Continuous functions

CONTINUOUS FUNCTION

 

The limit of a function as x approaches a can often be found simply by calculating the value of the function at a. The functions with this property are called continuous at a.

 

Definition of continuity:

A function f is a continuous at a number a if

lim_x→α f(x) = f(a)

Note:

The above definition requires three things if f is a continuous at a.

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

(2) lim _x→α f(x) exists

(3) lim _x→α f(x) = f(a)

The definition says that f is continuous at a if f(x) approaches f(a) as x approaches a. Thus a continuous function f has the property that a small change in x produces only a small change in f(x).

In fact, the change in f(x) can be kept as small as we please by keeping the change in x sufficiently small.

Physical phenomena are usually continuous. For example, height of a person varies continuously with time, velocity of a vehicle varies continuously with time.

Geometrically the graph of the function can be drawn without removing your pen from the paper, so, we can say the function is continuous.

If f is defined near a (in other words, f is defined on an open interval containing a, except perhaps at a), we say that ƒ is discontinuous at a (or f is discontinuity at a) if f is not continuous at a.

Physically, discontinuous occurs in such situations electric current.

There are three type of discontinuity namely

1. Removable discontinuity.

2. Jump discontinuity.

3. Infinite discontinuity.

 

 

Left and Right Continuous

A function f is a continuous from the right at a point a if

lim_x→a+ f(x) = f(a)

A function f is a continuous from the left at a point a if

lim_x→a‒ f(x) = f(a)

Theorem

The following types of functions are continuous in their domain

(1) Polynomial function is continuous for all x ∈ R.

(2) Rational function is continuous for all x ∈ R.

(3) Root function is continuous for x≥ 0.

(4) Trigonometric functions sinx and cos x are continuous for all x ∈ R, where tan x is discontinuous at x == π/2.

(5) Inverse trigonometric functions for all x ∈ R.

(6) Exponential functions for all x ∈ R.

(7) Logarithmic functions continuous for x>0.

Theorem:

If ƒ and g are continuous at a point a, then

(1)f+g is continuous at a

(2) f g is continuous at a

(3) cf is continuous at a, where c is constant

(4) fg is continuous at a

(5) f/g is continuous at a if g(a) 0

 

Example 75. Discuss the continuity of the function f(x) = 

Solution: A function f is a continuous at a point a if

lim_x→af(x) = f(a)

The given function 1/x²

is defined for all real value of x and also ƒ(0) = 1

lim_x→0 f(x) = lim_x→0 1/x² = ∞

lim_x→a f(x) does not exists.

Hence f(x) is discontinuous at x = 0.

 

Example 76. Discuss the continuity of the function f(x) = 

Solution: A function f is a continuous at a point a if

lim_x→a f(x) = f(a)

The given function (x²‒x‒2)/(x‒2) is defined for all value of x and also f(2) = 1.

lim_x→2 f(x) = 3 ≠ 1 = f(2). Hence f(x) is discontinuous at x = 2.

 

Example 77. Discuss the continuity of the function f(x) = 

Solution: A function f is a continuous at a point a if

lim_x→a f(x) = f(a)

The given function (x²‒x‒2)/(x‒2) is defined for all value of x and also f(2) = 3.

lim_x→2 f(x) = 3 = f(2). Hence f(x) is continuous at x = 2.

 

Example 78. Use the definition of continuity and the properties of limits to show that the function is continuous at a given number a.

f(x) = (x + 2x³)⁴; a = −1

Solution: A function f is a continuous at a point a if

lim_x→α f(x) = f(a)

The given function

f(x) = (x + 2x³)⁴

lim_x→α f(x) = lim_x→‒1 f(x) = (lim_x→‒1 (x + 2x³)⁴)

= (‒1+2(‒1)³)⁴

= (−1 + 2(−1))⁴ = (−3)⁴

= 81 ... (1)

f(a) = f(‒1)

= (‒1+2(‒1)³)⁴

= 81 ... (2)

From (1) and (2),

lim_x→α f(x) = f(a)

f(x) is continuous at a = ‒1

 

Example 79. Use the definition of continuity and the properties of limits to show that the function is continuous at a given number a.

f(x) = 3x⁴ − 5x + ³√[x² + 4], a = 2

Solution: A function f is a continuous at a point a if

lim _x→α f(x) = f(a)

The given function

f(x) = 3x⁴ ‒ 5x + ³√[x² + 4]

lim_x→α f(x) = lim_x→2 f(x)

= 3(2)⁴ ‒ 5(2) + ³√(2²+4) = 3(16) ‒ 10 + ³√[4 + 4]

= 48‒10+³√8 = 38+2

= 40   ... (1)

f(a) = f(2)

= 3(2)⁴‒5(2) + ³√(2²+4) = 3(16) ‒ 10 + ³√(4 + 4)

= 48‒10+ ³√8 = 38+2

= 40 ... (2)

From (1) and (2),

lim_x→a f(x) = f(a)

f(x) is continuous at a = 2

 

Example 80. Show that f is continuous on (‒∞, ∞), f(x) = 

Solution: 1‒x² is a polynomial which is continuous for x≤ 1 and logx is continuous for x > 1

 f(x) is continuous in the domain (‒∞,1) U (1,∞).

At x = 1

To find f(1):

Since x = 1 lies in x ≤ 1

f(x) = 1‒x²

ƒ(1) = 1‒1² = 0

lim_x→1f(x) = f(1)

Hence f(x) is continuous at x = 1

f(x) is continuous on (‒∞, ∞).

 

Example 81. Let f(x) =  Evaluate each if the (x‒3)2 if x >3

following limits, if they exist

Also, find where f(x) is continuous.

Solution:

Also f is continuous at x = 3 since by (4) and (6)

 

Example 82. Show that ƒ is continuous on (‒∞, ∞), f(x) = .

Solution: sin x is continuous for x > π/4 and cos x is continuous for x≤ π/4.

ƒ(x) is continuous in the domain (‒∞,π/4)U (π/4,∞).

At x = π/4

Hence f(x) is continuous at x = π/4.

f(x) is continuous on (‒∞, ∞).

 

Example 83. For what values of constant c is the function f continuous on (‒∞,∞)

Solution: Given f is continuous in the domain (‒∞, ∞).

Since f(x) is continuous at x = 2, we have

lim _x→2‒ f(x) = lim_x→2 f(x) ... (1)

lim _x→2‒ f(x) = lim _x→2 (cx² + 2x)

c(2)² + 2(2) = 4c + 4

lim _x→2+ f(x) = lim _x→2 (x³‒ cx).

 = (2)³‒ c(2) = 8‒2c

Since f is continuous in the domain (‒∞, ∞),

lim _x→2‒ f(x) = lim _x→2+ f(x)

4c+4=8‒2c

4c+2c=8‒4

6c = 4

 c = 2/3

 

Example 84. For what values of constant a and b is f(x) =  is continuous at every x?

Solution: Given f is continuous in the domain (‒∞, ∞).

 (3) a ‒ b = 3... (4)

Adding (2)and (4), we get

2a = 5

 a=5/2

Substitute a = 5/2 in (2), we get

5/2 + b = 2

B = ‒ 1/2

 

Example 85. For what values of constant a, b is the function f continuous on (‒∞, ∞)

Solution: Given f is continuous in the domain (‒∞, ∞).

Since f(x) is continuous at x = 2, we have

lim f(x) = lim f(x) ... (1)

 (3) ⇒ 9a‒3b+3 = 6‒a+b

⇒ 10a‒4b = 3... (4)

Solving for (2) and (4), we ge

 a = ‒ 15/2 and b= ‒ 39/2

 

Example 86. Where is the function f(x) = Inx+tan^‒1x / x²‒1 continuous?

Solution: y = ln x is continuous for x > 0 i.e(0, ∞) and y = tan^‒1x is is continuous on R.

.. y = ln x + tan^‒1x is continuous on (0,∞). The denominator,

 y = x² ‒1 is a polynomial, so it is continuous for all x. However, the function is continuous where x²‒1 = 0 ↔ x = ±1. Since the domain of the numerator is (0,∞), take x = 1. Hence f(x) is not continuous at x = 1.

 f(x) = Inx+tan^‒1x / x²‒1 is continuous on the interval (0,1) U (1,∞).

 

Example 87. Show that function f(x) = 1 − √[1−x²] is the continuous in the interval [‒1,1].

Solution:

= ƒ(1)

f(x) is continuous at x = 1

f(x) is continuous on [‒1,1]

 

Example 88. Use the definition of continuity and the properties of limits to show that the functions f(x) = x+√[x‒4] is continuous on the interval [4,∞]

Solution: Let ‒4 < a < 1

lim_x→a f(x) = lim_x→a (x + √[x‒4]) = a + √[a‒4]

 = f(a)

f(x) is continuous on (4, ∞]

lim _x→4 f(x) = lim _x→4 (x + √[x‒4])

= 4+√[4‒4] = 4

f(4) = 4+√[4‒4] = 4

lim _x→4 f(x) = f(4)

Hence f(x) is continuous on x = 4

f(x) is continuous on [4, ∞]

 

Example 89. Evaluate lim _x→π [ sinx / 2+cosx ] using continuity

Solution: The function y sinx is continuous for all values of x. The function y = 2+ cos x is sum of two continuous functions and therefore continuous for all values of x. Notice that 2 + cos x is never zero because

cosx ≥ ‒1⇒ 2+ cos x ≥ 2 −1 = 1

2 + cosx ≥ 1

Thus the ratio

f(x) = sinx / 2+cosx

is continuous for all values of x. Hence by definition of continuous function

0 / 2‒1 = 0.

 

Example 90. Discuss the continuity of f(x) = tan x.

Solution: The function f(x) = tan x = sinx / cosx is continuous except when cosx = 0.

When x = ± [ (2n‒1)π /2 ],  n = 1,2,3,..., cos x = 0. Therefore, f(x) = tan x has infinite discontinuities when x = ±π/2, ± зπ/2, ± 5π/2, ... and the same is illustrated in the following figure.

 

 

EXERCISE

 

24. Use the definition of continuity and the properties of limits to show that the function is continuous on the given interval.

(a) f(x) = x² + √[7 –x], a = 4

(b) f(x) = 2x‒3x² / 1+x³ , a=1

(c) f(x) = 2x+3 / x‒2 , (2,∞)

(d) f(x) = 2√[3‒x] , (‒∞,3]

(e) f(x) =  x²+5x / 2x+1 , a = 2

(f) f(x) = 2√[3x² + 1], a = 1

 

25. Use continuity to evaluate the limit.

(a) lim_x→4 [ 5+√x / √5+x ]

(b) lim _x→1 e^x2-x

 

26. Where are each of the following functions discontinuous?

(d) f(x) = [x] Ans: discontinuities at all of the integers.