Lecture_07_2026_27_AB

1.

Question 1. (November 2013 Exam, Q19)
5x − 2 x
is
lim 5
3
x → 0 3 x − 20 x
3
5
1
2
5
1
a) −
b) −
c)
d) −
10
3
3
4
e) The limit does not exist
Solution. To find limits of rational functions we
divide the numerator and denominator by the
dominant term.
But, contrary to the case when x → ,
now the dominant term is not the highest power,
but the lowest power of x.

2.

On dividing by x 3 we obtain
0
2
1
5x − 2 x
5 − 2x
lim 5
=− .
= lim 2
3
x → 0 3 x − 20 x
x →0 3 x − 20
4
0
What happens if we divide by x 5 ?
+
−2
3
5
5x − 2
5x − 2 x
=
lim
lim 5
x →0 3 − 20 x − 2
x → 0 3 x − 20 x 3
+
5
The division by x does not help.
3
5
It is still hard to tell what the value of the limit is.
1
2
5
1
a) −
b) −
c)
d) −
10
3
3
4
e) The limit does not exist

3.

The Left Limit of a Function
Definition. The number l is the left limit of the
function f at a point a, if for any 0 there
exists a number 0 , such that | f ( x ) − l |
if a − x a .
Example. lim sgn( x − 1)
x 1

4.

First, we draw the graph of the function
f (x) = sgn(x – 1) and try to guess the value of
the left limit lim sgn( x − 1)
x 1
f ( x ) = sgn( x − 1)
1
−1
x
1
−1
Looking at the left part of the function f (x) we can
guess that the left limit
lim sgn( x − 1)
exists and it is equal to – 1. x 1

5.

The Right Limit of a Function
Definition. The number l is the right limit of the
function f at a point a, if for any 0 there
exists a number 0 , such that
| f ( x ) − l | if a x a + .
Example. lim sgn x
x 0

6.

Again, we draw the graph of the function
f (x) = sgn x and try to guess the value of the
right limit lim sgn x
x 0
f ( x ) = sgn x
1
x
−1
1
−1
It looks like the right limit lim sgn x exists and is
x 0
equal to + 1.

7.

What do the left and right limits tell us
about the limit of the function?
Theorem. lim f ( x ) = l if and only if
x→ a
lim f ( x ) = lim f ( x ) = l .
x a
x a

8. Calculating Complicated Limits: Equivalent Functions

Functions f (x) and g(x) are equivalent as x → a
f ( x)
if lim
= 1.
x→a g ( x)
If the functions f (x) and g(x) are equivalent
we write f (x) ~ g(x) as x → a .
For example, the First Special Limit,
sin x
lim
= 1,
x→0
x
tells us that sin x ~ x, as x → 0 .

9. Calculating Complicated Limits Equivalent Functions

Another example, 1 – cos x ~ ½ x 2, as x → 0 ,
because
2 sin ( x )
1 − cos x
.
L = lim 1 2 = lim
x→ 0
x→0
x
x
2
2 1
2
1 2
2
Introduce a new variable y = ½ x to obtain
sin y sin y
sin y
sin y
L = lim
= lim
lim
= 1.
y→0
y→0
y
y
y y→0 y

10.

Equivalent polynomials for elementary functions
(only) as x → 0 .
b
f ( x ) = x – do not look for equivalent functions
f ( x ) = sin( c x ) ~ c x
1 2 2
f ( x ) = 1 − cos( c x ) ~ 2 c x
f ( x ) = ln( 1 + c x ) ~ c x
x
f ( x ) = b ~ 1 + x ln b
c
f ( x ) = (1 + x ) ~ 1 + c x
c
c
c
f ( x ) = ( b + x ) ~ b (1 + b x )
f ( x ) = arctan( c x ) ~ cx
f ( x ) = arcsin( c x ) ~ cx
−1
tan ( cx ) ~ cx
−1
sin ( cx ) ~ cx

11. Continuous Functions

A function f (x) is continuous at a point a iff (if and
only if) lim f ( x ) = f ( a ).
x→ a
The definition of continuity is short, however it
contains three essential conditions:
1. The function f (x) is defined at x = a.
That is a D [ f ].
2. lim f ( x ) exists.
x→ a
3. lim f ( x ) = f ( a ).
x→ a

12. Continuous Functions

A function f (x) is continuous at a point a iff
0, 0 : | f ( x ) − f (a ) |
when | x − a | .
Example. Any linear function f (x) = b x + c is
continuous at any point x = a.
Let =
, then for any x from | x − a |
1+ | b |
we obtain 0 , 0 : | f ( x ) − f ( a ) | =
| bx + c − ba − c | = | b ( x − a ) | = | b | | x − a | | b |
|b|
Therefore f (x) = b x + c is
=
.
1+ | b |
continuous at any point a.

13. Continuous Functions

If a function f (x) is continuous at a point x = a then
lim f ( x ) = f ( a ) = f ( lim x ).
x→ a
x→ a
This property is important because it helps us to
calculate limits of composite functions
f (x) = g(h(x)).
If we know that the function g(.) is continuous, then
lim g ( h ( x )) = g ( lim h ( x )).
x→ a
x→ a
For example,
sin x
sin x 1
lim exp
= exp lim
=e.
x→0
x
x→0 x

14. Continuous Functions

Theorem. All elementary function f (x) are
continuous at all points x = a of their domains
D[ f ].

15. Discontinuous Functions

A function f (x) has a removable discontinuity at a
point a if lim f ( x ) exists, but
x→ a
either lim f ( x ) f ( a ),
x→ a
or the function f (x) is not defined at x = a.

16.

Example: a removable discontinuity
f10((xx)) ==
sin x
, if x 0;
x
01?,, if x = 0.
0.4
0 .2
−2
2
x

17.

Another example f ( x ) = sgn
2
( x − 4)
f ( x ) = sgn ( x − 4 )
2
1
x
−1
lim sgn ( x − 4 ) =
2
x→ 4
= lim 1 = 1 f ( 4 ).
1
−1
x→4
The function f (x) = sgn2(x – 4) has a removable
discontinuity at the point x = 4.

18. Discontinuous Functions

A function f (x) has a Type I (or Jump) discontinuity
at a point a if both the left and the right limits
exist, but they are not equal:
lim f ( x ) lim f ( x ).
x a
x a
Example. The sign function, defined by
− 1, if x 0
sgn x = 0, if x = 0
1, if x 0
has a Type I discontinuity at the point x = 0.

19.

lim sgn x = lim − 1 = − 1
x 0
x→0
At the point x = 0, the
left and right limits
exist, but they are
not equal to each
other.
lim sgn x = lim 1 = 1
x 0
f ( x ) = sgn x
x→0
1
x
−1
1
−1
The function f (x) = sgn x has a Type I (Jump)
discontinuity at the point x = 0.

20. Discontinuous Functions

A function f (x) has a Type II discontinuity at a point
a if either left or right limit does not exist at that
point.

21.

x +1
lim
x 0
x
1
= lim x +
x 0
x
= 0 − = −
2
1
x +1
= x+
f ( x) =
x
x
2
5
x
−1
1
The function
2
x +1
f ( x) =
x
has a Type II discontinuity
at the point x = 0.
−5
x +1
lim
x 0
x 1
= lim x +
x 0
x
= 0 + = +
2