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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).
168 A TEXTBOOK OF ENGINEERING MATHEMATICS [VOL. I]
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.
Example 3: Using definition show that 0
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
LIMIT, CONTINUITY AND DIFFERENTIABILITY 169
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
170 A TEXTBOOK OF ENGINEERING MATHEMATICS [VOL. I]
This is symbolically expressed as
)(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→
∞−=
Example 1: Using definition show that 20
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.
Example 2: Using definition show that 2
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
LIMIT, CONTINUITY AND DIFFERENTIABILITY 171
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
This is symbolically expressed as
)(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→
+ − =
172 A TEXTBOOK OF ENGINEERING MATHEMATICS [VOL. I]
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.
LIMIT, CONTINUITY AND DIFFERENTIABILITY 173
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.
ILLUSTRATIVE EXAMPLES
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
→ + → ++ = = = → + ⇒ < <
174 A TEXTBOOK OF ENGINEERING MATHEMATICS [VOL. I]
� ).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.
Example 4: Show that the function
)(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
≤ ∴ ≤
LIMIT, CONTINUITY AND DIFFERENTIABILITY 175
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.
Example 5: Show that the function
)(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.
176 A TEXTBOOK OF ENGINEERING MATHEMATICS [VOL. I]
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
LIMIT, CONTINUITY AND DIFFERENTIABILITY 177
� )(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.
ILLUSTRATIVE EXAMPLES
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 >� )
178 A TEXTBOOK OF ENGINEERING MATHEMATICS [VOL. I]
= 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
LIMIT, CONTINUITY AND DIFFERENTIABILITY 179
and )(lim1
xfx +→
= 1
lim ( 1)x
x→ +
− > <�( 1 1)x
= 1 – 1 = 0
� )(lim1
xfx −→
= )(lim1
xfx +→
= 0 = )1(f
Hence ƒ(x) is continuous at x = 1
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
Hence ƒ(x) is continuous at x = 0
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.
180 A TEXTBOOK OF ENGINEERING MATHEMATICS [VOL. I]
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.
LIMIT, CONTINUITY AND DIFFERENTIABILITY 181
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
Hence ƒ(x) is continuous at x = 0
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.
182 A TEXTBOOK OF ENGINEERING MATHEMATICS [VOL. I]
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
Hence ƒ(x) is continuous at x = 1
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)]
LIMIT, CONTINUITY AND DIFFERENTIABILITY 183
= )(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 ��
184 A TEXTBOOK OF ENGINEERING MATHEMATICS [VOL. I]
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
LIMIT, CONTINUITY AND DIFFERENTIABILITY 185
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
186 A TEXTBOOK OF ENGINEERING MATHEMATICS [VOL. I]
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
LIMIT, CONTINUITY AND DIFFERENTIABILITY 187
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).
188 A TEXTBOOK OF ENGINEERING MATHEMATICS [VOL. I]
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.
LIMIT, CONTINUITY AND DIFFERENTIABILITY 189
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+
190 A TEXTBOOK OF ENGINEERING MATHEMATICS [VOL. I]
(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.
LIMIT, CONTINUITY AND DIFFERENTIABILITY 191
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.