1. Concept of Continuity and Algebra of continuous functions

Chapter 5

Continuity and Differentiability

Concept of Continuity and Algebra of continuous functions:

Definition-1:

A function f(x) is said to be continuous at x=a if

limx→af(x)=f(a)

A function is said to be continuous on the interval [a,b]

if it is continuous at each point in the interval.

Definition-2:

A function f(x) is said to be continuous at a point x = a, in its domain if the following three conditions are satisfied:

1. f(a) exists (i.e. the value of f(a) is finite)
2. Lim_x→a f(x) exists (i.e. the right-hand limit = left-hand limit, and both are finite)
3. Lim_x→a f(x) = f(a)

The function f(x) is said to be continuous in the interval I = [x₁,x₂] if the three conditions mentioned above are satisfied for every point in the interval I.

If LHL=RHL ,then limit exists.

If LHL=RHL=f(a), then function is continuous at x = a.

Example 1 Given the graph of f(x), shown below, determine if f(x) is continuous at x=−2, x=0, and x=3.

First x=−2.

f(−2)=2   , limx→-2f(x) doesn't exist.

The function value and the limit aren’t the same and so the function is not continuous at this point. This kind of discontinuity in a graph is called a jump discontinuity. Jump discontinuities occur where the graph has a break in it as this graph does and the values of the function to either side of the break are finite (i.e. the function doesn’t go to infinity).

Now x=0

f(0)=1
limx→0f(x) =1

The function is continuous at this point since the function and limit have the same value.

Finally x=3.

f(3)=−1

limx→3f(x) =0

The function is not continuous at this point. This kind of discontinuity is called a removable discontinuity. Removable discontinuities are those where there is a hole in the graph as there is in this case.

Discontinuity Conditions:

The function “f” will be discontinuous at x = a in any of the following cases:

• f (a) is not defined.
• limx→a-f(x) And  limx→a+f(x)  exist but are not equal.
• limx→a-f(x) And  limx→a+f(x) exist and are equal but not equal to f (a).

Types of Discontinuity

The four different types of discontinuities are:

• Removable Discontinuity
• Jump Discontinuity
• Infinite Discontinuity

Removable Discontinuity

A function which has well- defined two-sided limits at x = a, but either f(a) is not defined or f(a) is not equal to its limits. limx→af(x)≠f(a)

Example:

This type of discontinuity can be easily eliminated by redefining the function .

Jump Discontinuity

It is a type of discontinuity, in which the left-hand limit and right-hand limit for a function x = a exists, but they are not equal to each other.

Infinite Discontinuity

The  function diverges at x =a to give a discontinuous nature. It means that the function f(a) is not defined. Since the value of the function at x = a does not approach any finite value or tends to infinity, the limit of a function x → a are also not defined.

Example : Discuss the continuity of the function f defined by

f (x) =  , x ¹0.

Solution Fix any non zero real number c, we have

=   = = f( c )

and hence, f is continuous

at every point in the domain of f. Thus f is a continuous function.

Definition -3:

The (ε, δ)-definition of continuity. We recall the definition of continuity: Let f : [a, b] → R and x0 ∈ [a, b]. f is continuous at x0 if for every ε > 0 there exists δ > 0 such that |x − x0| < δ implies |f(x) − f(x0)| < ε. We sometimes indicate that the δ may depend on ε by writing δ(ε).

Example:

The function defined by f (x) = √x is continuous.

Proof:
Given ε > 0 we must show that |√x - √p| < ε provided that x, p are close enough.
Now |√x - √p| = |x - p|/|√x + √p| < |x - p| /√p and so choosing δ = ε/√p will do.

Definition

If f and g are functions from R to R, we define the function f + g by (f + g)(x) = f (x) + g(x) for all x in R.
Similarly we may define the difference, product and quotient of functions.

Theorem

If f and g are continuous a point p of R, then so are f + g, f - g, f.g and (provided g(p) ≠ 0) ^f /g .

Proof

This follows directly from the corresponding arithmetic properties of sequences.
For example: to prove that f + g is continuous at p ∈ R
Suppose (x_n)→ p. We are told that (f (x_n))→ f (p) and (g(x_n))→ g(p) and we must prove that (f + g)(x_n))→ (f + g)(p).
But the LHS of this expression is f (x_n) + g(x_n) and the RHS is f (p) + g(p) and so the result follows from the arithmetic properties of sequences

Theorem

The composite of continuous functions is continuous.

Proof

Suppose f: R→ R and g: R→ R. Then the composition g  f is defined by g  f (x) = g(f (x)).
We assume that f is continuous at p and that g is continuous at f (p). So suppose that (x_i)→ p. Then (f (x_i))→ f (p) and then (g(f (x_i)))→ g(f (p)) which is what we need.

Examples

1. Clearly the identity function which x ↦ x is continuous.
   Hence, using the above, any polynomial function is continuous and hence any rational function (a ratio of polynomial functions) is continuous at any point where the denominator is non-zero.
2. We will prove later that functions like √x, sinx, cosx, exp(x), logx, ... are continuous. It follows that , for example sin²(x + 5), exp(-x²), √(1 + x⁴), ... are continuous since they are made by composing continuous functions.

Hence, the function f(x) is continuous at x =0.

Algebra of continuous functions:

Theorem : Suppose f and g be two real functions continuous at a real number c.

Then

(1) f + g is continuous at x = c.

(2) f – g is continuous at x = c.

(3) f . g is continuous at x = c.

(4) f/g is continuous at x = c, (provided g(c) ¹0).

(5) the algebraic operations between two functions are also continuous.

(6) f(g(x)) and g(f(x)) are continuous at x = a (Composite Function is continuous)

(7) Trigonometric  Function is Continuous

(8) Exponential  Function is Continuous

(9) Logarithm function is continuous .

(10) Polynomial Function is continuous.

(11) Modulus  Function is Continuous.

(12) All rational functions are continuous.

Proof: (1) Given,

lim_x→a f(x) = f(a)

lim_{x→ a} g(x) = g(a)

Now as per the theorem,

lim_{x → a} (f+g)(x) ⇒ lim _{x → c} [f(x) + g(x)]

⇒ lim_{x → c} f(x) + lim_{x → c} g(x)

⇒ f(a) + g(a)

⇒ (f + g)(a)

Therefore,

lim_{x → a} (f+g)(x) = (f + g)(c)

Hence, f+g is continuous at x = a.

(3)

Proof: Given,

lim_x→a f(x) = f(a)

lim_{x→ a} g(x) = g(a)

So, the limit of product of two functions, f and g at x is given by:

lim_{x → a} (f . g)(x) ⇒ lim _{x → c} [f(x) . g(x)]

⇒ lim_{x → c} f(x) . lim_{x → c} g(x)

⇒ f(a) . g(a)

⇒ (f . g)(a)

Therefore,

lim_{x → a} (f . g)(x) = (f . g)(c)

Hence, f . g is continuous at x = a.

Question :Find the values of k so that the function f is continuous at the indicated point.

ANSWER:

The given function f is

The given function f is continuous at x = 5, if f is defined at x = 5 and if the value of f at x = 5 equals the limit of f at x = 5

It is evident that f is defined at x = 5 and f(5)=kx+1=5k+1

Therefore, the required value of 9/5.

Problem:

Prove that the function defined by f(x) = | cos x | is a continuous function.

Solution: Given, f(x) = |cos x|

f(x) is the real function for all real numbers ‘x’ and the domain of f(x) is the real number

Let g(x) = cos x and h(x) = |x|

g(x) and h(x) are cosine functions and modulus functions are continuous for all real numbers.

Now,

(goh) (x) = g (h(x)) = g(|x|) = cos |x| is a composite function, hence is a continuous function. But it is not equal to f(x).

Again,

(hog) (x) = h(g(x)) = g(cos x) = |cos x| = f(x) [Given]

Hence,

f(x) = |cos x| = hog (x) is a composition function of two continuous functions. Therefore, it is also a continuous function.

Continuity in open interval (a, b)

f(x) will be continuous in the open interval (a,b) if at any point in the given interval the function is continuous.

Continuity in closed interval [a, b]

A function f(x) is said to be continuous in the closed interval [a,b] if it satisfies the following three conditions.

1) f(x) is be continuous in the open interval (a, b)

2) f(x) is continuous at the point a from right i.e. 

3) f(x) is continuous at the point b from left i.e.