are continuous at a If, in addition, g(a) ≠ 0 then f/g is also continuous

at a

Suppose that g is continuous at a and f is continuous at g(a). Then f(g(x)) is continuous at a.

Which functions are continuous?

Theorem

Polynomials, rational functions, root functions, power functions, trigonometric functions, exponential functions, logarithmic functions are continuous on their domains

All functions that can be obtained from the functions listed above using addition, subtraction, multiplication, division, and composition, are also continuous on their domains

Example

Determine, where is the following function continuous:

x

xxf2

1cos12)(

Solution

According to the previous theorem, we need to find domain of f

Conditions on x: x – 1 ≥ 0 and 2 – x >0 Therefore x ≥ 1 and 2 > x So 1 ≤ x < 2 Thus f is continuous on [1,2)

x

xxf2

1cos12)(

Intermediate Value Theorem

River and Road

River and Road

Definitions

A solution of equation is also calleda root of equation

A number c such that f(c)=0 is calleda root of function f

Intermediate Value Theorem (IVT)

f is continuous on [a,b] N is a number between f(a) and f(b)

i.e f(a) ≤ N ≤ f(b) or f(b) ≤ N ≤ f(a)

then there exists at least one c in [a,b] s.t. f(c) = N

x

y

a

y = f(x)

f(a)

f(b)

b

N

c

Intermediate Value Theorem (IVT)

f is continuous on [a,b] N is a number between f(a) and f(b)

i.e f(a) ≤ N ≤ f(b) or f(b) ≤ N ≤ f(a)

then there exists at least one c in [a,b] s.t. f(c) = N

x

y

a

y = f(x)

f(a)

f(b)

b

N

c1 c2c3

Equivalent statement of IVT

f is continuous on [a,b] N is a number between f(a) and f(b), i.e

f(a) ≤ N ≤ f(b) or f(b) ≤ N ≤ f(a) then f(a) – N ≤ N – N ≤ f(b) – N

or f(b) – N ≤ N – N ≤ f(a) – N so f(a) – N ≤ 0 ≤ f(b) – N

or f(b) – N ≤ 0 ≤ f(a) – N Instead of f(x) we can consider g(x) = f(x) – N so g(a) ≤ 0 ≤ g(b)

or g(b) ≤ 0 ≤ g(a) There exists at least one c in [a,b] such that g(c) = 0

Equivalent statement of IVT

f is continuous on [a,b] f(a) and f(b) have opposite signs

i.e f(a) ≤ 0 ≤ f(b) or f(b) ≤ 0 ≤ f(a)

then there exists at least one c in [a,b] s.t. f(c) = 0

x

y

a

y = f(x)

f(a)

f(b)

bN = 0

c

Continuity is important!

Let f(x) = 1/x Let a = -1 and b = 1 f(-1) = -1, f(1) = 1 However, there is no c

such that f(c) = 1/c =0

x

y

0-1

-1

1

1

Important remarks

IVT can be used to prove existence of a root of equation

It cannot be used to find exact value of the root!

Example 1

Prove that equation x = 3 – x5 has a solution (root)

Remarks Do not try to solve the equation! (it is impossible

to find exact solution) Use IVT to prove that solution exists

Steps to prove that x = 3 – x5 has a solution Write equation in the form f(x) = 0

x5 + x – 3 = 0 so f(x) = x5 + x – 3

Check that the condition of IVT is satisfied, i.e. that f(x) is continuous f(x) = x5 + x – 3 is a polynomial, so it is continuous on (-∞, ∞)

Find a and b such that f(a) and f(b) are of opposite signs, i.e. show that f(x) changes sign (hint: try some integers or some numbers at which it is easy to compute f) Try a=0: f(0) = 05 + 0 – 3 = -3 < 0 Now we need to find b such that f(b) >0 Try b=1: f(1) = 15 + 1 – 3 = -1 < 0 does not work Try b=2: f(2) = 25 + 2 – 3 =31 >0 works!

Use IVT to show that root exists in [a,b] So a = 0, b = 2, f(0) <0, f(2) >0 and therefore there exists c in [0,2]

such that f(c)=0, which means that the equation has a solution

x = 3 – x5 ⇔ x5 + x – 3 = 0

x

y

0

-3

31

2N = 0

c (root)

Example 2

Find approximate solution of the equationx = 3 – x5

Idea: method of bisections

Use the IVT to find an interval [a,b] that contains a root

Find the midpoint of an interval that contains root: midpoint = m = (a+b)/2

Compute the value of the function in the midpoint

If f(a) and f (m) are of opposite signs, switch to [a,m] (since it contains root by the IVT),otherwise switch to [m,b]

Repeat the procedure until the length of interval is sufficiently small

f(x) = x5 + x – 3 = 0

0 2

f(x)≈

x

-3 31

We already know that [0,2] contains root

Midpoint = (0+2)/2 = 1

-1< 0 > 0

f(x) = x5 + x – 3 = 0

0 2

f(x)≈

x

-3 31

1

-1

1.5

6.1

Midpoint = (1+2)/2 = 1.5

f(x) = x5 + x – 3 = 0

0 2

f(x)≈

x

-3 31

1

-1

1.5

6.1

Midpoint = (1+1.5)/2 = 1.25

1.25

1.3

1

-1

1.25

1.3

1.125

-.07

f(x) = x5 + x – 3 = 0

0 2

f(x)≈

x

-3 31

1.5

6.1

Midpoint = (1 + 1.25)/2 = 1.125

By the IVT, interval [1.125, 1.25] contains root

Length of the interval: 1.25 – 1.125 = 0.125 = 2 / 16 = = the length of the original interval / 24

24 appears since we divided 4 times

Both 1.25 and 1.125 are within 0.125 from the root!

Since f(1.125) ≈ -.07, choose c ≈ 1.125

Computer gives c ≈ 1.13299617282...

Exercise

Prove that the equation

sin x = 1 – x2

has at least two solutions

Hint:

Write the equation in the form f(x) = 0 and find three numbers x1, x2, x3,such that f(x1) and f(x2) have opposite signs AND f(x2) and f(x3) haveopposite signs. Then by the IVT the interval [ x1, x2 ] contains a root ANDthe interval [ x2, x3 ] contains a root.