Algebra of LimitsAssume that both of the following limits exist and c and is a real number:
Then:
0)(lim
),(lim/)(lim)(/)(lim..4
)(lim)(lim)()(lim..3
)(lim)(lim)()(lim..2
)(lim)(lim..1
xgthatprovided
xgxfxgxf
xgxfxgxf
xgxfxgxf
xfcxcf
ax
axaxax
axaxax
axaxax
axax
Calculating LimitsFinding the limit of a function f a point x = a.
Distinguishing the following cases:1. The case when f is continuous a x = a.2. The case 0/0.3. The case ∞/ ∞4. The case of an infinite limit5. The case c/∞, where c is a real number.6. The case, when it is possible to use the
squeeze theorem.
1. The case when f is continuous at x = a
If f is continues at x=a, then:
Notice:1. Polynomial functions and the cubic root function ( & all functions of its two families) are everywhere continuous.2. Rational, trigonometric and root functions are continuous at every point of their domains.3. If f and g are continuous a x=a, then so are cf, f+g, f-g, fg and f/g (provided that he limit of f at x=a is not zero)
Examples for the case when f is continuous at x = a
04
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Examples for the case when f is continuous at x = a
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Examples for the case when f is continuous at x= a
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2. The case 0/0
Suppose we want to find:
For the case when:
Then this is called the case 0/0. Caution: The limit is not equal 0/0. This is just a name that classifies the type of limits having such property.
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)(lim
xh
xgax
.0)(lim&)(lim arexhxgaxax
Examples for the case 0/0
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Examples for the case 0/0
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Examples for the case 0/0
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Question: Simplify the formula of f and graph it!
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3. The case ∞/ ∞
Suppose we want to find:
For the case when the limits of both functions f and g are infinite
Then this is called the case ∞/ ∞. Caution: The limit is not equal ∞/∞. This is just a name that classifies the type of limits having such property.
)(
)(lim
)(
)(lim
xh
xgOr
xh
xgxax
Limits at infinity
A function y=f(x) may approach a real number b as x increases or decreases with no bound.When this happens, we say that f has a limit at infinity, and that the line y=b is a horizontal asymptote for f.
Limit at infinity: The Case of Rational Functions
A rational function r(x) = p(x)/q(x) has a limit at infinity if the degree of p(x) is equal or less than the degree of q(x).
A rational function r(x) = p(x)/q(x) does not have a limit at infinity (but has rather infinite right and left limits) if the degree of p(x) is greater than the degree of q(x).
Example (1)Let
Find
Solution:Since the degree of the polynomial in the numerator, which is 9, is equal to the degree of the polynomial in the denominator, then
146
325)(
79
29
xx
xxxf
)(lim xfx
To show that, we follow the following steps:
6
5
)(lim9
9
inatordenomtheinxofcofficientThe
numeratortheinxofcofficientThexf
x
6
5
0)0(46
)0(3)0(25
1lim
1lim46lim
1lim3
1lim25lim
1146lim
3125lim
1146
3125
lim146
325lim)(lim
92
97
92
97
92
97
79
29
xx
xx
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xxxx
xxxf
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x
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Example (2)Let
Find
Solution:Since the degree of the polynomial in the numerator, which is 9, is less than the degree of the polynomial in the denominator, which is 12, then
146
325)(
712
29
xx
xxxf
)(lim xfx
To show that, we follow the following steps:
0)(lim
xfx
06
0
0)0(46
)0(3)0(2)0(5
1lim
1lim46lim
1lim3
1lim2
1lim5
1146lim
312
5lim
1146
312
5
lim146
325lim)(lim
125
12103
125
12103
125
12103
712
29
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Example (3)Let
Find
Solution:Since the degree of the polynomial in the numerator, which is 12, is greater than the degree of the polynomial in the denominator, which is 9, then
146
325)(
79
212
xx
xxxf
)(lim xfx
They are infinite limits. To show that, we follow the following steps:
3
9
12
79
212
lim
6
5lim
146
325lim)(lim
xassamethearewhich
x
xassametheare
xx
xxxf
x
x
xx
.)(lim existnotdoxfx
Example (4)Let
Find
Solution:Since the degree of the polynomial in the numerator, which is 12, is greater than the degree of the polynomial in the denominator, which is 9, then
146
325)(
79
212
xx
xxxf
)(lim xfx
They are infinite limits. To show that, we follow the following steps:
)(lim
6
5lim
146
325lim)(lim
3
9
12
79
212
xassamethearewhich
x
xassametheare
xx
xxxf
x
x
xx
.)(lim existnotdoxfx
Example (5)Let
Find
Solution:Since the degree of the polynomial in the numerator, which is 12, is greater than the degree of the polynomial in the denominator, which is 8, then
146
325)(
78
212
xx
xxxf
)(lim xfx
They are infinite limits. To show that, we follow the following steps:
.)(lim existnotdoxfx
4
8
12
78
212
lim
6
5lim
146
325lim)(lim
xassamethearewhich
x
xassametheare
xx
xxxf
x
x
xx
Example (6)Let
Find
Solution:Since the degree of the polynomial in the numerator, which is 12, is greater than the degree of the polynomial in the denominator, which is 8, then
146
325)(
78
212
xx
xxxf
)(lim xfx
They are infinite limits. To show that, we follow the following steps:
.)(lim existnotdoxfx
)(lim
6
5lim
146
325lim)(lim
4
8
12
78
212
xassamethearewhich
x
xassametheare
xx
xxxf
x
x
xx
Limits & Infinity
Problems Involving Roots
Introduction
We know that:
√x2 = |x|, which is equal x is x non-negative and equal to – x if x is negative
For if x = 2, then √(2)2 = √4 = 2 = |2|& if x = - 2, then √(-2)2 = √4 = -(-2) =|-2|
Example
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4. The case of infinite limit
Infinite Limits
A function f may increases or decreases with no bound near certain values c for the independent variable x. When this happens, we say that f has an infinite limit, and that f has a vertical asymptote at x = c The line x=c is called a vertical asymptote for f.
Infinite Limits
A function has an infinite one-sided limit at a point x=c if at that point the considered one-sided limit of the denominator is zero and that of the numerator is not zero. The sign of the infinite limit is determined by the sign of both the numerator and the denominator at values close to the considered point x=c approached by the variable x (from the considered side).
Infinite Limits- The Case of Rational Functions
A rational function has an infinite one-sided limit at a point x=c if c a zero of the denominator but not of the numerator. The sign of the infinite limit is determined by the sign of both the numerator and the denominator at values close to the considered point x=c approached by the variable x (from the considered side).
Example (1)Let
Find
Solution:First x=0 is a zero of the denominator which is not a zero of the numerator.
xxf
1)(
)(lim.0
xfax
)(lim.0
xfbx
a. As x approaches 0 from the right, the numerator is always positive ( it is equal to 1) and the denominator approaches 0 while keeping positive; hence, the function increases with no bound. Thus:
)(lim0
xfx
The function has a vertical asymptote at x = 0, which is the line x = 0 (see the graph in the file on basic algebraic functions).
b. As x approaches 0 from the left, the numerator is always positive ( it is equal to 1) and the denominator approaches 0 while keeping negative; hence, the function decreases with no bound.
)(lim0
xfx
Example (2)Let
Find
Solution:First x=1 is a zero of the denominator which is not a zero of the numerator.
)(lim.1
xfax
)(lim.1
xfbx
1
5)(
x
xxf
)(lim1
xfx
The function has a vertical asymptote at x = 1, which is the line x = 1
)(lim1
xfx
a. As x approaches 1 from the right, the numerator approaches 6 (thus keeping positive), and the denominator approaches 0 while keeping positive; hence, the function increases with no bound. Thus,
b. As x approaches 1 from the left, the numerator approaches 6 (thus keeping positive), and the denominator approaches 0 while keeping negative; hence, the function decreases with no bound. The function has a vertical asymptote at x=1, which is the line x = 1. Thus:
Example (3)Let
Find
Solution:First x=3 is a zero of the denominator which is not a zero of the numerator.
)(lim.3
xfax
)(lim.3
xfbx
)3)(1(
)4)(1(
34
45)(
2
2
xx
xx
xx
xxxf
)(lim
)033,3(
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;14343
3
4lim
)3)(1(
)4)(1(lim
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)(lim
3
3
3
2
2
3
3
xfThus
xsoandxhavewexasbecause
positivekeepingwhilexand
negativekeepingthusxxAs
x
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x
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3
3
3
2
2
3
3
xfThus
xsoandxhavewexasbecause
negativekeepingwhilexand
negativekeepingthusxxAs
x
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x
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x
x
x
6. The case constant/∞Suppose we want to find:
For the case when:
In this case, no mater what the formulas of g and h are, we will always have:
Then this is called the case c/∞. Caution: The limit is not equal c/ ∞. This is just a name that classifies the type of limits having such property. This limit is always equal zero
)(
)(lim
xh
xgax
)(lim&)(lim xhRcxgaxax
0)(
)(lim
xh
xgax
Example on the case constant/∞
01
1lim
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)(lim
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11lim)(lim1)(
:
:
1
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,
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xxxh
xg
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haveWe
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Find
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x
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x
6. Using the Squeeze Theorem
The Squeeze (Sandwich or Pinching)) Theorem
Suppose that we want to find the limit of a function f at a given point x=a and that the values of f on some interval containing this point (with the possible exception of that point) lie between the values of a couple of functions g and h whose limits at x=a are equal. The squeeze theorem says that in this case the limit of f at x=a will equal the limit of g and h at this point.
The Squeeze Theory
lxf
Then
xhlxg
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adcxxhxfxg
Let
ax
axax
)(
:
)()(
&
),(
),(;)()()(
:
lim
limlim
Example (1)
1)(
,,
)4,0(;)(12,)4,0(1
&
1)1(&112)12(
:
:
)(
)4,0(;)(12
:
lim
limlim
lim
1
2
22
11
1
2
xf
theoremsqueezethebyThus
xxxfx
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haveWe
Soluion
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Find
xxxfx
Let
x
xx
x
Example (2)
7)(
,,
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&
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lim
lim
lim
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4
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2
4
4
4
2
xf
theoremsqueezethebyThus
xxxfx
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haveWe
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Find
xxxxfx
Let
x
x
x
x
Example (3)
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sin(
,
)(0)(
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0)5,5(;1
sin
)...(
?0)5,5(;11
sin1
:
:
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sin(
2
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2
0
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Example (4)
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cos(
,
)(0)(
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0)5,5(;2
cos
)...(
?0)5,5(;12
cos1
:
:
:
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cos(
4
0
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0
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4
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lim
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Example (5)
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Example (6)
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2)3,0(;)(112
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2)3,0(;)(112
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lim
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lim
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2
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xf
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and
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Find
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Let
x
x
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x
x
x
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Example (7)
1)(
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0)2
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)2
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0)2
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:
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lim
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x
and
xxfxand
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Find
xxfx
Let
x
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Example (8)
1sin
:),8(,
0)2
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cos
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:
sin
lim
lim
0
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x
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atarriveweExampleinascontinuingThus
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x
x
Find
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Example (9)
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:
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5:
:
5
)5sin(
limlim
limlim
lim
05
55
5
t
t
x
x
Thus
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xtLet
Soluion
x
x
Find
tx
xx
x