`star` Algebra of continuous function

Q 3185012867

Discuss the continuity of the function f defined by `f (x) = 1/x , x ≠ 0` .

Class 12 Chapter 5 Example 9

Class 12 Chapter 5 Example 9

Fix any non zero real number c, we have

` lim_(x->c) f(x) = lim_(x->c) 1/x = 1/c`

Also, since for `c ≠ 0, f (c) = 1/c`, we have ` lim_(x->c) f(x) = f(c)` and hence, f is continuous

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

We take this opportunity to explain the concept of infinity. This we do by analysing

the function `f (x) =1/x` near x = 0. To carry out this analysis we follow the usual trick of

finding the value of the function at real numbers close to 0. Essentially we are trying to

find the right hand limit of f at 0. We tabulate this in the following (Table 5.1).

We observe that as x gets closer to 0 from the right, the value of f (x) shoots up

higher. This may be rephrased as: the value of f (x) may be made larger than any given

number by choosing a positive real number very close to 0. In symbols, we write

`lim_(x->0^+) f(x) = + oo`

(to be read as: the right hand limit of f (x) at 0 is plus infinity). We wish to emphasise

that + ∞ is NOT a real number and hence the right hand limit of f at 0 does not exist (as

a real number).

Similarly, the left hand limit of f at 0 may be found. The following table is self

explanatory.

From the Table 5.2, we deduce that the

value of f (x) may be made smaller than any

given number by choosing a negative real

number very close to 0. In symbols,

we write

`lim_(x-> 0^-) f(x) = - oo`

(to be read as: the left hand limit of f (x) at 0 is

minus infinity). Again, we wish to emphasise

that – ∞ is NOT a real number and hence the

left hand limit of f at 0 does not exist (as a real

number). The graph of the reciprocal function

given in Fig 5.3 is a geometric representation

of the above mentioned facts.

Q 3175112966

Discuss the continuity of the function f defined by

`f(x) = { tt ( ( x+2, text (if) x le 1 ), ( x-2, text (if) x > 1) )`

Class 12 Chapter 5 Example 10

`f(x) = { tt ( ( x+2, text (if) x le 1 ), ( x-2, text (if) x > 1) )`

Class 12 Chapter 5 Example 10

The function f is defined at all points of the real line.

Case 1 If c < 1, then f (c) = c + 2. Therefore, `lim_(x->c) f(x) = lim_(x->c) f(x+2) = c+2`

Thus, f is continuous at all real numbers less than 1.

Case 2 If c > 1, then f (c) = c – 2. Therefore,

`lim_(x->c) f(x) = lim_(x->c) (x-2) = c-2 = f(c)`

Thus, f is continuous at all points x > 1.

Case 3 If c = 1, then the left hand limit of f at

x = 1 is

`lim_(x->1^-) f(x) = lim_(x->1^-) (x+2) = 1+2 = 3`

The right hand limit of f at x = 1 is

`lim_(x->1^+) f(x) = lim_(x->1^+) (x-2) = 1-2 = -1`

Since the left and right hand limits of f at x = 1

do not coincide, f is not continuous at x = 1. Hence

x = 1 is the only point of discontinuity of f. The graph of the function is given in Fig 5.4.

Q 3125123061

Find all the points of discontinuity of the function f defined by

`f(x) = { tt ( ( x+2, text (if) x <1 ), ( 0, text (if) x=1 ), ( x-2, text (if) x> 1 ) )`

Class 12 Chapter 5 Example 11

`f(x) = { tt ( ( x+2, text (if) x <1 ), ( 0, text (if) x=1 ), ( x-2, text (if) x> 1 ) )`

Class 12 Chapter 5 Example 11

As in the previous example we find that f

is continuous at all real numbers x ≠ 1. The left

hand limit of f at x = 1 is

`lim_(x->1^-) f(x) = im_(x-> 1^-) (x+2) = 1+2 =3`

The right hand limit of f at x = 1 is

`lim_(x->1^+) f(x) = lim_(x->1^+) (x-2) = 1-2 = -1`

Since, the left and right hand limits of f at x = 1

do not coincide, f is not continuous at x = 1. Hence

x = 1 is the only point of discontinuity of f. The

graph of the function is given in the Fig 5.5.

Q 3175123066

Discuss the continuity of the function defined by

`f(x) = { tt ( ( x+2, text (if) x< 0 ), ( -x+2 , text(if) x > 0 ) )`

Class 12 Chapter 5 Example 12

`f(x) = { tt ( ( x+2, text (if) x< 0 ), ( -x+2 , text(if) x > 0 ) )`

Class 12 Chapter 5 Example 12

Observe that the function is defined at all real numbers except at 0. Domain

of definition of this function is

`D_1 ∪ D_2` where `D_1 = {x ∈ R : x < 0}` and

`D_2 = {x ∈ R : x > 0}`

Case 1 If `c ∈ D_1`, then `lim_(x->c) f(x) = lim_(x->c) (x+2)`

`= c + 2 = f (c) `and hence f is continuous in `D_1`

Case 2 If `c ∈ D_2`, then `lim_(x->c) f(x) = lim_(x->c) (-x+2) `

`= – c + 2 = f (c)` and hence f is continuous in `D_2` .

Since f is continuous at all points in the domain of f,

we deduce that f is continuous. Graph of this

function is given in the Fig 5.6. Note that to graph

this function we need to lift the pen from the plane

of the paper, but we need to do that only for those points where the function is not

defined.

Q 3115223160

Discuss the continuity of the function f given by

`f(x) = { tt ( ( x, text (if) x ge 0 ), ( x^2 , text (if) x < 0 ) )`

Class 12 Chapter 5 Example 13

`f(x) = { tt ( ( x, text (if) x ge 0 ), ( x^2 , text (if) x < 0 ) )`

Class 12 Chapter 5 Example 13

Clearly the function is defined at

every real number. Graph of the function is

given in Fig 5.7. By inspection, it seems prudent

to partition the domain of definition of f into

three disjoint subsets of the real line.

Let `D_1 = {x ∈ R : x < 0}, D_2 = {0}` and

`D_3 = {x ∈ R : x > 0}`

Case 1 At any point in `D_1`, we have `f (x) = x^2` and it is easy to see that it is continuous

there (see Example 2).

Case 2 At any point in `D_3`, we have f (x) = x and it is easy to see that it is continuous

there (see Example 6).

Case 3 Now we analyse the function at x = 0. The value of the function at 0 is f (0) = 0.

The left hand limit of f at 0 is

`lim_(x->0^-) f(x) = lim_(x->0^-) x^2 = 0^2 = 0`

The right hand limit of f at 0 is

`lim_(x->0^+) f(x) = lim_(x->0^+) x= 0`

Thus `lim_(x->0) f(x) = 0 = f(0) ` and hence f is continuous at 0. This means that f is

continuous at every point in its domain and hence, f is a continuous function.

Q 3145223163

Show that every polynomial function is continuous.

Class 12 Chapter 5 Example 14

Class 12 Chapter 5 Example 14

Recall that a function p is a polynomial function if it is defined by

`p(x) = a_0 + a_1 x + ... + a_n x^n` for some natural number n, an ≠ 0 and ai ∈ R. Clearly this

function is defined for every real number. For a fixed real number c, we have

`lim_(x->c) p (x) = p (c)`

By definition, p is continuous at c. Since c is any real number, p is continuous at

every real number and hence p is a continuous function.

Q 3165223165

Find all the points of discontinuity of the greatest integer function defined

by f (x) = [x], where [x] denotes the greatest integer less than or equal to x.

Class 12 Chapter 5 Example 15

by f (x) = [x], where [x] denotes the greatest integer less than or equal to x.

Class 12 Chapter 5 Example 15

First observe that f is defined for all real numbers. Graph of the function is

given in Fig 5.8. From the graph it looks like that f is discontinuous at every integral

point. Below we explore, if this is true.

Case 1 Let c be a real number which is not equal to any integer. It is evident from the

graph that for all real numbers close to c the value of the function is equal to [c]; i.e.,

`lim_(x->c) f(x) = lim_(x->c) [x] = [c]`.

Also `f (c) = [c]` and hence the function is continuous at all real

numbers not equal to integers.

Case 2 Let c be an integer. Then we can find a sufficiently small real number

r > 0 such that [c – r] = c – 1 whereas [c + r] = c.

This, in terms of limits mean that

`lim_(x->c^-) f(x) = c-1 , lim_(x->c^+) f(x) = c`

Since these limits cannot be equal to each other for any c, the function is

discontinuous at every integral point.

`color{green} ✍️` `text(Theorem 1 :)` Suppose f and g be two real functions continuous at a real number `c`.

(1) `color{red}{f + g}` is continuous at `x = c`.

(2) `color{red}{f - g}` is continuous at `x = c`.

(3) `color{red}{f * g}` is continuous at `x = c`.

(4) `color{red}(f/g)` is continuous at `x = c`, (provided `g (c) ≠ 0`).

`"Proof :"` We are investigating continuity of `(f + g)` at `x = c`. Clearly it is defined at `x = c`. We have

`=> lim_(x-> c) (f+g) (x) =lim_(x-> c) [f(x) +g(x)]` (by definition of `f + g`)

`=> lim_(x-> c) f(x) + lim_(x-> c) g(x)` (by the theorem on limits)

`=> f(c) +g(c)` (as `f` and `g` are continuous)

`=> (f + g) (c)` (by definition of `f + g`)

Hence, `f + g` is continuous at `x = c`. Proofs for the remaining parts are similar.

`color{blue}{"E.g., : Prove that the function defined by" f (x) = tan x "is a continuous function"}`

The function `f (x) = tan x= (sin x)/(cos x)`

● This is defined for all real numbers such that `cos x ≠ 0`, i.e., `x ≠ (2n+1) pi/2`

● We know that `s i n e` and `c o s i n e` functions are continuous. Thus ` t a n x` being a quotient of two continuous functions is

continuous wherever it is defined.

So , If `f(x) = (p(x))/(q(x))` where p and q are polynomial functions. The domain of f is all real numbers except points at which `q` is zero. So f is continuous except the points where `q(x) = 0`.

(1) `color{red}{f + g}` is continuous at `x = c`.

(2) `color{red}{f - g}` is continuous at `x = c`.

(3) `color{red}{f * g}` is continuous at `x = c`.

(4) `color{red}(f/g)` is continuous at `x = c`, (provided `g (c) ≠ 0`).

`"Proof :"` We are investigating continuity of `(f + g)` at `x = c`. Clearly it is defined at `x = c`. We have

`=> lim_(x-> c) (f+g) (x) =lim_(x-> c) [f(x) +g(x)]` (by definition of `f + g`)

`=> lim_(x-> c) f(x) + lim_(x-> c) g(x)` (by the theorem on limits)

`=> f(c) +g(c)` (as `f` and `g` are continuous)

`=> (f + g) (c)` (by definition of `f + g`)

Hence, `f + g` is continuous at `x = c`. Proofs for the remaining parts are similar.

`color{blue}{"E.g., : Prove that the function defined by" f (x) = tan x "is a continuous function"}`

The function `f (x) = tan x= (sin x)/(cos x)`

● This is defined for all real numbers such that `cos x ≠ 0`, i.e., `x ≠ (2n+1) pi/2`

● We know that `s i n e` and `c o s i n e` functions are continuous. Thus ` t a n x` being a quotient of two continuous functions is

continuous wherever it is defined.

So , If `f(x) = (p(x))/(q(x))` where p and q are polynomial functions. The domain of f is all real numbers except points at which `q` is zero. So f is continuous except the points where `q(x) = 0`.

Q 3185223167

Prove that every rational function is continuous.

Class 12 Chapter 5 Example 16

Class 12 Chapter 5 Example 16

Recall that every rational function f is given by

`f(x) = (p (x) )/( q(x)) , q(x) ≠ 0`

where p and q are polynomial functions. The domain of f is all real numbers except

points at which q is zero. Since polynomial functions are continuous (Example 14), f is

continuous by (4) of Theorem 1.

Q 3115223169

Discuss the continuity of sine function.

Class 12 Chapter 5 Example 17

Class 12 Chapter 5 Example 17

To see this we use the following facts

`lim_(x->0) sin x = 0`

We have not proved it, but is intuitively clear from the graph of sin x near 0.

Now, observe that f (x) = sin x is defined for every real number. Let c be a real

number. Put x = c + h. If x → c we know that h → 0. Therefore

`lim_(x->c) f(x) = lim_(x->c) sin x`

`= lim_(h->0) sin (c+ h )`

`= lim_(h->0) [ sin c cos h + cos c sin h ]`

`= lim_(h->0) [ sin c cos h ] + lim_(h->0) [ cos c sin h ]`

`= sin c + 0 = sin c = f(c)`

Thus `lim_(x->c) f(x) = f(c)` and hence f is a continuous function.

Q 3125323261

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

Class 12 Chapter 5 Example 18

Class 12 Chapter 5 Example 18

The function f (x) = tan x = `(sin x)/(cos x)` This is defined for all real numbers such

that cos x ≠ 0, i.e., x ≠ (2n +1) `pi/2` . We have just proved that both sine and cosine

functions are continuous. Thus tan x being a quotient of two continuous functions is

continuous wherever it is defined.

An interesting fact is the behaviour of continuous functions with respect to

composition of functions. Recall that if f and g are two real functions, then

`(f o g) (x) = f ( g(x) )`

is defined whenever the range of g is a subset of domain of f. The following theorem

(stated without proof) captures the continuity of composite functions.

●`color{red}{" If f and g are real valued functions such that" \ \ (f o g) \ \ "is defined at c."}`

`color{red}{" If g is continuous at c and if f is continuous at g (c) , then (f o g) is continuous at c"}`.

`color{red}{" If g is continuous at c and if f is continuous at g (c) , then (f o g) is continuous at c"}`.

Q 3145323263

Show that the function defined by `f (x) = sin (x^2)` is a continuous function.

Class 12 Chapter 5 Example 19

Class 12 Chapter 5 Example 19

Observe that the function is defined for every real number. The function

f may be thought of as a composition g o h of the two functions g and h, where

`g (x) = sin x` and `h (x) = x^2` . Since both g and h are continuous functions, by Theorem 2,

it can be deduced that f is a continuous function.

Q 3165323265

Show that the function f defined by

`f (x) = | 1 – x + | x | |`,

where x is any real number, is a continuous function.

Class 12 Chapter 5 Example 20

`f (x) = | 1 – x + | x | |`,

where x is any real number, is a continuous function.

Class 12 Chapter 5 Example 20

Define g by `g (x) = 1 – x + | x|` and h by `h (x) = | x|` for all real x. Then

`(h o g) (x) = h (g (x))`

`= h (1– x + | x |)`

`= | 1– x + | x | | = f (x)`

In Example 7, we have seen that h is a continuous function. Hence g being a sum

of a polynomial function and the modulus function is continuous. But then f being a

composite of two continuous functions is continuous