1.2 – Algebraic Limits and Continuity. Find the following limits using the Limit Properties: 1.2 – Algebraic Limits and Continuity.

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