Types of Mathematical Limits
In mathematics, the concept of limits is fundamental when studying mathematical functions. It is important to understand the definition of a limit and the different types of limits that exist. The exercises below will help you practice and understand the concept of limits in mathematics.
Exercises for Defining Limits
The following exercises aim to help you understand the definition of a limit and its application in mathematical functions. Practicing these exercises will provide a clearer understanding of the concept of limits.
Exercises for Solving Limit Problems
Solving limit problems can be challenging. The following exercises will help you practice solving various types of limit problems in mathematical functions.
Exercises for Limit Functions with Indeterminations
The concept of limit functions with indeterminations is important in mathematics. The exercises provided below aim to help you practice solving limit functions with indeterminations in mathematical functions.
Advanced Exercises on Limits and Functions
These exercises are designed to challenge your understanding of limits and functions in mathematics. Practicing these advanced exercises will help solidify your knowledge and understanding of limit functions.
Now, let's take a look at the corrected exercises content:
JUNE 94
f(x)=
a) Domf(x) - IR
We study ch x=0
(05x
f(0) = a +0=a
e a+ 2x² = a
x→+
e
e cos x = 1
690
We study at xad
f(1) =a+2
e
b
スライ
X-1
b
K
JUNE 15
- f(x) = b
a + 2ײ = a+2
2x-4
2
#
six 20
Si0EX≤A
Sixz1
f(0)
A.V.
fini= e
A.H
e
A.O
a=1
X+1
a + 2 = b
1 + 2 = b = 3
ze
×9
Domf(x) = R-124 x-532 31-4
Vertical asymptote at
X→+∞0 2x-
X4-x 2v4
n= e
y=mx+n
e x 2-3
84400
y = //x+1
2x²-4x
x ²-3
2x-44
= +2x-3
2x-4
1/22
a=1
b=3
T 2
No horizontal asymptote exists.
SEPTIEMBRE 96
- ca?
f(x)=
e
X-1 x+1
e
We study at x=1
f(1) = x= 1
X+1-
SEPTIEMBRE 97
- 8(x) = -x +
Α. Η
e
X48
e
8174
A.O.
a = 2
x = 1
X40
JUNE 98
f(x)=
a) casíntotas?
A.V Domf(x) = R-104
Vertical Asymptote at x=0
e
-X+
m= e
X+1
n: e
4
y=-x
x²
y=mx+n
O
ax³ + bx
11x-16
x+
8444
3/²x 51²x
8448
six 41
đi xi
Ч
¿a yb?
We study at x=2
f(1) ==
X-1*
-x3+4
응
-x3+4
x2
(3+4
X²
2-∞
ON=0.0
↓
51 x 4-1
if -1 xe 2
If 25x
= -1
= +∞
+x 2
Domfix) - IP
IR
e
fc2):22-16-6
2²²11x-16 = 6 e₂ax²+bx=8a+2b
e
e
メッペ
19=2
*f
X-0¹
e
x-1-
e
X-11
X-0
X4+∞
-x+4 :
-x³+4+x³
x²
0+
Domf(x) = IR
We study at xid
8(-1) = 0
No horizontal asymptote exists.
0 = 0
8+=
:44
= 0
ax³ + bx = a+b
farbe
b = 0
8a+2b = 64a+b = 3
a = -b
-4b+b = 3
-3b = 3
6 = -1
a = 1
JUNE 99
1 gears fin.
1.
e
X-1
We study at x=1
f(1) = 2+a+b
e in X-1
X-11
origin f(x) = 0 →
coord. I X=0
JUNE 11
e
Ln x-1
A.C
e
A.H
2x².ax+b six 41
4
y=x+ (x-1)²
A.V at x=1
x8
2x+ax+b = 2+a+b
2x² + ax+b = 0 → 0+0+b = 0
a+b = -3 → a = -3-0 = -3
e
844
e
8118
JUNE 06
f(x)=
n = e
= 2n1
y=mx+n
e
m= ×440
√ax
Six 21
0 = 2x² + ax +b
x → +∞
X4
x+
514
v6x JSa
We study at x=8.
(8) = √80
2-32
e
X-1t
e
-1:-1
328
4
u
(x-1)²
x2-32 đi xao
x-4
Dom 8(x) = IR-114
X³ -X
04x48
4
(x-1)²
x+
X4
4
Iusa
(x-1)
4
(x-1)2
X³-1+4
x²-1
G.N=GD
Dom&= R
8 1
√5a = 8
2+a+b=-1
b = 0
a = -3
1:1
4
1+
A+
५
0
0+
30.
:0
318
=+∞
No horizontal asymptote exists.
Dom &(x) = [0, +∞0)
8a = 64 160
y = x
64
8
:8
SEPTIEMBRE 07
4
f(x) =
We study at x=0
f(0) = 0+20 cos (0) = 2a
x² + 2a cos (0) = 2a
e
X-90T
e 3x+2=2
X-O
e
X-T
LIMITES
We study at x=T
f(π) = ax² +b = aπT² +b = πT² +b
e ax²+b= π₁² +b
X-771
e (1+²)Ã'
8448
x-1
€
3x+2
ax²+b
e
e
e
X93
e √x²-5-2
x-3
X-3
8440
e
x ² + 2a cos (πT) = πT² - 20
₁)
six<0
S
if 04x4T
6
Si ㅅㅠ
Hoa
olo
응
(√x²-5-2) (√x²-5+2)
(x-3)(√x²-5 +2)
( ₁ - 4 - ) ² = 1'00
×(1-4-1)
2a-2a=1
e √ √4x² + 1 -√4x²-3x+2)
8444
e***
+
2400 *
Domf(x) = R
x²_q
^
x²-5-4
(x-3)(√x²-5 +2)
Ty² +6=7²²-2a
63-2
:888
T
P
(x-8)(x+3)
(x5)(√x²-5+2)
e = 1
(√4x²+1 - √4x²-3x+2) (√4x²+1 + √4x²-3x+2)
(√4x²+1
[4x2-3x+2)
4/²+1-4x² + 3x - 2
3x-4
(√√4x²+1 + √√√x²-3x+2
(√4x²+1+√4x²-3x+2)
3
4
2
6
داف
WIN
12x²
grade 1
x3-7x
and 15x
J
11
+∞
J
12
√5
12
