1
Chapter Two
Limit and Continuity
Lec. Hind Y. Saleh
2
Distance between points and Distance between point and
the line
Distance between two points A=(𝑥1 , 𝑦1 ) and B=(𝑥2 , 𝑦2 ) is defined by
𝑑 𝐴, 𝐵 =
(𝑥2 − 𝑥1 )2 +(𝑦2 − 𝑦1 )2
Example.
The distance between two points (-1,2) and (3,4) is
𝑑 = (3 − (−1))2 +(4 − 2)2 = 20
Distance between point and the line defined by
𝐴𝑥 + 𝐵𝑦 + 𝑐
𝑑=
𝐴2 + 𝐵 2
Example.
The distance between P=(3,4) and the line 2𝑥 + 3𝑦 + 1 is
2 3 +3 4 +1
19
𝑑=
=
2
2
13
2 +3
Lec. Hind Y. Saleh
Increasing and Decreasing Functions
3
Definition: If 𝑓(𝑥) is a given function for 𝑥 in the interval 𝑎, 𝑏 ,
𝑎≤𝑥≤𝑏
𝑓 is an increasing function if the value of 𝑓 increasing as 𝑥 increasing
in this interval.
𝑖. 𝑒 𝑓 𝑥1 < 𝑓(𝑥2 ) when
𝑥1 < 𝑥2 , 𝑥1 , 𝑥2 ∈ ,𝑎, 𝑏-
𝑓 is an decreasing function if the values of 𝑓 decrease as 𝑥
increasing in this interval.
𝑖. 𝑒 𝑓 𝑥1 > 𝑓(𝑥2 ) when
𝑥1 < 𝑥2 ,
𝑥1 , 𝑥2 ∈ ,𝑎, 𝑏-
Example: let 𝑓 𝑥 = 𝑥 2 on ,0,4- the function is increasing because
𝑓 𝑥1 < 𝑓(𝑥2 ) when 𝑥1 < 𝑥2
let 𝑓 𝑥 = 𝑥 2 on ,−4, 0- the function is decreasing because
𝑓 𝑥1 > 𝑓(𝑥2 ) when 𝑥1 < 𝑥2
Lec. Hind Y. Saleh
4
 Exercise: find the intervals that make the functions increasing and
decreasing of the following:
 𝑓 𝑥 = sin 𝑥
on ,−4𝜋, 4𝜋 𝑓 𝑥 = cos 𝑥
on ,−3𝜋, 5𝜋1
2
𝑓 𝑥 = 𝑥+1
on ,2.2, 6.1-
𝑓 𝑥 = 𝑥
on ,0, 10-
𝑓 𝑥 =
1
𝑥 2 :1
on ,−5, 10-
𝑓 𝑥 = 𝑥+2
 𝑓 𝑥 = 𝑥3 + 2
on ,−5, 10on ,−2, 2-
 𝑓 𝑥 = ln(𝑥 + 1)
 𝑓 𝑥 = 𝑒 ;𝑥 + 3
 𝑓 𝑥 = cosh 𝑥
on ,0,5on ,−10, 10on ,−10, 10-
Lec. Hind Y. Saleh
5
Slope.
The Slope of nonvertical line is
𝑟𝑖𝑠𝑒
∆𝑦 𝑦2 − 𝑦1
𝑚=
=
=
𝑟𝑢𝑛
∆𝑥 𝑥2 − 𝑥1
The slope of a line is the tangent of the lines angle of inclination. If 𝑚
denotes the slope and ∅ is the angle, then:
𝑚 = 𝑡𝑎𝑛∅
Example.
𝜋
Find the slope of the line that crosses the x-axis with angle ∅ = .
4
Solution.
𝑃2 L
𝑚 = 𝑡𝑎𝑛∅ = 𝑡𝑎𝑛 45° = 1
∆𝑦
𝑃1
Lec. Hind Y. Saleh
∅
∆𝑥
6
Lines that are parallel or perpendicular
Parallel lines have equal angles of inclination. Hence, if they are not
vertical, parallel lines have the same slope.
Conversely is true if neither of two perpendicular lines 𝐿1 and 𝐿2 is
vertical, their slope 𝑚1 𝑎𝑛𝑑 𝑚2 are related by the equation
𝑚1 . 𝑚2 = −1.
If 𝑚1 = 𝑚2 , then ∅1 = ∅2 and the lines are parallel
𝑆𝑙𝑜𝑝𝑒 𝑚1
𝐿1
𝑆𝑙𝑜𝑝𝑒 𝑚2 𝐿2
𝑚1
∅1
1
Lec. Hind Y. Saleh
𝑚2
∅2
1
7
Point – Slope equation
The equation 𝑦 − 𝑦1 = 𝑚 𝑥 − 𝑥1 is the point- slope equations of the
line that passes through the point 𝑥1 , 𝑦1 with slope 𝑚.
Examples. Write the equation for the line that passes the point
3
2
i- (2,3) with the slope − .
Solution:
𝑦 − 𝑦1 = 𝑚 𝑥 − 𝑥1
3
𝑦−3=− 𝑥−2
2
3
𝑦 =− 𝑥+6
2
6
0
Lec. Hind Y. Saleh
3
𝑦 =− 𝑥+6
2
4
8
Slope –Intercept Equations
If we take the point (𝑥1 , 𝑦1 ) = 0, 𝑏 in the point slope equation for the
line , we find that
𝑦−𝑏 =𝑚 𝑥−0
Hence 𝑦 = 𝑚𝑥 + 𝑏
Hence, the equation 𝑦 = 𝑚𝑥 + 𝑏 is the slope – intercept equation of
the line with slope 𝑚 𝑎𝑛𝑑 𝑦 −intercept 𝑏 .
Example.
1-The slope –intercept equation of line with slope 2 and 𝑦 − intercept
5 is 𝑦 = 2𝑥 + 5
2-Find the slope and 𝑦 −intercept of the line 8𝑥 + 5𝑦 = 20
Solution: H.W
Lec. Hind Y. Saleh
9
Limit and Law of Limit
Definition: if the values of a function 𝑓 of 𝑥 approach the value 𝐿 as
𝑥 approaches 𝑐 we say 𝑓 has limit 𝐿 as 𝑥 approaches to 𝑐 and we
write
lim 𝑓 𝑥 = 𝐿
𝑥→𝑐
Example. As table and Fig. suggest
lim 2𝑥 + 1 = 5
𝑥→2
𝑦
𝑦 = 2𝑥 + 1
5
1
0
2
Lec. Hind Y. Saleh
10
The Limit Law
If 𝐿, 𝑀, 𝑐, and 𝑘 are real numbers and 𝑙𝑖𝑚 𝑓(𝑥) = 𝐿and 𝑙𝑖𝑚 𝑔(𝑥) = 𝑀,
𝑥→𝑐
𝑥→𝑐
then:
1. Sum Rule: 𝑙𝑖𝑚( 𝑓(𝑥) + 𝑔 𝑥 ) = 𝐿 + 𝑀
𝑥→𝑐
2.
Difference Rule: 𝑙𝑖𝑚(𝑓(𝑥) − 𝑔 𝑥 ) = 𝐿 − 𝑀
𝑥→𝑐
3. Constant Multiple Rule: 𝑙𝑖𝑚(𝐾. 𝑓(𝑥)) = 𝐾. 𝑀 , 𝑎𝑛𝑦 𝑛𝑢𝑚𝑏𝑒𝑟 𝐾
𝑥→𝑐
4. Product Rule: 𝑙𝑖𝑚 𝑓 𝑥 . 𝑔 𝑥
𝑥→𝑐
5. Quotient Rule:
6.
𝑓(𝑥)
𝑙𝑖𝑚
𝑥→𝑐 𝑔(𝑥)
Power Rule: 𝑙𝑖𝑚 𝑓 𝑥
𝑥→𝑐
Lec. Hind Y. Saleh
𝑟
𝑛
=
𝐿
,
𝑀
= 𝐿
= 𝐿. 𝑀
𝑀 ≠0
𝑟
𝑛
,𝑛≠0
1. If 𝑓 𝑥 = 𝑘 , (𝑓 is constant function) , then 𝑙𝑖𝑚 𝑓(𝑥) = 𝑙𝑖𝑚 𝑘 = 𝑘.
𝑥→𝑐
11
𝑥→𝑐
2. If 𝑓 𝑥 = 𝑥 (𝑓 is identity function) , then 𝑙𝑖𝑚 𝑓(𝑥) = 𝑙𝑖𝑚 𝑥 = 𝑐.
𝑥→𝑐
𝑥→𝑐
3. If 𝑓 𝑥 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑘 (𝑓 is quadratic function) , then
4. 𝑙𝑖𝑚 𝑓(𝑥) = 𝑙𝑖𝑚 𝑎𝑥 2 + 𝑙𝑖𝑚𝑏𝑥 + 𝑙𝑖𝑚𝑘 = 𝑎𝑐 2 + 𝑏𝑐 + 𝑘
𝑥→𝑐
𝑥→𝑐
𝑥→𝑐
𝑥→𝑐
Examples. Find
1). 𝑙𝑖𝑚 𝑥 2 (2 − 𝑥)
𝑥→3
2).
𝑥 2 :4𝑥:4
𝑙𝑖𝑚
𝑥→2 𝑥:2
Solution.
1) 𝑙𝑖𝑚 𝑥 2 (2 − 𝑥) = 32 . 2 − 3 = −9
𝑥→3
2)
or
𝑥 2 :2𝑥:4
𝑙𝑖𝑚
𝑥→2 𝑥:2
=
4:8:4
4
=4
𝑥 2 :4𝑥:4
(𝑥:2)2
𝑙𝑖𝑚
= 𝑙𝑖𝑚
𝑥→2 Lec.𝑥:2
𝑥→2 𝑥:2
Hind Y. Saleh
= 𝑙𝑖𝑚 𝑥 + 2 = 4
𝑥→2
12
Right-hand and Left-hand limit
Right hand limit:
lim+ 𝑓 𝑥 = 𝐿, means that 𝑓 𝑥 → 𝐿 as 𝑥 → 𝑐 from right .
𝑥→𝑐
 Left hand limit:
lim 𝑓 𝑥 = 𝐿, means that 𝑓 𝑥 → 𝐿 as 𝑥 → 𝑐 from left.
𝑥→𝑐 −
 Limit exist:
lim 𝑓 𝑥 exist and equal to 𝐿 , if and only if lim+ 𝑓 𝑥 = lim− 𝑓 𝑥 = 𝐿.
𝑥→𝑐
𝑥→𝑐
𝑥2 − 2 , 𝑥 ≥ 2
Example 1. If 𝑓 𝑥 =
then
𝑥,
𝑥<2
lim+ 𝑓 𝑥 = lim+ (𝑥 2 − 2) = 2, and lim− 𝑓 𝑥 = 2,
𝑥→2
𝑥→2
𝑥→2
𝑖. 𝑒 Limit exist. Since lim+ 𝑓 𝑥 = lim− 𝑓 𝑥 = 2.
Lec. Hind Y. Saleh
𝑥→2
𝑥→2
𝑥→𝑐
𝑥
13
 Example 2. Find lim
𝑥→0 𝑥
𝑥, 𝑥 ≥ 0
Solution: since 𝑥 =
−𝑥, 𝑥 < 0
𝑥
𝑥
then lim+
= lim+
= 1, and
𝑥
𝑥→0
𝑥
lim
𝑥→0− 𝑥
𝑖. 𝑒 lim
𝑥→0
=
𝑥
𝑥
𝑥→0
;𝑥
lim
𝑥→0− 𝑥
𝑥
= −1,
does not exist.
Because lim+ 𝑓 𝑥 = 1 ≠ −1 = lim− 𝑓 𝑥 .
𝑥→0
 Example 3 . Find lim 𝑥 .
𝑥→0
𝑥→0
Solution: H.W
Note. To find Limits, use substitution
a) If you get real number, then the limit exist. Or ∞, −∞, then the limit does not
exist.
0
∞
b) If you get , , 1∞ , ∞0 , ∞ − ∞, you have some operations before decided
0 ∞
the limit exist or not (this operations : factors analysis, multiply with
conjugate,
simplification).
Lec. Hind Y. Saleh
 Example 4. Find
14
 Solution.
𝑥 2 ;4
lim
𝑥→2 𝑥;2
𝑥 2 ;4
lim
𝑥→2 𝑥;2
=
.
0
0
𝑥2 − 4
(𝑥 + 2)(𝑥 − 2)
∴ lim
= lim
= lim 𝑥 + 2 = 4
𝑥→2 𝑥 − 2
𝑥→2
𝑥→2
𝑥−2
𝑥;1
𝑥;1
 Example 5 Find lim
𝑥→1
 Solution. lim
𝑥→1
∴ lim
𝑥→1
𝑥;1
𝑥;1
0
0
𝑥−1
( 𝑥 + 1)( 𝑥 − 1)
= lim
= lim
𝑥→1
𝑥→1
𝑥−1
𝑥−1
Example 6.

=
.
1
Find lim 2
𝑥→2 𝑥 ;4
1
1
Solution. lim 2
= =∞
0
𝑥→2 𝑥 ;4
Lec. Hind Y. Saleh
𝑥 + 1 =2
𝑥:9;3
𝑥
Example 7. Find lim
15
Solution:lim
𝑥→0
𝑥→0
𝑥:9;3
𝑥
=
.
0
0
𝑥+9−3
( 𝑥 + 9 − 9)( 𝑥 + 9 + 3)
= lim
𝑥→0
𝑥→0
𝑥
𝑥 𝑥+9+3
𝑥+9−9
1
1
= lim
= lim
= .
𝑥→0 𝑥 𝑥 + 9 + 3
𝑥→0
6
𝑥+9+3
Example 8. find lim− 𝑥 − 𝑥
∴ lim
𝑥→3
Solution. lim− 𝑥 − 𝑥
= lim− 𝑥 − lim− 𝑥
𝑥→3
𝑥→3
Example 9 Find lim 𝑥 − 1 .
𝑥→3
Solution:
∴ lim+ 𝑥 − 1
𝑥→3
= lim+ 2 = 2
And lim− 𝑥 − 1
𝑥→3
𝑥→3
= lim− 1 = 1
𝑥→3
∴ lim 𝑥 −
not exist.
Lec.1
Hind Y.does
Saleh
𝑥→3
𝑥→3
= 3−2=1
Example 10. Evaluate
16
Solution.
Lec. Hind Y. Saleh
Exercise Find
17
1-
𝑡𝑎𝑛𝑥;𝑠𝑖𝑛𝑥
lim
𝑠𝑖𝑛3 𝑥
𝑥→0
𝑥 2 ;25
2− lim
𝑥→5 3𝑥;15
3-Find
.
.
1;𝑐𝑜𝑠𝑥
lim
𝑠𝑖𝑛𝑥
𝑥→0
3 − 𝑥,
4- if 𝑓 𝑥 = 𝑥 + 1,
2
Lec. Hind Y. Saleh
.
𝑥<2
lim 𝑓 𝑥 exist?
𝑥 > 2 , Is 𝑥→2
Limit involving infinity
18
 Limit as 𝑥 → ∞ or 𝑥 → −∞,
for example
1
𝑥→∞ 𝑥
lim
1
𝑥→;∞ 𝑥
= 0,
lim
Examples.
 lim 5 +
𝑥→∞
4
𝑥→;∞ 𝑥 2
 lim
𝑦=
1
𝑥
= lim 5 +
𝑥→∞
4
𝑥→;∞ 𝑥 2
= lim
1
𝑥
=0
1
𝑥→∞ 𝑥
= lim 5 + lim
𝑥→∞
1
𝑥→;∞ 𝑥
= lim 4 . lim
𝑥→;∞
=5+0=5
1
𝑥→;∞ 𝑥
. lim
= (4)(0)(0) = 0
Limits of Rational Functions as 𝑥 → ±∞
1.
𝑓(𝑥)
𝑥→±∞ 𝑔(𝑥)
lim
=0
if deg 𝑓 < deg(𝑔)
𝑓(𝑥)
𝑥→±∞ 𝑔(𝑥)
= 𝑘 (𝑘 is finite number) if deg 𝑓 = deg(𝑔)
𝑓(𝑥)
𝑥→±∞ 𝑔(𝑥)
=∞
2. lim
3. lim
Lec. Hind Y. Saleh
if deg 𝑓 > deg(𝑔)
1
𝑥
Examples
19
1-
;𝑥
lim
𝑥→∞ 7𝑥:4
=
;𝑥
lim
𝑥→∞ 7𝑥:4
= lim
𝑥→∞
5𝑥:2
𝑥→∞ 2𝑥 3 ;1
2- lim
3-. lim
𝑥 2 :1
𝑥→∞
𝑥;1
Lec. Hind Y. Saleh
5𝑥:2
𝑥→∞ 2𝑥 3 ;1
= lim
= lim
𝑥→∞
𝑥 2 :1
𝑥;1
𝑥
= lim
𝑥→∞
;𝑥
7𝑥 4
:
𝑥 𝑥
−1
−1
−1
=
=
4
7+0
7
7+
𝑥
= lim
𝑥→∞
= lim
𝑥→∞
5𝑥 2
:
𝑥3 𝑥3
2𝑥3 1
;
𝑥3 𝑥3
𝑥2 1
:
𝑥2 𝑥2
𝑥
1
;
𝑥2 𝑥2
= lim
𝑥→∞
5
2
:
𝑥2 𝑥3
1
2; 3
𝑥
1
= lim
𝑥→∞
1: 2
𝑥
1 1
;
𝑥 𝑥2
=
=
1:0
0;0
0:0
2;0
= 0.
= ∞.
20
Some important theorem on limit
 Theorem 1: if 𝑓(𝑥) is bounded in deleted neighborhood of 𝑥°
and if 𝑥° lim 𝑔(𝑥) = 0, then lim 𝑓 𝑥 𝑔(𝑥) = 0
𝑥→𝑥°
𝑥→𝑥°
Examples1-Find lim 𝑥 2 𝑠𝑖𝑛
Solution: 𝑓 𝑥 =
since
∴
1
𝑠𝑖𝑛
𝑥
𝑥→0
1
𝑠𝑖𝑛
𝑥
1
𝑥
is bounded ,
≤ 1 and lim 𝑔( 𝑥) = lim 𝑥 2 = 0
1
2
lim 𝑥 . 𝑠𝑖𝑛
𝑥
𝑥→0
𝑥→0
𝑥→0
=0
2-Find lim 𝑥. 𝑥
𝑥→0
Solution: Since 𝑥 is bounded in a delated neighborhood of 0
and lim 𝑥 = 0
𝑥→0
∴ lim 𝑥 . 𝑥 =0
𝑥→0
Lec. Hind Y. Saleh
21
Theorem 2. (Sandwich theorem)
if 𝑓(𝑥) ≤ 𝑔(𝑥) ≤ 𝑕 𝑥 and if lim 𝑓(𝑥) = lim 𝑕(𝑥) = 𝐿, then
𝑥→𝑥°
lim 𝑔(𝑥) = 𝐿
𝑥→𝑥°
𝑥→𝑥°
Example. Given 𝑔 𝑥 − 5 < 𝑥 2 , find lim 𝑔 𝑥
𝑥→0
2
Solution. Since 𝑔 𝑥 − 5 < 𝑥 2 ⟹ −𝑥 ≤ 𝑔 𝑥 − 5 ≤ 𝑥 2
⟹ 5 − 𝑥2 ≤ 𝑔 𝑥 ≤ 𝑥2 + 5
∴ lim 5 − 𝑥 2 = 5 and lim 𝑥 2 + 5 = 5
𝑥→0
𝑥→0
∴ lim 𝑔( 𝑥) = 5
𝑥→0
Theorem 3. (Limit of composite function)
If lim 𝑓 𝑥 = 𝐿 and 𝑔(𝑥) is continuous at 𝐿, then lim 𝑔 𝑓 𝑥
𝑥→𝑥°
𝑥→𝑥°
Example
Lec. Hind Y. Saleh
=𝑔 𝐿 .
Theorem 3. (Limit of composite function)
22
If lim 𝑓 𝑥 = 𝐿 and 𝑔(𝑥) is continuous at 𝐿, then lim 𝑔 𝑓 𝑥
𝑥→𝑥°
𝑥→𝑥°
=𝑔 𝐿 .
Example1. if 𝑓 is continuous function and 𝑓 5 = 3, lim 𝑔 𝑥 = 5, find
𝑥→1
lim 𝑓 𝑔 𝑥
𝑥→1
Solution. Since 𝑓 is continuous and lim 𝑔 𝑥 = 5, then
𝑥→1
lim 𝑓 𝑔 𝑥 = 𝑓 5 = 3.
𝑥→1
Example2. find lim
𝑥→2
Solution.
4;𝑥 2
2;𝑥
using limit of a composite function.
4;𝑥 2
(2:𝑥)(2;𝑥)
lim
lim
2;𝑥
𝑥→2 2;𝑥 𝑥→2
= lim 2 + 𝑥 = 4
𝑥→2
𝑥 is continuous at 𝑥 = 4, then lim
𝑥→2
4;𝑥 2
2;𝑥
= 4 = 2.
Example3. find lim sin(𝑐𝑜𝑠𝑥) using limit of a composite function.
𝑥→0
Solution. H.W
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Objectives
23
i.
𝑥 𝑛 ;𝑎𝑛
lim
𝑥→𝑎 𝑥;𝑎
ii.
lim 𝑠𝑖𝑛𝑥 = 0 𝑎𝑛𝑑 lim 𝑐𝑜𝑠𝑥 = 1
𝑥→0
iii.
𝑠𝑖𝑛𝑥
lim
𝑥→0 𝑥
iv.
𝑒 𝑥 ;1
lim
𝑥→0 𝑥
v.
= 𝑛𝑎𝑛;1
𝑥→0
=1
=1
1
𝑥
lim (1 + 𝑥) = 𝑒
𝑥→0
log(1:𝑥)
𝑥
𝑥→0
vi. lim
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=1
(𝑥 ≠ 𝑎)
Example
24
Solution.
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25
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26
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27
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Continuity
28
 Definition.
The function 𝑦 = 𝑓(𝑥) is continuous at 𝑥 = 𝑐 if and only if all three of
the following statements are true
1. 𝑓(𝑐) exist
2.
3.
(𝑐 lies in domain of 𝑓)
lim 𝑓 𝑥 𝑒𝑥𝑖𝑠𝑡 (𝑓 has a limit as 𝑥 → 𝑐)
𝑥→𝑐
lim 𝑓 𝑥 = 𝑓(𝑐)
𝑥→𝑐
(the limit equals the function value ).
 Definition.
The function 𝑓(𝑥) is continuous if it is continuous at each point of
its Domain.
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29
 Examples.
1-Show that 𝑓 𝑥 = 𝑥 2 + 𝑥 + 1 is continuous at 𝑥 = −1.
Solution. 1. 𝑓 −1 = (−1)2 + −1 + 1 = 1 exist
2. lim (𝑥 2 +𝑥 + 1) = 1
exist
𝑥→;1
3. lim 𝑓 𝑥 = 𝑓 −1 = 1
𝑥→;1
∴ 𝑓(𝑥) is continuous at 𝑥 = −1
2
𝑖𝑓 𝑥 = 1
2- Let 𝑓 𝑥 = 𝑥 2 ;1
, Is 𝑓(𝑥) continuous at 𝑥 = 1.
𝑖𝑓 𝑥 ≠ 1
𝑥;1
Solution. 1. 𝑓 1 = 2
2. lim 𝑓(𝑥) =
𝑥→1
𝑥 2 ;1
lim
𝑥→1 𝑥;1
exist
(𝑥:1)(𝑥;1)
𝑥;1
𝑥→1
= lim
3. lim 𝑓 𝑥 = 𝑓 1 = 2
𝑥→1
∴ 𝑓(𝑥) is continuous at 𝑥 = 1.
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= lim 𝑥 + 1 = 2 exist
𝑥→1
3-Let 𝑓 𝑥 =
30
2𝑥
𝑥 2 ;9
𝑥;3
𝑖𝑓 𝑥 ≤ 3
,
𝑖𝑓 𝑥 > 3
A-Is 𝑓(𝑥) continuous at 𝑥 = 3.
B- Is 𝑓(𝑥) continuous at 𝑥 = 4.
C- Is 𝑓(𝑥) continuous at 𝑥 = 2.
Solution. H.W
4- Let 𝑓 𝑥 =
at 𝑥 = 3.
𝑥 2 ;9
𝑥;3
𝑘𝑥 − 3
𝑖𝑓 𝑥 < 3
𝑖𝑓 𝑥 ≥ 3
, find the value of 𝑘 that makes 𝑓(𝑥) is continuous
Solution. Since 𝑓(𝑥) is continuous at x=3 then lim 𝑓 𝑥 = 𝑓(3)
𝑥2
𝑥→3
−9
lim
= 3𝑘 − 3
𝑥→3 𝑥 − 3
(𝑥 + 3)(𝑥 − 3)
lim
= 3𝑘 − 3
𝑥→3
𝑥−3
lim 𝑥 + 3 = 3𝑘 − 3
𝑥→3
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6 = 3𝑘 − 3 ⟹ 3𝑘 = 9 ⟹ 𝑘 = 3.
31
𝑥
5- Let 𝑓 𝑥 = 𝑘𝑥 − 𝑕
−2𝑥
𝑖𝑓 𝑥 ≤ 1
𝑖𝑓 1 < 𝑥 < 4 , is continuous , find the value
𝑖𝑓 𝑥 ≥ 4
of h, 𝑘 .
Solution. since 𝑓(𝑥) is continuous, then the limit exist.
lim+ 𝑓 𝑥 = lim− 𝑓 𝑥 ⟹ lim+ 𝑘𝑥 − 𝑕 = lim− 𝑥
𝑥→1
𝑥→1
𝑥→1
𝑥→1
∴𝑘−𝑕 =1
… 1
lim+ 𝑓 𝑥 = lim− 𝑓 𝑥 ⟹ lim+ −2𝑥 = lim− 𝑘𝑥 − 𝑕
𝑥→4
𝑥→4
∴ −8 = 4𝑘 − 𝑕
𝑥→4
𝑥→4
… 2
Solving the two equations (1),(2), we get
𝑘 = −3 and 𝑕 = −4.
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Exercise.
32
2𝑥 + 1
1- Let 𝑓 𝑥 =
3𝑥 − 1
2- if 𝑓 𝑥 =
𝑖𝑓 𝑥 > 2
, Is 𝑓 𝑥 continuous.
𝑖𝑓 𝑥 < 2
2𝑥
𝑖𝑓 𝑥 > 3
𝑥2
𝑖𝑓 𝑥 ≤ 3
3- Show that f x =
𝑥
𝑥 2 :1
, Is 𝑓 𝑥 continuous or not.
is continuous at x=1
4- Show that f x = 𝑥 − 2 is continuous at x=2
5- Let 𝑓 𝑥 =
𝑥+2
3𝑥 − 2
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𝑖𝑓 𝑥 ≥ 2
𝑖𝑓 𝑥 < 2
, is continuous.
33
6- Let 𝑓 𝑥 =
𝑠𝑖𝑛2𝑥
𝑥
2
𝑖𝑓 𝑥 ≠ 0
, is continuous.
𝑖𝑓 𝑥 = 0
7- Let 𝑓 𝑥 = 𝑎𝑥 + 5 is continuous at x=1, and f(1)=4 find 𝑎.
8- Prove that 𝑓 𝑥 = 𝑡𝑎𝑛𝑥 is continuous at 𝑥 ∈ 0,
𝜋
2
.
9- Prove that 𝑓 𝑥 = 𝑐𝑜𝑠𝑥 is continuous at 𝑥 ∈ 𝑹
10- Prove that 𝑓 𝑥 = 𝑒 𝑥 is continuous at 𝑥 ∈ 𝑹
11- Let 𝑓 𝑥 =
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k𝑥 2
𝑖𝑓 𝑥 ≤ 2
3
𝑖𝑓 𝑥 > 2
, is continuous at x=2, find k.
34
 The polynomials. 𝑝 𝑥 = 𝑎° + 𝑎1 𝑥 + 𝑎2 𝑥 2 + ⋯ + 𝑎𝑛 𝑥 𝑛 are continuous at
every point of it’s domain of definition.
 The rational function are continuous at every point where it’s
denominator is nonzero.
1.
𝑓 𝑥 =
2.
𝑓 𝑥 =
3.
𝑓 𝑥 =
2𝑥
is continuous at 𝑅\ 2
𝑥;2
1
is continuous at 𝑅\*0+
𝑥
3;𝑥
is continuous at 𝑅,0,1)
,𝑥-
Algebraic operation on continuous
If 𝑓(𝑥) and 𝑔(𝑥) are continuous functions at 𝑥 = 𝑥° then:
1.
2.
3.
4.
5.
𝑓 𝑥 + 𝑔(𝑥) is continuous at 𝑥° .
𝑓 𝑥 − 𝑔(𝑥) is continuous at 𝑥° .
𝑓 𝑥 . 𝑔(𝑥) is continuous at 𝑥° .
𝑓 𝑥 /𝑔(𝑥) is continuous at 𝑥° , where 𝑔 𝑥° ≠ 0.
𝑘𝑓 Lec.
𝑥 HindisY.continuous
at 𝑥° , where 𝑘 is constant .
Saleh
Example. Show that 𝑓 𝑥 = 𝑥 + |𝑥| is continuous at 𝑥 = 2.
35
Solution. Clearly that 𝑓1 𝑥 = 𝑥 is continuous at 𝑥 = 2 (since 𝑓1 (𝑥) is
𝑥, 𝑥 ≥ 0
polynomial) and 𝑓2 𝑥 = 𝑥 =
, is continuous at 𝑥 = 2
−𝑥, 𝑥 < 0
∴ 𝑓 𝑥 = 𝑓1 𝑥 + 𝑓2 𝑥 = 𝑥 + |𝑥| is continuous at 𝑥 = 2.
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36
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37
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Signum Function.
38
The function f(x)=sgn(x) is defined by
−1 𝑥 < 0
𝑓 𝑥 = 0 𝑥=0
1 𝑥>1
Is discontinuous at x=0 because 𝑓 0 = 0 exist but
lim+ 𝑓(𝑥) = 1 ≠ −1 = lim− 𝑓 𝑥
𝑥→1
𝑥→1
Great Integer Function.
1
The 𝑓(𝑥) = 𝑥 is continuous at 𝑥 = but discontinuous at x=1.
2
lim+ 𝑥 = 1.1 = 1 ≠ 0 = 0.9 = lim− 𝑥
𝑥→1
𝑥→1
lim+ 𝑥 = 0.6 = 0 = 0.4 = lim− 𝑥
1
𝑥→2
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1
𝑥→2
Types of Discontinuous Functions
39
 1- Jump Discontinuity.
A function f has a jump discontinuity at x=c if lim+ 𝑓(𝑥) ≠ lim− 𝑓 𝑥
𝑥→𝑐
Example. If 𝑓 𝑥 =
𝑥+1
𝑥→𝑐
𝑖𝑓 𝑥 ≤ 0
,
𝑖𝑓 𝑥 > 0
𝑥2
Is 𝑓 𝑥 continuous at 𝑥 = 0.
Solution.
lim+ 𝑓(𝑥) = lim+ 𝑥 2 = 0 ≠ 1 = lim− 𝑓 𝑥 = lim− (𝑥 + 1)
𝑥→0
𝑥→0
𝑥→1
𝑥→1
𝑓 is discontinuous at x=0 because the left not equal right hand limit.
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40
 2- Infinite Discontinuity
A function f has an infinite discontinuity at x=c if it is unbounded in
every interval about c.
Example. If 𝑓 x =
1
𝑥2
Is 𝑓 continuous.
Solution.
1
1
lim 𝑓(𝑥) = lim 2 = = ∞
𝑥→0
𝑥→0 𝑥
0
𝑓 has infinite discontinuous at x=0.
Example. If 𝑓 x = 𝑡𝑎𝑛𝑥 Is 𝑓 continuous.
Solution.
lim𝜋 𝑓(𝑥) = lim𝜋 𝑡𝑎𝑛𝑥 = ∞
𝑥→
2
𝑥→
2
𝑓 has infinite discontinuous
at 𝑥 =
(2𝑛:1)𝜋
±
.
2
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 3-Oscillatory Discontinuity.
41
A function f has a Oscillatory discontinuity at x=c that has more
complicate type
Example. If 𝑓 x =
1
sin
𝑥
Is 𝑓 continuous.
Solution.
1
𝑥
1
0
lim 𝑓(𝑥) = lim sin = sin = sin∞
𝑥→0
𝑥→0
Limit not exists
𝑓 has Oscillatory discontinuous at x=0.
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 4-Removable Discontinuity
42
A function f has a removable discontinuity at x=c if lim 𝑓(𝑥) exists and
𝑥→0
equal 𝐿. and 𝑓 either dose not exists or 𝑓 𝑐 ≠ 𝐿.
2
𝑥
𝑥 < 1 Is 𝑓 continuous.
Example 1. If 𝑓 x =
2−𝑥 𝑥 >1
Solution.
lim+ 𝑓(𝑥) = lim+ 2 − 𝑥 = 1 = lim− 𝑓 𝑥 = lim− 𝑥 2
𝑥→0
𝑥→1
𝑥→0
𝑥→1
Limit exists and equal 1. but f(1) not exists
𝑓 has Removable discontinuous at x=1.
Example 2. If 𝑓 x =
1;𝑐𝑜𝑠𝑥
𝑥
𝑥≠0
3
𝑥=0
Is 𝑓 continuous.
Solution.
1 − 𝑐𝑜𝑠𝑥
lim 𝑓(𝑥) = lim
=0
𝑥→0
𝑥→0
𝑥
Limit exists and equal 0. but f(0)=3 not equal limit
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𝑓 has Removable
discontinuous at x=0.