Limit, Continuity and Differentiability › UoN › Eng3A › LimitContunityDiff.pdfLimit, Continuity and Differentiability 4.1 INTRODUCTION The basic concepts of the theory of calculus

CHAPTER 4*

Limit, Continuity and Differentiability

4.1 INTRODUCTION

The basic concepts of the theory of calculus of real variables are limit, continuity and differentiabilityof a function of real variables. Here we give an intuitive idea of limit and then the analytical definitionof it. The rest of this chapter deals with the continuity and differentiability of a real valued function.

4.2 LIMIT

Meaning of x � a (x tends to a): Let ‘x’ be a real variable and ‘a’ be a fixed real number. Supposethat ‘x’ assumes successive values a + 0.1, a + 0.01, a + 0.001, a + 0.0001, ...(Fig. 4.1)

Fig. 4.1 (x � a +)

Obviously, as x passes through these successive values, the numerical difference between x andthe real number a, i.e., | x – a | becomes less and less gradually and becomes so small that we can write| x – a | < ��for every given � > 0. We express this situation symbolically as x � a + to mean “x tends(or approaches) to a from R.H.S.” Thus, x � a +, whenever of the successive values of x ultimatelysatisfy a < x < a + �, � being any positive real number, no matter however small.

Again, suppose the real variable x assumes successive values a – 0.1, a – 0.01, a – 0.001,a – 0.0001, … (Fig. 4.2). As x passes through these successive values, the numerical difference between

Fig. 4.2 (x � a –)

x and the real number a, i.e., | x – a | becomes less and less gradually and becomes so small that we canwrite | x – a | < � for every pre-assigned �� > 0. We express this situation symbolically as x � a – andread as “ x tends (or approaches) to a from L.H.S.” Thus, x � a –, whenever, the successive values ofx ultimately satisfy a – � < x < a, � being any positive real number, no matter however small.

* This chapter is not included in the WBUT syllabus but is essential for the study.

166 A TEXTBOOK OF ENGINEERING MATHEMATICS [VOL. I]

By the expression “x tends (or approaches) to a” or symbolically by x � a, we mean, given any� > 0, no matter however small, the successive values of x ultimately satisfy the inequality 0 < | x – a |< �. Note that x � a implies x � a.

Meanings of x � � and x � – �: If a real variable x, assuming positive values only, increaseswithout limit, i.e., it assumes values in such a way that the successive values ultimately become andremain greater than any pre-assigned arbitrary positive real number M, no matter however large, thenwe say that x tends (or approaches) to infinity. This is denoted symbolically as x � + �, or x � �.

If a real variable x, assuming negative values only, increases numerically without limit, i.e., itassumes values in such a way that the successive values become and remain less than – M, where M isany pre-assigned arbitrary positive real number, no matter however large, then we say that x tends (orapproaches) to minus infinity. This is denoted symbolically as x � – �.

It is noted that in reality, there exists no number such as + � or – � towards which x approaches.These are only symbols to mean that the value of x increases (for + �) or decreases (for –�) withoutlimit.

4.3 NEIGHBOURHOOD OF A POINT ON THE REAL LINE

Let a be a point on the real line (i.e., a line on which points are denoted by real numbers and vice versain an orderly manner) and � > 0 be a real number. The �-neighbourhood of a point a, denoted by N�(a)or simply by N (a), is defined as the interval:

,a x a− δ < < + δ or | |x a− < δ .

The deleted �-neighbourhood of the point a is obtained by deleting the point a from N�(a). It isdenoted by N�(a) – {a}, and is defined by

0 | | , . ., ,x a i e a x a x a< − < δ − δ < < + δ ≠ .

4.4 LIMIT OF A FUNCTION

Definition: A real valued function ƒ(x) of a real variable x is said to have a limit l if for any pre-assigned arbitrary positive number �, no matter however small, there corresponds a positive number �such that | ( ) | , . ., ( )f x l i e l f x l− < ε − ε < < + ε , whenever, 0 | | , ,x a x a< − < δ ≠ . ., ,i e a x a− δ < < + δ

.x a≠ This situation is denoted by writing

→lim ( )x a

f x = l, or ( ) asf x l x a→ →

The meaning is that for every neighbourhood (l – �, l + �) of l, there exists a deleted neighbourhood(a – �, a + �) of a such that ƒ(x) is in (l – �, l + �) for every x in (a – �, a + �) excluding x = a. Thisdefinition does not require the behaviour of ƒ(x) at x = a.

Geometrically this means that for every x in the two open intervals

a x a− δ < < and ,a x a< < + δ

the graph of f (x) lies between the horizontal lines

y = l − ε and y = l + ε

LIMIT, CONTINUITY AND DIFFERENTIABILITY 167

Fig. 4.3

Notes:(i) Here the graph of y = ƒ(x) has been assumed to be without any break in the interval under

consideration. For the determination of ),(lim xfax→

we are not at all concerned with the point

on the graph corresponding to x = a. The point on the graph corresponding to x = a may notbelong to the curve y = ƒ(x) or even may not exist at all.

(ii) A limit, if exists, is necessarily finite and unique.(iii) To prove the existence of limit of ƒ(x) it will be sufficient if we can show that the inequality

0 | |x a< − < δ follows from the inequality | ( ) | ;f x l− < ε � is given and � can be found. Ifwe can find � > 0 for given � > 0 which satisfy the previous two inequalities, we write

lim ( )x a

f x→

= l.

(iv) The existence of )(lim xfax→

depends completely on the values of ƒ(x) for x near a (not for x

at a).

A function ƒ(x) is said to have a limit l1 as x � a from the left, if for given � > 0, no matterhowever small, we can find a � > 0 such that

1| ( ) |f x l− < ε for a x a− δ < < .

This is denoted by

lim ( )x a

L f x→

= → −1 or lim ( )

x al f x = l1 or )0( −af = 1l

and say that l1 is the left-hand limit of ƒ(x).

Similarly, if for given � > 0, no matter however small, we can find a � > 0 such that

2| ( ) |f x l− <∈ for a x a< < + δ

then we say that ƒ(x) has the limit l2 as x � a from the right.

This is symbolically expressed as

limit ( )x a

R f x→ 2 2 2or lim ( ) or ( 0)

x al f x l f a l

→ += = + =

where l2 is called the right-hand limit of ƒ(x).


From the definition of limit, it follows that lxfax

=→

)(limit if and only if

lim ( )

x af x

→ −1 2 lim ( )

x al l l f x

→ += = = =

ILLUSTRATIVE EXAMPLES

Example 1: Using definition show that 63lim2

=→

xx

.

Solution: Given � > 0, we have to find out � > 0, such that

| 3 6 |x − < ε for 0 | 2 |x< − < δ

i.e., | 2 |3

xε− < for 0 | 2 |x< − < δ

Therefore, if we choose ,3

εδ = our definition is satisfied, since

| 3 6 |x − < ε for 0 | 2 |3

xε< − <

Hence the result.

Example 2: Using definition show that 2

2

4lim 4

2x

x

x→

− =−

.

Solution: Given � > 0, we have to find out � > 0 such that

2 44

2

x

x

− − < ε− for 0 | 2 |x< − < δ

i.e.,( 2) ( 2)

42

x x

x

− + − < ε− for 0 | 2 |x< − < δ

Since we are considering the limit when x � 2, then x – 2 � 0 and hence we may cancel the factor(x – 2) from both numerator and denominator. Thus

| 2 4 |x + − < ε for 0 | 2 |x< − < δ

i.e., | 2 |x − < ε for 0 | 2 |x< − < δ

Therefore, if we choose � = �, our definition is fulfilled, since2 4

42

x

x

− − < ε− for 0 | 2 |x< − < ε

Hence the result.


1lim sin 0x

xx→

= .

Solution: Given � > 0, we have to find out a � > 0, such that

1sin 0x

x− < ε for 0 | 0 |x< − < δ


or1

| | sinxx

< ε for 0 | 0 |x< − < δ

or | |x < ε for 0 | 0 |x< − < δ .

Therefore, if we choose � = �, our definition is satisfied, since

| |x < ε1 1

| | sin (Since sin 1, where 0, but 0)⇒ < ε ≤ → ≠x x xx x

i.e.,1

sin 0xx− < ε for 0 | 0 |x< − < δ

Hence the result.

Example 4: Show that 0

| |limx

x

x→ does not exist.

Solution: Let

)(xf| |

, 0x

xx

= ≠

� )(lim0

xfx −→ 0 0

lim ( 0) lim ( 1) 1x x

xx

x→ − → −

−= < = − = −�

and )(lim0

xfx +→ 0 0

lim ( 0) lim (1) 1x x

xx

x→ + → += > = =�

�0

| |lim

x

x

x→ − 0

| |lim

x

x

x→ +≠

Therefore, 0

| |limx

x

x→ does not exist.

Example 5: Does ),(lim2

xfx→

where ƒ(x) = 2x + 1, x > 2 and ( ) 3 1, 2 exist?f x x x= + ≤Solution: we have,

)(lim2

xfx −→ 2

lim (3 1) 7 ( 2)x

x x→ −

= + = <�

and )(lim2

xfx +→ 2

lim (2 1) 5 ( 2)x

x x→ +

= + = >�

� )(lim2

xfx −→

)(lim2

xfx +→

≠

Hence, )(lim2

xfx→

does not exist.

4.5 DIFFERENT TYPES OF LIMITS

Definition 1: A function ƒ(x) is said to tend to � as x � a, if for any pre-assigned positive number N,however large it may be, there exists a � > 0, such that

Nxf >)( δ<−< ||0for ax



)(lim xfax→

∞=

Definition 2: A function ƒ(x) is said to tend to – � as x � a, if for any pre-assigned positive numberN, however large it may be, there exists a � > 0, such that

Nxf >− )( δ<−< ||0for ax .

This is symbolically written as

lim ( )x a

f x→

∞−=


1lim .x x→

= ∞

Solution: Given N > 0, no matter however large, we have to find out a � > 0, such that

2

1N

x> δ<−< |0|0for x

or2 1

xN

< δ<< ||0for x

Therefore, if we choose 1

,N

δ = our definition is satisfied, since

2

1N

x>

1for 0 | 0 |x

N< − <

Hence the result.

Definition 3: A function ƒ(x) is said to have a limit l as x tends to �, if for any pre-assigned positivenumber �, no matter however small, there exists a positive number m, however large it may be, suchthat

| ( ) |f x l− < ε mx >for

This is denoted by lxfx

=∞→

)(lim .

Definition 4: A function ƒ(x) is said to have a limit l as x tends to –�, if for any pre-assigned positivenumber �, however small it may be, there exists a positive number m, however large, such that

| ( ) |f x l− < ε mx >−for

This is expressed by writing

)(lim xfx ∞−→

= l.


1lim 0

x x→ ∞= .

Solution: Given � > 0, no matter however small, we have to find out a m > 0, such that

2

10

x− < ε mx >for


or 2

1

| |x< ε mx >for

or2 1

| |x >ε

mx >for

Therefore, if we choose 1

,m =ε

our definition is satisfied, since

2

10

x− < ε 1

for x >ε

Hence the result.

Definition 5: A function ƒ(x) is said to tend to � as x tends to �, if, for any given positive number Nthere exists a positive number m such that ƒ(x) > N for x > m


)(lim xfx ∞→

∞=

Example 3: Show that ∞=∞→

2lim xx

.

Solution: Given N > 0 we have to find out a m > 0 such that

Nx >2 mx >for

or Nx >|| mx >for

Therefore, if we choose ,Nm = our definition is satisfied, since

Nx >2 Nx >for

Hence the result.

4.6 SOME STANDARD LIMITS

(i)0

sinlim 1x

x

x→= (ii)

1lim 1

x

xe

x→ ∞

+ =

(iii) ex x

x=+

→

1

0)1(lim (iv)

0

1lim log (1 ) 1ex

xx→

+ =

(v)0

1lim 1

x

x

e

x→

− = (vi)0

1lim log , 0

x

ex

aa a

x→

− = >

(vii) 1limn n

n

x a

x an a

x a−

→

− =−

(viii)0

(1 ) 1lim

n

x

xn

x→

+ − =


4.7 FUNDAMENTAL THEOREMS

Let ƒ(x) and g(x) be two real single valued functions of real variable x and ,)(lim,)(lim mxglxfaxax

==→→

where l and m are finite.

Then (i) lim ( ) lim ( ) , where is a constantx a x a

cf x c f x cl c→ →

= =

(ii) 1 2 1 2lim{ ( ) ( )} ,x a

c f x c g x c l c m→

+ = +

where c1, c2 are constants.

(iii) lim{ ( ) . ( )}x a

f x g x l m→

= ⋅

(iv)( )

lim , provided 0( )x a

f x lm

g x m→

= ≠

(v) If ( ) ( ) ( )x f x xϕ ≤ ≤ ψ in (a – h, a + h), h > 0 and ,)(lim)(lim lxxaxax

=ψ=ϕ→→

then

lxfax

=→

)(lim

(vi) If )()( xgxf ≤ in 0),,( >+− hhaha and ,)(lim,)(lim mxglxfaxax

==→→

then l � m

(vii) If )()(limand)(lim bfyfbxbyax

==ϕ→→

then lim { ( )} lim ( ) ( )x a x a

f x f x f b→ →

ϕ = ϕ =

4.8 CONTINUITY OF A FUNCTION

Continuity at a Point

A real valued function ƒ(x) of a real variable x is said to be continuous at x = a provided thefollowing three requirements are fulfilled:

(i) ƒ(a) exists, i.e., the function has a value at x = a

(ii) )(lim xfax→

exists and

(iii) )()(lim afxfax

=→

More precisely, we may come to the following analytical definition.

Definition: A function ƒ(x) defined in a neighbourhood of a including a itself is said to becontinuous at x = a, if given � > 0, no matter however small, there exists a � > 0, such that

| ( ) ( ) |f x f a− < ε δ<− ||for ax

i.e., ( ) ( ) ( )f a f x f a− ε < < + ε for a x a− δ < < + δ

Note: A point where ƒ(x) is not continuous is called a point of discontinuity.


Continuity from the Left

Definition: A function ƒ(x) is said to be continuous from the left at a point x = a if

),()(lim–

afxfax

=→

i.e., given � > 0, there exists a � > 0 such that

| ( ) ( ) |f x f a− < ε axa ≤<δ−for

Continuity from the Right

Definition: A function ƒ(x) is said to be continuous from the right at a point x = a if

),()(lim afxfax

=+→

i.e., given � > 0, there exists a � > 0 such that

| ( ) ( ) |f x f a− < ε δ+<≤ axafor

Note: It is obvious that if a function ƒ(x) is continuous both from the left and from the right at apoint x = a, then it is continuous at x = a.

Continuity in an Interval

A function ƒ(x) is said to be continuous in an interval if it is continuous at every point of theinterval.

Note: Geometrically, the function ƒ(x) is said to be continuous at the point x = a if there is nobreak in the graph of the function at the point whose abscissa is a and in an arbitrarily smallneighbourhood of a.


Example 1: A function ƒ(x) is defined as follow:

)(xf 10,2 <<= xx

21, <≤= xx

21, 2 3

4x x= ≤ <

Show that it is continuous at x = 1 and discontinuous at x = 2.Solution:

(i) Here, 1 1

(1 0) lim ( ) lim 1 (as 1 1 2)x x

f f x x x x→ + → +

+ = = = → + ⇒ < <

and2

1 1(1 0) lim ( ) lim 1 (as 1 0 1)

x xf f x x x x

→ − → −− = = = → − ⇒ < <

Also ƒ(1) = 1

� )(.,.),01()1()01( xfeifff +==− is continuous at x = 1.

(ii) Now, 2 2

(2 0) lim ( ) lim 2 (as 2 – 1 2)x x

f f x x x x→ − → −

− = = = → ⇒ < <

and 2

2 2

1(2 0) lim ( ) lim 1 (as 2 2 3)

4x xf f x x x x

→ + → ++ = = = → + ⇒ < <


� ).02()02( +≠− ff

So ƒ(x) is discontinuous at x = 2.

Example 2: Show that the function ƒ(x) = x – [x], where [x] denotes the greatest integer notgreater than x, is discontinuous at x = 0.

Solution: Here

)(xf 01when1)1( <≤−+=−−= xxx

10when0 <≤=−= xxx

�)00( −f

0 0lim ( ) lim ( 1) 1 ( 1 0)

x xf x x x

→ − → −= = + = − < <�

and )00( +f 0lim)(lim00

===+→+→

xxfxx

� )00( −f )00( +≠ f

Therefore ƒ(x) is discontinuous at x = 0

Note: This function is discontinuous at all integral values of x.

Example 3: Show that the function

)(xf1

cos , when 0= ≠x xx

0when,0 == x

is continuous at x = 0.

Solution: Here

|)0()(| fxf − 1cos 0x

x= −

1 1| | cos | | cos 1 for all non zero real x x x

x x

= ≤ ≤ �

Therefore, given � > 0, there exists � = � > 0, such that | ( ) (0) | whenever | 0 |f x f x− < ε − < δ

Hence ƒ(x) is continuous at x = 0.


)(xf 2 1sin , when 0x x

x= ≠

0when,0 == x

is continuous at x = 0.

Solution: Here

|)0()(| fxf − = 2 1sin 0x

x−

= 2 21 1| | sin | | sin 1 for all real x x x

x x

≤ ∴ ≤


Therefore, given � > 0, there exists a 0,δ = ε > such that

| ( ) (0) | whenever | 0 |f x f x− < ε − < δ .

Therefore )(xf is continuous at x = 0.


)(xf| |

, when 0x

xx

= ≠

0when,1 == x

is right-continuous at x = 0 but not continuous there.Solution: Since

)(lim0

xfx +→ 0 0

| |lim lim ( 0)

x x

x xx

x x→ + → += = >�

0lim 1 1 (0) ( 0),

xf x

→ += = = ≠�

ƒ(x) is right-continuous at x = 0, but it is not continuous at x = 0, since

)(lim–0

xfx→ 0 – 0 – 0 – 0

| |lim lim ( 0) lim ( 1) 1 lim ( ) 1

x x x x

x xx f x

x x→ → → → +

−= = < = − = − ≠ =� .

4.9 THEOREM OF CONTINUITY

Let ƒ(x) and g(x) be both continuous at x = a, then

(i) axxgxfxgxfxgxf at continuous are)()(and)()(),()( =−+

(ii)( )

( )

f x

g x is continuous at x = a provided g (a) � 0

(iii) |ƒ(x)| and |g(x)| are continuous at x = a(iv) Any constant function is continuous at any point.

(v) Let ,0,)( 011

10 ≠++…++= −− aaxaxaxaxf nn

nn be a polynomial in x of degree n, thenƒ(x) is continuous for all real values of x.

(vi) If g(x) is continuous at ƒ(a) then g{ƒ(x)} is continuous at x = a

4.10 DERIVABILITY OF A FUNCTION

Definition: Let y = ƒ(x) be a real single valued function defined in the closed interval [a,b] and c be a

point in (a, b), i.e., a < c < b. If 0

( ) ( )limh

f c h f c

h→

+ − exists, and finite, then the limit is called the

derivative of ƒ(x) at x = c denoted by ƒ' (c) or dx

dy at x = c and we say that ƒ is derivable or differentiable

at x = c.


Right-hand and Left-hand Derivatives

If 0

( ) ( )lim

h

f c h f c

h→ +

+ − exists, is called the right-hand derivative of ƒ(x) at x = c denoted by

( ) or ( 0).Rf c f c′ ′ + Also if 0 –

( ) ( )lim

h

f c h f c

h→

+ − exists, is called the left-hand derivative of ƒ(x) at

x = c denoted by ( ) or ( 0).Lf c f c′ ′ −

Notes: (i) ( )f c′ exists if and only if ( )Lf c′ and ( )Rf c′ both exist and are equal.

(ii) If any one fails to exist or if both exist but are unequal, then ( )f c′ does not exist.

Derivability in an Interval

Let ƒ(x) be a function defined in [a, b]. Then ƒ(x) is said to be derivable in [a, b] if

(i) ( )f c′ exists for all c such that a < c < b

(ii) )0( +′ af exists

(iii) )0( −′ bf exists

Theorem: If a function ƒ(x) has a finite derivative at x = c, then it is continuous at x = c.

Proof: Let ƒ(x) has finite derivative ( )f c′ at x = c.

� ( ) ( )f c h f c+ − = ( ) ( )+ −

f c h f ch

h

�0

lim{ ( ) ( )}h

f c h f c→

+ − =0

( ) ( )lim→

+ − h

f c h f ch

h

= 0 0

( ) ( )lim limh h

f c h f ch

h→ →

+ −

(since both the limits exist)

= ( ) 0f c′ × = 0

Thus for any � > 0, there exists a � > 0, such that

( ) ( ) whenever 0f c h f c h+ − < ε − < δ

or ( ) ( ) wheneverf x f c x c− < ε − < δ ( putting x = c + h )

This proves that ƒ(x) is continuous at x = c.

Remark: The converse of the above theorem is not true in general, that is, a function may becontinuous at a point, but may not be derivable there. It is illustrated in the following example.

Example: Show that the function ƒ(x) = | x |, x is real, is continuous but not derivable at x = 0.Solution: Here

)(xf =

<−≥

0if

0if

xx

xx


� )(lim0

xfx +→

= xx +→ 0lim = 0 ( 0)x >�

and )(lim0

xfx −→

= )(lim0

xx

−−→

= 0 (� x < 0)

Also, ƒ(0) = 0

Thus, )(lim0

xfx −→

= )(lim0

xfx +→

= )0(f

Hence ƒ(x) is continuous at x = 0

Now, )00( −′f = 0

( ) (0)lim

h

f h f

h→ −

−=

0

0lim

h

h

h→ −

− − )0( <h�

= 0

lim ( 1)h→ −

− = – 1 ( 0)h ≠�

Also, )00( +′f = 0

( ) (0)lim

h

f h f

h→ +

− =

0

0lim ( 0)

h

hh

h→ +

− >�

= 0

lim 1h→ +

= 1 ( 0)h ≠�

� )00()00( +′≠−′ ff ,

therefore, )0(f ′ does not exist though ƒ(x) is continuous at x = 0. Hence the result.


Example 1: Letƒ(x) = x, 0 < x < 1

= 2 – x, 1 � x � 2

= x – x2, x > 2.

Show that ƒ(x) is discontinuous at x = 2 and also verify that )2(f ′ does not exist.Solution: Here

)(lim–2

xfx→

= 2 –

lim (2 )x

x→

− = 0 ( 1 2)x< <�

and )(lim2

xfx +→

= 2

2lim ( )

xx x

→ +− = 2− ( 2)x >�

Hence )(lim)(lim22

xfxfxx +→−→

≠ and consequently ƒ(x) is discontinuous at x = 2

Now, (2)Lf ′ = 0

(2 ) (2)lim

h

f h f

h→ −

+ −=

0

2 (2 ) 0lim

h

h

h→ −

− + − ( 0)h <�

= 0

limh

h

h→ −

−=

0lim ( 1)

h→ −− = 1− ( 0)h ≠�

and (2)Rf ′ = 0

(2 ) (2)lim

h

f h f

h→ +

+ − =

2

0

(2 ) (2 ) 0lim

h

h h

h→ +

+ − + − ( 0h >� )


= 2

0

( 3 2)lim

h

h h

h→ +

− + +=

0

2lim 3

hh

h→ +

− + +

This limit does not exist (since it tends to – �). Hence )2(f ′ does not exist.

Example 2: Let ƒ(x) be the function defined by ƒ(x) = 2 | x | + | x – 2 |, x is real. Show that ƒ(x)is not derivable at x = 0, 2 and is derivable at every other points.

Solution: Hereƒ(x) = – 3x + 2, when x < 0

= x + 2, when 0 � x � 2

= 3x – 2, when x > 2

Now, (0)Lf ′ = 0

(0 ) (0)lim

h

f h f

h→ −

+ −=

0

3 2 2lim

h

h

h→ −

− + −( 0)h <�

= 0

lim 3h→ −

− = – 3 ( 0)h ≠�

and (0)Rf ′ = 0

(0 ) (0)lim

h

f h f

h→ +

+ −=

0

2 2lim

h

h

h→ +

+ −( 0 2)h< <�

= 0

lim 1h→ +

= 1 ( 0)h ≠�

� (0) (0)Lf Rf′ ′≠

Hence ƒ'(0) does not exist, i.e., ƒ(x) is not derivable at x = 0.

Again, (2)Lf ′ = 0

(2 ) (2)lim

h

f h f

h→ −

+ −=

0

2 2 4lim

h

h

h→ −

+ + −( 0 2 2)h< + <�

= 0

lim 1h→ −

= 1 ( 0)h ≠�

and )2(fR ′ = 0

(2 ) (2)lim

h

f h f

h→ +

+ −=

0

3 (2 ) 2 4lim

h

h

h→ +

+ − −

= 0

lim 3h→ +

= 3 ( 0)h ≠�

� (2) (2)Lf Rf′ ′≠

Therefore ƒ'(2) does not exist, i.e., ƒ(x) is not derivable at x = 2

It is obvious that ƒ(x) is derivable at every point other than x = 0, 2

Example 3: Prove that the function ƒ(x) = | x – 1 |, 0 < x < 2 is continuous at x = 1 but notdifferentiable there. (W.B.U.T. 2004)

Solution: Hereƒ(x) = – x + 1, when 0 < x < 1

= x – 1, when 1 � x < 2

Now, )(lim1

xfx −→

= 1

lim ( 1) ( 0 1)x

x x→ −

− + < <�

= – 1 + 1 = 0


and )(lim1

xfx +→

= 1

lim ( 1)x

x→ +

− > <�( 1 1)x

= 1 – 1 = 0

� )(lim1

xfx −→

= )(lim1

xfx +→

= 0 = )1(f


Again (1)Lf ′ = 0

(1 ) (1)lim

h

f h f

h→ −

+ −=

0

(1 ) 1 0lim

h

h

h→ −

− + + − ( 0 1 1)h< + <�

= 0

lim ( 1)h→ −

− = –1 ( 0)h ≠�

and (1)Rf ′ = 0

(1 ) (1)lim

h

f h f

h→ +

+ −=

0

(1 ) 1 0lim

h

h

h→ +

+ − −( 1 1 2)h< + >�

= 0

lim 1h→ +

= 1 ( 0)h ≠�

� (1) (1)Lf Rf′ ′≠

Therefore we conclude that ƒ(x) is continuous at x = 1 but not differentiable there.

Example 4: Examine the continuity and differentiability of the function

)(xf =

>+≤

0when,sin1

0when,1

xx

x(W.B.U.T. 2003)

at x = 0.Solution: Here,

)00( −f = )(lim0

xfx −→

= 0

lim 1x→ −

= 1 ( 0)x <�

)00( +f = )(lim0

xfx +→

= 0

lim (1 sin )x

x→ +

+ = 1 ( 0)x >�

and ƒ(0) = 1

� )00( −f = )00( +f = )0(f


Again )00( −′f = 0

(0 ) (0)lim

h

f h f

h→ −

+ −=

0

1 1lim

h h→ −

− = 0, ( 0)h <�

and )00( +′f = 0

(0 ) (0)lim

h

f h f

h→ +

+ −=

0

1 sin 1lim

h

h

h→ +

+ −( 0)h >�

= 0

sinlim

h

h

h→ += 1

Since )(),00()00( xfff +′≠−′ is not derivable (or differentiable) at x = 0.


Example 5: A function ƒ(x) is defined by

)(xf = 1

sin if 0x xx

≠

= 0 if x = 0.

Prove that ƒ(x) is continuous but not derivable at x = 0 (W.B.U.T. 2006, 2009)Solution: Here

( ) (0)f x f− = 1

sin 0xx−

= 1 1

sin sin 1 for all non-zero real x x xx x

≤ ≤ �

Therefore, given � > 0, there exists a � = � > 0, such that

( ) (0) whenever 0f x f x− < ε − < δ .

Hence ƒ(x) is continuous at x = 0.

For the derivability of ƒ(x) at x = 0, we have from the definition

)0(f ′ = 0

(0 ) (0)limh

f h f

h→

+ −=

0

1sin 0

limh

hh

h→

−

=0

1lim sinh h→

,

which does not exist since ƒ(x) oscillates finitely near x = 0.

Therefore, ƒ(x) is continuous at x = 0 but not derivable at x = 0.

Example 6: Show that the function ƒ(x) defined by

)(xf = 2 1sin if 0x x

x≠

= 0 if x = 0 is derivable at x = 0.Solution: Here,

)0(f ′ = 0

(0 ) (0)limh

f h f

h→

+ −

=

2

0

1sin 0

limh

hh

h→

−=

0

1lim sin ( 0)h

h hh→

≠�

= 01

1 sin 1h

− ≤ ≤ �

Hence ƒ(x) is derivable at x = 0.


Example 7: If

)(xf = 1/, 0

1 x

xx

e≠

+= 0, x = 0

then show that ƒ(x) is continuous at x = 0 but not derivable at x = 0.Solution: We have

)00( +f = )(lim0

xfx +→

= 1/0

lim ( 0)1 xx

xx

e→ +≠

+�

= 1/

1/0lim

1

x

xx

xe

e

−

−→ + + = 0 1/( 0 as 0 )xe x− → → +�

and )00( −f =0

lim ( )x

f x→ −

= 1/0

lim ( 0)1 xx

xx

e→ −≠

+�

= 0 1/( 0 as 0 )xe x→ → −�

Also ƒ(0) = 0

� )00( −f = )00( +f = )0(f


Now, )00( −′f = 0

(0 ) (0)lim

h

f h f

h→ −

+ − =

1/

0

1lim ( 0)h

h

h

e hh→ −

+ ≠�

= 1/0

1lim

1 hh e→ − + = 1

1/1as 0 and so 0 as 0hh e h

h → −∞ → − → → − �

Also, )00( +′f = 0

(0 ) (0)lim

h

f h f

h→ +

+ − =

1/

0

1limh

h

h

eh→ +

+

= 1/0

1lim ( 0)

1 hhh

e→ +≠

+�

= 01/1

as 0 and so as 0hh e hh

→ ∞ → + → ∞ → + � .

� )00()00( +′≠−′ ff .

Therefore ƒ(x) is not derivable at x = 0. Hence we conclude that ƒ(x) is continuous at x = 0 but notderivable at x = 0.


Example 8: Examine the continuity and derivability of the function ƒ(x) defined by

)(xf =

≤≤+<≤++

21for13

10for22

xx

xxx

at x = 1.Solution: Here

)01( −f = 1

lim ( )x

f x→ −

= 2

1lim ( 2) ( 0 1)

xx x x

→ −+ + < <�

= 1 + 1 + 2 = 4

)01( +f = )(lim1

xfx +→

= 1

lim (3 1) ( 1 2)x

x x→ +

+ < <�

= 3 + 1 = 4

and ƒ(1) = 3 + 1 = 4

� )01( −f = )01( +f = )1(f


Again )01( −′f =0

(1 ) (1)lim

h

f h f

h→ −

+ −

= 2

0

(1 ) (1 ) 2 4lim ( 0 1 1)

h

h hh

h→ −

+ + + + − < + <�

= 2

0

3lim

h

h h

h→ −

+ =

0lim ( 3) 3 ( 0)

hh h

→ −+ = ≠�

and )01( +′f = 0

(1 ) (1)lim

h

f h f

h→ +

+ − =

0

3(1 ) 1 4lim ( 1 1 2)

h

hh

h→ +

+ + − < + <�

= 0

3lim

h

h

h→ + =

0lim 3 3 ( 0)

hh

→ += ≠�

� )01( −′f = 3)01( =+′f

Therefore, ƒ(x) is derivable (or differentiable) at x = 1

Example 9: It is given that )( yxf + = ,0)(),()( ≠xfyfxf for all real x, y and f��(0) = 2. Prove

that for all real )(, xfx ′ ).(2 xf= Hence find the value of ƒ(x).

Solution: It is given that )()()( yfxfyxf =+ for all real x, y.

for x = y = 0, we get from (1), ..(1)

)0(f = )0()0()0( fff ⇒ = )0)0((1 ≠f� ...(2)

Now, )(xf ′ = 0

( ) ( )limh

f x h f x

h→

+ −=

0

( ) ( ) ( )limh

f x f h f x

h→

−[by (1)]

= 0

( ) 1( ) lim

h

f hf x

h→

−= ƒ(x)

0

( ) (0)limh

f h f

h→

−[by (2)]


= )(2)0()( xffxf =′ [since ƒ'(0) = 2]

Thus )(xf ′ = ),(2 xf for all real x.

�df

f= 2dx

Integrating both sides, we have

df

f∫ = 2 ,dx∫ or )(log xf = cx log2 +

When x = 0, log ƒ(0) = log c, or log c = 0 (� ƒ(0) = 1, by (2))

� )(log xf = 2 , or ( )x f x = xe2

Example 10: Let ƒ(x) be a continuous function and g(x) be a discontinuous function. Prove thatƒ(x) + g(x) be a discontinuous function.

Solution: Let ),()()( xgxfxF += where ƒ(x) is continuous and g(x) is discontinuous. Let ussuppose that F(x) is continuous so that g(x) = F(x) – ƒ(x) is also continuous, which is a contradiction as

g(x) is given to be discontinuous. Therefore )()()( xgxfxF += must be discontinuous.

Example 11: Let ƒ(x) be a function satisfying the condition )()( xfxf =− for all real x. If f�(0)exists, find its value.

Solution: It is given that ),()( xfxf =− for all real x ...(1)Also ƒ'(0) exists.

�0

(0 ) (0)limh

f h f

h→

+ −=

0

(0 ) (0)limh

f h f

h→

− −−

or0

( ) (0)limh

f h f

h→

−=

0

( ) (0)limh

f h f

h→

−− [by (1)]

or0

( ) (0)2 lim

h

f h f

h→

−= 0

or )0(2 f ′ = 0� ƒ'(0) = 0

4.11 DIFFERENTIAL OF A FUNCTION

If the derivative of a function ƒ(x) exists, the relation

0

( ) ( )limh

f x h f x

h→

+ −= )(xf ′

defining the derivative is equivalent to the relation

( ) ( )f x h f x

h

+ −= ( ) , or ( ) ( )f x f x h f x′ + ε + − = ( )hf x h′ + ε ...(1)

where � � 0 as h ��


Here f (x + h) – f (x) denotes the increment of the function ƒ(x). Therefore, when derivative of afunction ƒ(x) exists, the increment of ƒ(x) consists of two parts — one part hƒ'(x), called the principalpart or linear part and the other part �h, called error. The principal part hƒ'(x) is known as the differentialof the function ƒ(x), denoted by dƒ(x).

� )(xdf = )(xfh ′ = ( )df x

hdx

...(2)

In particular, if ƒ(x) = x, then (2) gives dx = h ...(3)

Therefore, )(xdf = ( )df x

dxdx

. ...(4)

Notes:(i) The differential of an independent variable is same as the increment of that variable.

(ii) The differential of a dependent variable is not identical with its increment (compare (1) and(4)).

Example: Obtain dx, dy, �y – dy, given, y = x2 + x, x = 1 and �x = 0.1.

Solution: dx = x∆ = y∆;1.0 = )11(}1.1)1.1{( 22 +−+ = 0.31

dy = dy

dxdx

= dxx )12( + = 1.0)12( ×+ = 3.0

� �y – dy = 0.31 – 0.3 = 0.01.

MULTIPLE CHOICE QUESTIONS

1. If )(xf = 1

1, 1, then lim ( )

1 →

− ≠− x

xx f x

x

(a) equal to 1 (b) equal to –1(c) equal to 0 (d) does not exist

2. 20

1 coslimx

x

x→

− is

(a) 0 (b)2

1(c)

4

1(d) 1

3.2

20

sin ( cos )limx

x

x→

π equals

(a) – � (b)2

π(c) � (d) 1

4. lim ,x x

x xx

a b

a b→ ∞

−+

where a > b > 1 is equal to

(a) 1 (b) –1 (c) (d) none of these


5.

2

2–

1lim

1

x

xx

e

e→ ∞

+

− is equal to

(a) 0 (b) 1 (c) –1 (d) does not exist

6.2lim

xx x x

→ ∞ − + is equal to

(a)2

1− (b)2

1(c) 0 (d) none of these

7. If )(xf = 2

3, 0 1

1, 1

ax x

xx

a

+ < <− + ≥

and )(lim1

xfx→

exists, then the value of a is

(a) 0 (b) 1 (c) (d) –1

8.2 2 2 2

3

1 2 3lim

n

n

n→ ∞

+ + +…+ is equal to

(a)1

2(b)

1

6(c)

1

3(d) none of these

9. If )(xf = )(limthen1when,4

1when,13

1xf

xx

xx

x→

>−≤+

(a) is equal to 4 (b) is equal to 3(c) is equal to 1 (d) does not exist

10. If )(xf =

1/

1 /

2, for 0

1 20, for 0

x

xx

x

≠

+ =

then )(lim0

xfx→

(a) is equal to 1 (b) is equal to 2(c) is equal to 0 (d) does not exist

11. If )(xf =

>+≤+

1for32

1for2 2

xx

xaxx is continuous at x = 1, then the value of a is

(a) 3 (b) –3 (c) � (d) none of these

12. If )(xf = 2

1(1 cos )x

x− , when x � 0, the value of ƒ(0) for which ƒ(x) is continuous at x = 0, is

(a) 1 (b)2

1(c)

4

1(d) 0


13. If )(xf = 2

2

4 3

1

x x

x

− +−

when x � 1, the value of ƒ(1) for which ƒ(x) is continuous at x = 1 is

(a) 2 (b) 1 (c) –1 (d) 0

14. If )(xf =

≥+<+

1,2

1,2

xx

xbax then the values of a and b for which ƒ(x) is continuous at x = 1 are

(a) a = –1, b =1 (b) a = 1, b = 1 (c) a = 0, b = 2 (d) a = 1, b = 2

15. The function )(xf =

=

≠

0,0

0,1

sin

x

xx

x is

(a) Continuous and differentiable at x = 0(b) Continuous but not differentiable at x = 0(c) Neither continuous nor differentiable at x = 0(d) None of these (W.B.U.T. 2009)

16. If )(xf = 1 sin

,sin 2 2

xx

x

− π≠ is continuous at x = 2

π then

2f

π

=

(a)1

2(b) 1 (c) �� (d) 0

(W.B.U.T. 2008)

17. The function )(xf = x

x is

(a) Continuous and derivable at x = 0(b) Continuous but not derivable at x = 0(c) Derivable but not continuous at x = 0(d) Neither continuous nor derivable at x = 0

18. The function )(xf = x is

(a) Continuous and differentiable at x = 0(b) Continuous but not differentiable at x = 0(c) Discontinuous but differentiable at x = 0(d) None of the above

19. The function ƒ(x) = | x – 1 | + x – 1 is(a) Continuous and differentiable at x = 1(b) Continuous but not differentiable at x = 1(c) Neither continuous nor differentiable at x = 1(d) None of the above


20. The function )(xf = 1 1

x

x+ − is

(a) Continuous and differentiable at x = 1(b) Discontinuous but differentiable at x = 1(c) Continuous but not differentiable at x = 1(d) None of the above

21. The function )(xf = 2 1

sin , for 0

0, for 0

x xx

x

≠ =

is

(a) Continuous and differentiable at x = 0(b) Continuous but not differentiable at x = 0(c) Neither continuous nor differentiable at x = 0(d) None of the above

22. The function )(xf =

2

, 0

0, 0

xx

x

x

≠

=

is

(a) Continuous and differentiable at x = 0(b) Continuous but not differentiable at x = 0(c) Discontinuous but differentiable at x = 0(d) None of the above

23. The function )(xf = x x is

(a) Continuous but not differentiable at x = 0(b) Discontinuous but differentiable at x = 0(c) Neither continuous nor differentiable at x = 0(d) Continuous and differentiable at x = 0

ANSWERS

1. (d) 2. (b) 3. (c) 4. (a)

5. (b) 6. (a) 7. (d) 8. (c)

9. (d) 10. (d) 11. (a) 12. (b)

13. (c) 14. (d) 15. (b) 16. (d)

17. (d) 18. (b) 19. (b) 20. (c)

21. (a) 22. (b) 23. (d).


PROBLEMS

1. Using � – � definition of limit, show that

(i)2

1

1lim

1x

x

x→

−−

= 2 (ii)0

1lim cosx

xx→

= 0

(iii)0

1lim ( 1) sinx

xx→

+ does not exist.

2. Is the function 1 1x x

x

+ − − defined for all values of x ? Indicate the values of x for which it

is defined and real. Find the limit as x � 0.

3. Show that the following limits do not exist:

(i) x

xe /1

0lim→

(ii)0

limx

x

x→

(iii)2

2lim

2x

x

x→

−−

(iv)0

sinlimx

x

x→

(v)1

sin( 1)lim

1x

x x

x→

−−

4. A function ƒ(x) is defined as under:

)(xf = 2

2, 1

4 1, 1 3

5, 3.

x x

x x

x x

+ <

− ≤ < + ≥

Show that )(lim1

xfx→

= 3 and )(lim3

xfx→

does not exist.

5. A function is defined as under:

)(xf =

sin, 0

, 0.

xx

xax b x

< + >

If )(lim0

xfx→

exists then find a, b and the limit.

6. A function is defined as under:

)(xf = 1, 24 <xx

= 1, 2 ≥xx .

Show that )(lim1

xfx→

= 1 and )(lim1

xfx −→

does not exist.


7. Discuss the left continuity, the right continuity and the continuity of the following functions atthe points indicated.

2( ) ( ) ( 4) /( 2), 2

4, 2

3, 2

i f x x x x

x

x x

= − − <

= = = + >

, at x = 2

21/( ) ( ) , 0

2, 0

xii f x e x

x

− = ≠

= = , at x = 0

1/

1/( ) ( ) , 0

10, 0

x

x

xeiii f x x

ex

= ≠

+= =

, at x = 0.

8. Discuss the continuity of the following functions at the points indicated:

(i) )(xf = xxx at/1− = 0, 1

(ii) )(xf = xxx at][][ 22 − = 0, 1 (where [ y ] denotes the greatest integer less than or equal to y)

(iii) ƒ(x) is defined in 0 � x � 1 by the followings:

( ) 0, 0

11 , 0 in 0 1

21

1, 12

f x x

x x x

x

= == − < < ≤ ≤= ≤ ≤

2

( ) ( ) 2 , 0

1 , 0 1

3 , 1 2

3 , 2

iv f x x x

x x

x x

x x x

= + ≤ = + < < = − ≤ ≤ = − >

, at x = 0, 1, 2.

9. Find ƒ(0) so that )(xf = xx /1)21( + for 0x ≠ may be continuous at x = 0.

10. Test for continuity of the following functions at x = 0:

(i) ƒ(x) = x x− (ii) ƒ(x) = 1

x

x+


(iii) ƒ(x) = 1/, 0

1 x

xx

e≠

+(iv) ƒ(x) = 1/

1, 0

1 2 xx ≠

+ = 0, x = 0 = 0, x = 0

(v)1/

1 /

2( ) , 0

1 21, 0

x

xf x x

x

= ≠+

= =

11. Examine continuity and differentiability of ƒ(x) at x = 0

where )(xf = 1

cos , 0x xx

≠

= 0, x = 0. (W.B.U.T. 2007)12. Show that the function ƒ(x) defined by

)(xf = 2 1

cos if 0x xx

≠

= 0 if x = 0is derivable at x = 0.

13. Let ƒ(x) = x2, if x � 1 and ƒ(x) = ax + b, if x > 1. Find the coefficients a and b at which thefunction is continuous and has a derivative at x = 1.

14. Show that the function ƒ(x) defined by

ƒ(x) = 3 + 2x for 3

02

x− < ≤

= 3 – 2x for 3

02

x< <

is continuous but not derivable at x = 0.15. Let

ƒ(x) = x, 0 < x < 1= 2 – x, 1 � x � 2

= 21, 2

2x x x− > .

Show that ƒ(x) is continuous at x = 1, x = 2 and that ƒ'(2) exists but ƒ'(1) does not exist.16. Show that the function ƒ(x) defined by

ƒ(x) = x3 for x2 < 1

= 1 for x2 � 1

is continuous but not derivable at x = 1.17. Show that the function ƒ(x) defined by

)(xf = 21

1 sin (log ) , 03

x x x + ≠

= 0, x = 0is continuous but not derivable at x = 0.


ANSWERS TO PROBLEMS

2. 1 5. )(lim0

xfx→ = 1, b = 1, a may be any real number.

7. (i) left continuous (ii) discontinuous (iii) continuous.8. (i) at x = 0 it is discontinuous and at x = 1 it is continuous

(ii) at x = 0 it is discontinuous and at x = 1 it is continuous

(iii) continuous everywhere except at x = 0 and x = 1

2(iv) at x = 0, 2, it is discontinuous and at x = 1, it is continuous.

9. ƒ(0) = e2.10. (i) continuous (ii) continuous

(iii) continuous (iv) discontinuous(v) discontinuous.

13. a = 2, b = –1.

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Limit, Continuity and Differentiability › UoN › Eng3A › LimitContunityDiff.pdfLimit, Continuity and Differentiability 4.1 INTRODUCTION The basic concepts of the theory of calculus