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Title: FUNCTIONS. LIMIT. CONTINUITY
Description: Comprehensive notes and practicals (Problems and Solutions) covering: Functions
Description: Comprehensive notes and practicals (Problems and Solutions) covering: Functions
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FUNCTIONS
...
CONTINUITY
Practical class №13
...
Solve inequalities:
а) x 4;
b) x 2 16;
d) 0 x 1 4;
c) x 3 1;
e) x 2 9;
f) x 2 4
...
2
...
2
3
...
2
4
...
5
...
6
...
ba
b) 0 , 1 , ,
Find the domain of the function:
7
...
10
...
y ln
x 1
...
x5
8
...
9
...
y
12
...
y
x 1 3 x
...
1
...
y 4 x
...
Determine if a Function is Even, Odd or Neithe
16
...
2 3
x 4 3x 2
17
...
x x 2
18
...
For function y f x and X find inverse function y f 1 x and its
domain
...
y 1 3x, x ;
...
y x 2 , а) x ; 0; b) x 0;
...
y x 2 6 x 5
...
y
x ; 1;
log 2 x, x 1;
...
x2
23
...
x
2 x 1,
25
...
24
...
y
x 3
27
...
x 25
28
...
2
x 2 7 x 12
...
y 1 2 x 4log3 1 x
...
For function y x 2 1, x 0; find inverse function and its
domain
...
31
...
2
3
32
...
2
33
...
34
...
1 x2 9
The solution of typical problems
Problem 1
...
The domain of this function is x , satisfying the inequality x 2 3x 0
x 3,
x x 3 0
т
...
The range is
x
0
,
y 0 , since radical expression can take any positive value, then
E f ; 0
...
For function y 3x 1 find inverse function
...
y
y3
From the equation y f x , y 3x 1
x 1
express x : x log3 y 1, т
...
Let argument of g be x , and function value be y ,
3
then we obtain y f 1 x log3 x 1 – inverse
х
3
x 1
y log3 x 1 function to the function y 3
...
3
...
Graphs of the inverse functions are symmetric with re-
spect to the straight line y x (рис
...
1)
...
Graph function y x 2 3
...
According to
this rule its necessary to move the parabola y x 2 on two units to the left
along the axis Ox and on three units
down along the axis Oy ,without changing its shape (pic
...
2)
...
3
...
а) 4; 4 ; b) 4; 4; c) 2; 4 ; d) 1; 5; e) ; 3 and 3; ;
1 3
f 1 1 4, f ;
2 4
2
1
x2 1
5
1 2x x
b) 0 3, 1 , 2
;
,
2
x 1 x 2x 3
x
c) g 1 1, g 2 4, g 0 1; d) b a
...
2;
...
3; 3
...
5;
...
1; 3
...
1; 3
...
; 1 2;
...
6
...
; 5 3;
...
; 3 3; 4
...
4; 0 0; 4
...
Odd
function
...
The function of the General form
...
Even function
...
1 x
, D f 1 ;
...
а) y x , D 1 0; ; b) y x , D 1 0;
...
; 5 5; 5 5;
...
; 4 3;
...
0;
...
0,5; 1
...
y x 1, D 1 1;
...
а) 2; 2;
f
b) ; 3 3; ; c) ; 1 1; ; d) ; 1 1;
...
Limit of the function
...
x 2 x 1
1
...
lim
x
1
2
4x 2
x3 1
1 sin 2 x
...
x 2 x 2
4
...
lim
...
6
...
lim 2
7
...
8
...
x 2 x 2
x 1 0 x 1
x 3 x 1
x 2 0
x 1 x 6
2 x2 5x 3
x
9
...
10
...
11
...
x 3
x 1
x0 1 3 x 1
x2 1
x2 9
2 x3
2 x 2 x3 1
2x 7
12
...
13
...
14
...
x7 x 49
x 4 x3 2
x 5 x 2
3n 4 2
n
n2
15
...
16
...
17
...
3
n
n n 2
n 3n 4
n 1
x4
6
1
2
18
...
20
...
19
...
x x 2 x
x3 x 3 x 9
x
Choose the graph of function y f x , satisfying following conditions:
lim f x ,
lim f x ,
x
2
0
21
...
lim
f
x
0
...
x 0
x 2 0
x 0
1
y
y
y
O
x
а)
O
1
2
б)
х
1
1
O
х
в)
63
f x 0,
x lim
22
...
x
y
2) lim f x 1
...
y
O
y
х
O
x
O
1
х
2
-1
-1
а)
б)
в)
Are these functions continuous at the given points:
23
...
f x
, x0 2
...
25
...
lim 2
...
lim
...
lim 2
...
lim
...
lim 2
...
lim
...
lim 2
...
lim
...
lim 2
...
Find points of discontinuity for the function and graph:
x 12 , x 1;
y 1 x , 1 x 2;
2
x 2
...
lim x x 2 x 1
...
1 2x 1
x
37
...
1
...
lim ln x 1
...
lim
...
lim
...
x 1
1 5x
n 1
43
...
n 2n
x2 x 2
46
...
x 1
x3 1
5n 3
...
lim
2n
5n 4
44
...
lim
...
lim
...
lim 2
...
lim
...
lim n 2
...
lim
...
Find points of discontinuity: а) y
б) y
...
x 1 4 x 2
Problem 1
...
The point x0 1 D f , then by definition of
2 2
3x 5
3 1 5
continuous function lim
f 1
4
...
If irrationality in-
Since f x
terferes with a given reduction, it is first eliminated
...
Calculate lim
x4
x 2 16
х 2 7 x 12
...
To cancel the fraction on x 4 factor it:
0
x 4x 4 lim x 4 8 8
...
Calculate lim
...
Multiply numerator and denominator times
0
expression: 3 x 9
...
x 0 x 3 x 9
x 0 3 x 9
6
To calculate an indeterminate form cancel the fraction
...
Calculate lim x
...
Since 7 x 2 x , we need to cancel the
x1
fraction on 7 , we obtain
x 1
2
x
7
x 1
x
2 7
7
2
lim
7, since 0 when
lim
x
x 2 x 7 x 1 x
7
2
7 1
7
x
...
a1 x a0 and
Qm ( x) bm x m bm1 x m1
...
Problem 5
...
Since the numerator and denominator are polynomials of the same degree, we
obtain:
8 х3 2 x 2 5 х 2 8
lim
4
...
0
terminate form
1
4
...
Calculate lim
1
4
x2 4 x 2
4 x 2 0
2 x
1
1
lim
lim
...
Calculate lim n n 2 n
...
Then lim n n2 n lim
lim
2
2
n
n
n n n
n n n n
n
1
1
lim
...
Find points of discontinuity for the piecewise function
x 2 , x 1;
f x
2 x, x 1
...
1
Рис
...
3
x10
Because one-sided limits are finite but not equal, then
in the point x0 1 the function has the jump
d
O
x10
f 1 0 lim f x lim 2 x 2, f 1 1
...
1
...
0
...
1
...
5
...
6
...
0
...
4
...
3,5
...
7
...
2
...
1
...
0,4
...
0,25
...
16
...
17
...
1
...
0
...
1
...
1) б; 2) а; 3) в
...
1) в; 2) а; 3) б
...
No
...
Yes
...
а) x 1 ; b) x 2 ;
c) x 2; x 2
...
0,5
...
0
...
0,5
...
0
...
31
...
32
...
0
...
0,5
...
0,5
...
x 1(jump)
...
38
...
40
...
41
...
42
...
43
...
44
...
45
...
1
...
2,5
...
11
...
4
...
25
...
3
0,5
...
а) x 0 (asymptotic discontinuity); б) x 2 (removable)
Practical class №15
...
x 0
0
x 0
x 0
tan x 0
lim
1;
x 0 x 0
The second special limit:
e ( e 2,71828182
...
x0
The third special limit:
log a 1 x 0
ln 1 x 0
1
,
lim
1
...
x 0
x 0
x 0
x 0
lim
1 x 1 0
lim
...
If lim
x x0
x
1, then x and x are called equivalent
x
infinitesimal functions
...
Equivalence of infinitesimal functions are: x 0 x х0 :
sin x ~ x
...
tan x ~ x
...
log a 1 x ~
e x 1 ~ x
...
ln 1 x ~ x
...
а x 1 ~ x ln a
...
Problems
Calculate limits:
sin 5 x
1
...
x 0
x
1 e5 x
4
...
x 0 4 x
tg 2 x
2
...
x0 arcsin 3 x
1 2x 1
5
...
x 0
x
32 x 1
3
...
x 0 3x
ln 1 5 x
6
...
x 0
2x
69
7
...
arctg 2 3x
10
...
lim
...
lim
...
lim n2 arctg
tg
...
lim 2 2n
...
...
lim n 1 sin
n
p
2n
...
lim
...
lim 1
...
x 0 sin 5 x
ln 1 2sin 2 x
9
...
x 0
n
n 3
18
...
n n 1
Material for individual work
Calculate limits:
19
...
20
...
lim n 2 1 tg 2
...
lim
...
n
n
x2 2 x
24
...
x 2 sin x 2
21
...
1 cos x
25
...
x 0
x
x3
28
...
x
x
26
...
lim
...
lim
...
lim
x1
3
1 x 1
ln 11 10 x
e5 x 5 1
The solution of typical problems
Problem 1
...
sin x tg 3x 0
x0
x2
0
Since when x 0 : sin x ~ x, tg 3x ~ 3x , then lim
lim
x 3x
x0
x2
3
...
Calculate limit: lim
ln 1 10 x
...
x 0 e5 x 1 0 x 0 5 x
x 0
70
...
Problem 3
...
x 1
Here x
/ 0 , but x 1 0 when x 1 use the equivalent:
sin x 1 ~ x 1, x 1
...
lim
lim
lim
x 1 x 2 1 0 x 1 x 2 1 0 x 1 x 1 2
x 1
2
Problem 4
...
а)
1
15
e
5
lim(1 3x) x
x 0
1 lim(1 3x)
1
( 53)
3x
x 0
1
lim (1 3 x) 3 x
x 0
15
...
x
x
1
c) lim(2 x 3) x
x 2
2
1
4
lim(1 2 x 2 ) x
2
4
x 2
lim 1 2 x 2
x 2
1
2 x 2
2
x2
1
e2
e
...
5
...
2
...
2ln 3
...
1,25
...
1
...
2,5
...
0
...
0,3
...
0,4
...
18
...
12
...
13
...
14
...
15
...
16
...
17
...
18
...
19
...
2
2
2
20
...
21
...
p
...
1
...
2
...
0
...
2
...
90
...
3
...
e
...
2
Title: FUNCTIONS. LIMIT. CONTINUITY
Description: Comprehensive notes and practicals (Problems and Solutions) covering: Functions
Description: Comprehensive notes and practicals (Problems and Solutions) covering: Functions