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1.2 – Algebraic Limits and Continuity
lim𝑥→ 4𝑥2−3 lim
𝑥→ 4𝑥+ lim
𝑥→ 47
Find the following limits using the Limit Properties:
( 4 )2−3 ( 4 )+7
1 1
limx 4
x2 3x 7
16−12+7
limx 0
x2 3x 2
√ lim𝑥→ 0𝑥2−3 𝑥+2
√ lim𝑥→ 0𝑥2−3 lim
𝑥→0𝑥+ lim
𝑥→ 02
√ (0 )2−3 (0 )+2
√2
1.2 – Algebraic Limits and Continuity
Find the requested limits for the given function.
lim𝑥→ 0
𝑓 (𝑥 )=¿¿
lim𝑥→−1
𝑓 (𝑥 )=¿¿
lim𝑥→ 2
𝑓 (𝑥 )=¿¿
(0 )2−1=¿−1
(−1 )2−1=¿0
3(2 )2−1=¿
𝑓 (𝑥 )=𝑥2−1
1.2 – Algebraic Limits and Continuity
Theorem on Limits of Rational Functions
For any rational function , with in the domain of ,
lim𝑥→ 3
1
4−𝑥2 =¿¿
Find the following limits:
−15
lim𝑥→5
𝑥𝑥2− 𝑥
=¿¿ 14
lim𝑥→−4
𝑥−6
𝑥2−36=¿¿ 1
2
lim𝑥→ 6
𝑥+2
𝑥2+𝑥−10=¿¿ 1
4
1
4− (3 )2=¿
5
(5 )2−5=¿
−4−6
(4 )2−36=¿
6+2
(6 )2+6−10=¿
1.2 – Algebraic Limits and Continuity
Find the following limits:
lim𝑥→0
𝑥𝑥2−𝑥
=¿¿
−1
lim𝑥→ 6
𝑥−6
𝑥2−36=¿¿
112
lim𝑥→ 0
𝑥𝑥 (𝑥−1)
=¿¿
lim𝑥→ 6
𝑥−6(𝑥+6 ) (𝑥−6 )
=¿¿
lim𝑥→0
1(𝑥−1)
=¿¿
lim𝑥→ 6
1(𝑥+6 )
=¿¿
00𝑖𝑛𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑡𝑒 𝑓𝑜𝑟𝑚
00𝑖𝑛𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑡𝑒 𝑓𝑜𝑟𝑚
1.2 – Algebraic Limits and Continuity
Find the following limits:
−17
lim𝑥→−2
𝑥+2
𝑥2−3 𝑥−10=¿¿
lim𝑥→−2
𝑥+2(𝑥+2 ) (𝑥−5 )
=¿¿ lim𝑥→−2
1(𝑥−5 )
=¿¿
00𝑖𝑛𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑡𝑒 𝑓𝑜𝑟𝑚
1.2 – Algebraic Limits and Continuity
A continuous function is one that can be plotted without the plot being broken.
Is the graph of f(x) a continuous function on the interval [0, 4]? No
At what values of x is the function discontinuous and why?
𝑥=1 h𝑇 𝑒𝑟𝑒𝑖𝑠 𝑎 𝑗𝑢𝑚𝑝 .𝑥=2 h𝑇 𝑒𝑟𝑒𝑖𝑠 𝑎h𝑜𝑙𝑒 .𝑥=4 h𝑇 𝑒𝑟𝑒𝑖𝑠 𝑎 𝑗𝑢𝑚𝑝 .
Is the graph of f(x) continuous at ? Yes
1.2 – Algebraic Limits and Continuity
lim𝑥→ 3−
𝑓 (𝑥 )=¿¿2 lim𝑥→3+¿ 𝑓 (𝑥 )=¿¿ ¿
¿2
lim𝑥→ 3
𝑓 (𝑥 )=¿¿2 𝑓 (3 )=¿2
lim𝑥→ 1−
𝑓 (𝑥 )=¿ ¿0 lim𝑥→1+¿ 𝑓 (𝑥 )=¿ ¿¿
¿1
lim𝑥→ 1
𝑓 (𝑥 )=¿¿𝐷𝑁𝐸 𝑓 (1 )=¿1
lim𝑥→ 2−
𝑓 (𝑥 )=¿¿1 lim𝑥→2+¿ 𝑓 (𝑥 )=¿ ¿ ¿
¿1
lim𝑥→ 2
𝑓 (𝑥 )=¿¿1 𝑓 (2 )=¿2
lim𝑥→ 4−
𝑓 (𝑥 )=¿¿1 lim𝑥→4+¿ 𝑓 (𝑥 )=¿ ¿¿
¿𝑛𝑜𝑛𝑒
lim𝑥→ 4
𝑓 (𝑥 )=¿¿𝑛𝑜𝑛𝑒 𝑓 (4 )=¿0.5
What are the rules for continuity at a point?
1.2 – Algebraic Limits and Continuity
lim𝑥→𝑐
𝑓 (𝑥 )𝑒𝑥𝑖𝑠𝑡𝑠
∴ 𝑓 (𝑥 ) 𝑖𝑠𝑛𝑜𝑡 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠𝑎𝑡 𝑥=1.
lim𝑥→ 1
𝑓 (𝑥 )=𝐷𝑁𝐸
𝑓 (1 )=1𝑥=1
𝑓 (𝑐 )𝑒𝑥𝑖𝑠𝑡𝑠
1.2 – Algebraic Limits and Continuity
lim𝑥→𝑐
𝑓 (𝑥 )𝑒𝑥𝑖𝑠𝑡𝑠
∴ 𝑓 (𝑥 ) 𝑖𝑠𝑛𝑜𝑡 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠𝑎𝑡 𝑥=2.
lim𝑥→ 2
𝑓 (𝑥 )=1
𝑓 (2 )=2𝑥=2
𝑓 (𝑐 )𝑒𝑥𝑖𝑠𝑡𝑠
lim𝑥→𝑐
𝑓 (𝑥 )= 𝑓 (𝑐) 2≠1
1.2 – Algebraic Limits and Continuity
lim𝑥→𝑐
𝑓 (𝑥 )𝑒𝑥𝑖𝑠𝑡𝑠
∴ 𝑓 (𝑥 ) 𝑖𝑠𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠𝑎𝑡 𝑥=3.
lim𝑥→ 3
𝑓 (𝑥 )=2
𝑓 (2 )=2𝑥=3
𝑓 (𝑐 )𝑒𝑥𝑖𝑠𝑡𝑠
lim𝑥→𝑐
𝑓 (𝑥 )= 𝑓 (𝑐) 2=2
1.2 – Algebraic Limits and Continuity
1.2 – Algebraic Limits and Continuity
2012 Pearson Education, Inc. All rights reserved
Is the following function continuous at x = 3?
f (x) x2 5
lim𝑥→𝑐
𝑓 (𝑥 )𝑒𝑥𝑖𝑠𝑡𝑠𝑓 (3 )=¿ (3 )2−5=4𝑓 (𝑐 )𝑒𝑥𝑖𝑠𝑡𝑠
lim𝑥→ 3
𝑓 (𝑥 )=¿¿
∴
(3 )2−5=4
lim𝑥→𝑐
𝑓 (𝑥 )= 𝑓 (𝑐) 4=4
1.2 – Algebraic Limits and Continuity
2012 Pearson Education, Inc. All rights reserved
Is the following function continuous at x = –2?
g(x) 1
2x 3, for x 2
x 1, for x 2
lim𝑥→𝑐
𝑓 (𝑥 )𝑒𝑥𝑖𝑠𝑡𝑠
∴
lim𝑥→−2−
𝑔 (𝑥 )=¿¿
𝑔 (−2 )=¿ (−2 )−1=−3𝑓 (𝑐 )𝑒𝑥𝑖𝑠𝑡𝑠lim𝑥→−2
𝑔 (𝑥 )
2≠−3
12
(−2 )+3=2
lim𝑥→−2+¿𝑔 (𝑥 )=¿ ¿ ¿
¿(−2 )−1=−3
lim𝑥→−2
𝑔 (𝑥 )=𝐷𝑁𝐸
∴
1.3 – Average Rates of Change
The average rate of change of y with respect to x, as x changes from x1 to x2, is the ratio of the change in output to the change in input:
Definition:
y2 y1
x2 x1
, where x2 ≠ x1.
Examining the graph of the function, the average rate of change and the slope of the line from P(x1, y1) to Q(x2, y2) are the same. The line through P and Q, is called a secant line.
1.3 – Average Rates of Change
y2 y1
x2 x1
f (x2 ) f (x1)
x2 x1
,
1.3 – Average Rates of ChangeAverage Rate of Change or Slope of the Secant Line
3 3+2
h=2
𝑦= 𝑓 (𝑥 )
(3 , 𝑓 (3 ) )
(3+2 , 𝑓 (3+2 ) )
3 5
h=5−3=2
𝑦= 𝑓 (𝑥 )
(3 , 𝑓 (3 ) )
(5 , 𝑓 (5 ) )
𝑓 (5 )− 𝑓 (3 )2
𝑓 (3+2 )− 𝑓 (3 )2
Secant line
1.3 – Average Rates of ChangeAverage Rate of Change or Slope of the Secant Line
3+h
h
(3 , 𝑓 (3 ) )
(3+h , 𝑓 (3+h ) )
3
𝑦= 𝑓 (𝑥 )
Secant line
𝑥 𝑥+h
h
(𝑥 , 𝑓 (𝑥 ) )
(𝑥 , 𝑓 (𝑥+h ) )𝑓 (𝑥+h )
𝑓 (𝑥 )
𝑦= 𝑓 (𝑥 )
𝑓 (3+h )− 𝑓 (3 )h
𝑓 (𝑥+h )− 𝑓 (𝑥 )h
1.3 – Average Rates of Change
𝐴𝑣𝑒𝑟𝑎𝑔𝑒𝑅𝑎𝑡𝑒𝑜𝑓 h𝐶 𝑎𝑛𝑔𝑒=𝑓 (𝑥+h )− 𝑓 (𝑥 )
h
Average Rate of Change or Slope of the Secant Line
The Difference Quotient
Find the simplified difference quotient and the values of the difference quotient for the given values of x and h.
𝑓 (𝑥 )=𝑥2−𝑥𝑥=5 , h=0.1𝑎𝑛𝑑 𝑥=5 ,h=0.01𝑓 (𝑥+h )− 𝑓 (𝑥 )
h(𝑥+h )2− (𝑥+h )− (𝑥2−𝑥 )
h
𝑥2+2 h𝑥 +h2−𝑥−h−𝑥2+𝑥h
1.3 – Average Rates of Change
𝑥2+2 h𝑥 +h2−𝑥−h−𝑥2+𝑥h
2 h𝑥 +h2−hh
=¿h (2𝑥+h−1 )
h=¿2 𝑥+h−1
𝑥=5 , h=0.12 𝑥+h−1
2 (5 )+0.1−1
9 .1
𝑥=5 , h=0.012 𝑥+h−1
2 (5 )+0.01−1
9 .01
𝑎𝑣𝑒𝑟𝑎𝑔𝑒𝑟𝑎𝑡𝑒𝑜𝑓 h𝑐 𝑎𝑛𝑔𝑒𝑠𝑙𝑜𝑝𝑒𝑜𝑓 h𝑡 𝑒𝑠𝑒𝑐𝑎𝑛𝑡 𝑙𝑖𝑛𝑒
𝑣 𝑎𝑙𝑢𝑒𝑜𝑓 h𝑡 𝑒𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒𝑞𝑢𝑜𝑡𝑖𝑒𝑛𝑡𝑎𝑣𝑒𝑟𝑎𝑔𝑒𝑟𝑎𝑡𝑒𝑜𝑓 h𝑐 𝑎𝑛𝑔𝑒𝑠𝑙𝑜𝑝𝑒𝑜𝑓 h𝑡 𝑒𝑠𝑒𝑐𝑎𝑛𝑡 𝑙𝑖𝑛𝑒
𝑣 𝑎𝑙𝑢𝑒𝑜𝑓 h𝑡 𝑒𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒𝑞𝑢𝑜𝑡𝑖𝑒𝑛𝑡
1.3 – Average Rates of Change
𝐴𝑣𝑒𝑟𝑎𝑔𝑒𝑅𝑎𝑡𝑒𝑜𝑓 h𝐶 𝑎𝑛𝑔𝑒=𝑓 (𝑥+h )− 𝑓 (𝑥 )
h
Average Rate of Change or Slope of the Secant Line
The Difference Quotient
Find the simplified difference quotient and the values of the difference quotient for the given values of x and h.
𝑓 (𝑥 )= 9𝑥 𝑥=3 , h=0.01
𝑓 (𝑥+h )− 𝑓 (𝑥 )h
9𝑥+h
−9𝑥
h
𝐿𝐶𝐷 :𝑥 (𝑥+h )𝑥𝑥∙
9𝑥+h
−9𝑥∙𝑥+h𝑥+h
h9 𝑥
𝑥 (𝑥+h )−
9𝑥+9h𝑥 (𝑥+h )
h
1.3 – Average Rates of Change
𝑥=3 , h=0.01
9 𝑥𝑥 (𝑥+h )
−9𝑥+9h𝑥 (𝑥+h )
h
9 𝑥−9𝑥−9h𝑥 (𝑥+h )h
=¿−9h
h𝑥 (𝑥+h )=¿
−9𝑥 (𝑥+h )
−9𝑥 (𝑥+h )−9
3 (3+0.01 )=¿−99.03
=¿−0.9967 𝑎𝑣𝑒𝑟𝑎𝑔𝑒𝑟𝑎𝑡𝑒𝑜𝑓 h𝑐 𝑎𝑛𝑔𝑒𝑠𝑙𝑜𝑝𝑒𝑜𝑓 h𝑡 𝑒𝑠𝑒𝑐𝑎𝑛𝑡 𝑙𝑖𝑛𝑒
𝑣 𝑎𝑙𝑢𝑒𝑜𝑓 h𝑡 𝑒𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒𝑞𝑢𝑜𝑡𝑖𝑒𝑛𝑡
1.3 – Average Rates of Change
𝐼𝑛𝑡 h𝑜𝑢𝑟𝑠 ,𝑎 𝑡𝑟𝑢𝑐𝑘𝑡𝑟𝑎𝑣𝑒𝑙𝑠 𝑠 (𝑡 )𝑚𝑖𝑙𝑒𝑠 , h𝑤 𝑒𝑟𝑒 𝑠 (𝑡 )=10 𝑡 2 .𝑎 ¿𝐹𝑖𝑛𝑑𝑠 (5 )−𝑠 (2 ) . h𝑊 𝑎𝑡𝑑𝑜𝑒𝑠 𝑖𝑡 𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡?
.
𝑎 ¿𝑠 (5 )−𝑠 (2 )
10 (5 )2−10 (2 )2
250−40210210 miles were traveled from 2 to 5 hours.
𝑠 (5 )−𝑠 (2 )5−2
=¿250−405−2
=¿
2103
=¿70𝑚𝑖𝑙𝑒𝑠𝑝𝑒𝑟 h𝑜𝑢𝑟
1.3 – Average Rates of Change