big ideas constant rates of change rates of changeelyse/100/2016/limitsandvelocity.pdf · big ideas...

Big Ideas Constant Rates of Change

Rates of Change

Suppose the population of a small country was 1 million individuals in 1990, and isgrowing at a steady rate of 20,000 individuals per year.

1.2 mil

1.4 mil

Rates of Change

1.2 mil

1.4 mil

Rates of Change

1990 2000

1.2 mil

1.4 mil

Rates of Change

1990 2000

1.2 mil

1.4 mil

Rates of Change

1990 2000

1.2 mil

1.4 mil

Rates of Change

1990 2000

1.2 mil

1.4 mil

Rates of Change

1990 2000

1.2 mil

1.4 mil

Definition

The slope of a line that passes through the points (x1, y1) and (x2, y2) is “rise over run”

y2 − y1

x2 − x1.

This is also called the rate of change of the function.If a line has equation y = mx + b, its slope is m.

1990 2000

1.2 mil

1.4 mil

Rate of change:400, 000 people

20 years= 20, 000

people

year(doesn’t depend on the year)

Definition

y2 − y1

x2 − x1.

1990 2000

1.2 mil

1.4 mil

20 years= 20, 000

people

Definition

y2 − y1

x2 − x1.

1990 2000

1.2 mil

1.4 mil

20 years= 20, 000

people

(doesn’t depend on the year)

Definition

y2 − y1

x2 − x1.

1990 2000

1.2 mil

1.4 mil

20 years= 20, 000

people

Big Ideas Non-constant Rates of Change

Rates of Change

Suppose the population of a small country is given in the chart below.

1990 2000 2010 2020

0.1 mil

0.9 mil

2.5 mil

4.9 mil P(x)

0.8 mil ppl

10 years

2.4 mil ppl

10 years

Rate of change:∆ pop

∆ timedepends on time interval

Rates of Change

1990 2000 2010 2020

0.1 mil

0.9 mil

2.5 mil

4.9 mil P(x)

0.8 mil ppl

10 years

2.4 mil ppl

10 years

Rates of Change

1990 2000 2010 2020

0.1 mil

0.9 mil

2.5 mil

4.9 mil P(x)

0.8 mil ppl

10 years

2.4 mil ppl

10 years

∆ time

depends on time interval

Rates of Change

1990 2000 2010 2020

0.1 mil

0.9 mil

2.5 mil

4.9 mil P(x)

0.8 mil ppl

10 years

2.4 mil ppl

10 years

Rates of Change

1990 2000 2010 2020

0.1 mil

0.9 mil

2.5 mil

4.9 mil P(x)

0.8 mil ppl

10 years

2.4 mil ppl

10 years

Rates of Change

1990 2000 2010 2020

0.1 mil

0.9 mil

2.5 mil

4.9 mil P(x)

0.8 mil ppl

10 years

2.4 mil ppl

10 years

Rates of Change

1990 2000 2010 2020

0.1 mil

0.9 mil

2.5 mil

4.9 mil P(x)

0.8 mil ppl

10 years

2.4 mil ppl

10 years

Rates of Change

1990 2000 2010 2020

0.1 mil

0.9 mil

2.5 mil

4.9 mil P(x)

0.8 mil ppl

10 years

2.4 mil ppl

10 years

Definition

Let y = f (x) be a curve that passes through (x1, y1) and (x2, y2). Then the average rateof change of f (x) when x1 ≤ x ≤ x2 is

y2 − y1

x2 − x1

1990 2000 2010 20200.1 mil

0.9 mil

2.5 mil

4.9 mil P(x)

0.8 mil ppl

10 years

2.4 mil ppl

10 years

Average rate of change from 1990 to 2000: 80, 000 people per year.Average rate of change from 2010 to 2020: 240, 000 people per year.

Definition

y2 − y1

x2 − x1

1990 2000 2010 20200.1 mil

0.9 mil

2.5 mil

4.9 mil P(x)

0.8 mil ppl

10 years

2.4 mil ppl

10 years

Average rate of change from 1990 to 2000: 80, 000 people per year.

Average rate of change from 2010 to 2020: 240, 000 people per year.

Definition

y2 − y1

x2 − x1

1990 2000 2010 20200.1 mil

0.9 mil

2.5 mil

4.9 mil P(x)

0.8 mil ppl

10 years

2.4 mil ppl

10 years

Average rate of change from 1990 to 2000: 80, 000 people per year.Average rate of change from 2010 to 2020: 240, 000 people per year.

Average Rate of Change and Slope

What is the difference between average rate of change and slope?

Definition

y2 − y1

x2 − x1.

Definition

y2 − y1

x2 − x1

The average rate of change for a straight line is always the same, regardless of theinterval we choose. We call it the slope of the line. If a curve is not a straight line, itsaverage rate of change will differ over different intervals.

Definition

y2 − y1

x2 − x1.

Definition

y2 − y1

x2 − x1

Definition

y2 − y1

x2 − x1.

Definition

y2 − y1

x2 − x1

Big Ideas Instantaneous Rates of Change

Rates of Change

How fast is the population growing in the year 2010?

1990 2000 2010 2020

0.1 mil

0.9 mil

2.5 mil

4.9 mil P(x)

0.8 mil ppl

10 years

2.4 mil ppl

10 years

2005 2015

tangent line

secant line

Rates of Change

1990 2000 2010 2020

0.1 mil

0.9 mil

2.5 mil

4.9 mil P(x)

2005 2015

tangent line

secant line

Rates of Change

1990 2000 2010 2020

0.1 mil

0.9 mil

2.5 mil

4.9 mil P(x)

2005 2015

tangent line

secant line

Rates of Change

1990 2000 2010 2020

0.1 mil

0.9 mil

2.5 mil

4.9 mil P(x)

2005 2015

tangent line

secant line

Rates of Change

1990 2000 2010 2020

0.1 mil

0.9 mil

2.5 mil

4.9 mil P(x)

2005 2015

tangent line

secant line

Rates of Change

1990 2000 2010 2020

0.1 mil

0.9 mil

2.5 mil

4.9 mil P(x)

2005 2015

tangent line

secant line

Rates of Change

1990 2000 2010 2020

0.1 mil

0.9 mil

2.5 mil

4.9 mil P(x)

2005 2015

tangent line

secant line

Rates of Change

1990 2000 2010 2020

0.1 mil

0.9 mil

2.5 mil

4.9 mil P(x)

2005 2015

tangent line

secant line

Rates of Change

1990 2000 2010 2020

0.1 mil

0.9 mil

2.5 mil

4.9 mil P(x)

2005 2015

tangent line

secant line

Rates of Change

1990 2000 2010 2020

0.1 mil

0.9 mil

2.5 mil

4.9 mil P(x)

2005 2015

tangent line

secant line

Rates of Change

1990 2000 2010 2020

0.1 mil

0.9 mil

2.5 mil

4.9 mil P(x)

2005 2015

tangent line

secant line

Rates of Change

1990 2000 2010 2020

0.1 mil

0.9 mil

2.5 mil

4.9 mil P(x)

2005 2015

tangent line

secant line

Definition

The secant line to the curve y = f (x) through points R and Q is a line that passesthrough R and Q.

We call the slope of the secant line the average rate of change of f (x) from R to Q.

Qtangent line

secant line

Definition

The tangent line to the curve y = f (x) at point P is a line that

• passes through P and

• has the same slope as f (x) at P.

We call the slope of the tangent line the instantaneous rate of change of f (x) at P.

Definition

tangent line

secant line

Definition

tangent line

secant line

Definition

The secant line to the curve y = f (x) through points R and Q is a line that passesthrough R and Q.We call the slope of the secant line the average rate of change of f (x) from R to Q.

tangent line

secant line

Definition

tangent line

secant line

Definition

tangent line

secant line

Definition

Qtangent line

secant line

Definition

Qtangent line

secant line

Definition

Limits 1.1-1.2 Drawing Tangents and a First Limit

On the graph below, draw the secant lineto the curve through points P and Q.

On the graph below, draw the tangentline to the curve at point P.

ey = s(t)

8:00 8:30

1/2 hour

A. secant line to y = s(t) from t = 8 : 00 to

B. slope of the secant line to y = s(t) from t = 8 : 00 to

C. tangent line to y = s(t) at

D. slope of the tangent line to y = s(t) at

ey = s(t)

8:00 8:308:07

1/2 hour

A. secant line to y = s(t) from t = 8 : 00 to

B. slope of the secant line to y = s(t) from t = 8 : 00 to

C. tangent line to y = s(t) at

D. slope of the tangent line to y = s(t) at

ey = s(t)

8:00 8:30

1/2 hour

It takes me half an hour to bike 6 km. So, 12 kph represents the:

A. secant line to y = s(t) from t = 8 : 00 to t = 8 : 30

B. slope of the secant line to y = s(t) from t = 8 : 00 to t = 8 : 30

C. tangent line to y = s(t) at t = 8 : 30

D. slope of the tangent line to y = s(t) at t = 8 : 30

ey = s(t)

8:00 8:30

1/2 hour

ey = s(t)

8:00 8:30

1/2 hour

ey = s(t)

8:00 8:30

61/2 hour

ey = s(t)

8:00 8:30

61/2 hour

ey = s(t)

8:00 8:30

61/2 hour

ey = s(t)

8:00 8:30

1/2 hour

When I pedal up Crescent Rd at 8:25, the speedometer on my bike tells me I’m going5kph. This corresponds to the:

ey = s(t)

8:00 8:30

1/2 hour

ey = s(t)

8:00 8:30

1/2 hour

ey = s(t)

8:00 8:30

1/2 hour

Suppose the distance from the ground s (in meters) of a helium-filled balloon at time tover a 10-second interval is given by s(t) = t2, graphed below. Try to estimate how fastthe balloon is rising when t = 5.

yy = s(t)

One way:Estimate the slope of thetangent line to the curveat t = 5.

Another way:Calculate average rate of changefor intervals around 5that get smaller and smaller.

yy = s(t)

Let’s look for an algebraic way of determining the velocity of the balloon when t = 5.

y = s(t) = t2

Suppose the interval [5, ] has length h. What is the right endpoint of the interval?

y = s(t) = t2

Write the equation for the average (vertical) velocity from t = 5 to t = 5 + h.

y = s(t) = t2

vel =∆ height

∆ time=

s(5 + h)− s(5)

(5 + h)− 5=

(5 + h)2 − 52

y = s(t) = t2

vel =∆ height

∆ time=

s(5 + h)− s(5)

(5 + h)− 5=

(5 + h)2 − 52

What happens to the equation above when h is very, very small?

y = s(t) = t2

What do you think is the slope of the tangent line to the graph when t = 5?

Limit Notation

Average Velocity, t = 5 to t = 5 + h:

s(5 + h)− s(5)

=(5 + h)2 − 52

= 10 + h when h 6= 0

When h is very small,Vel ≈ 10

We write:limh→0

(10 + h)︸︷︷︸function

“The limit as h goes to 0 of (10 + h) is 10.”As h gets extremely close to 0, (10 + h) gets extremely close to 10.

Limit Notation

s(5 + h)− s(5)

=(5 + h)2 − 52

= 10 + h when h 6= 0

We write:limh→0

Limit Notation

s(5 + h)− s(5)

=(5 + h)2 − 52

= 10 + h when h 6= 0

We write:limh→0

Limit Notation

s(5 + h)− s(5)

=(5 + h)2 − 52

= 10 + h when h 6= 0

When h is very small,

Vel ≈ 10

We write:limh→0

Limit Notation

s(5 + h)− s(5)

=(5 + h)2 − 52

= 10 + h when h 6= 0

We write:limh→0

Limit Notation

s(5 + h)− s(5)

=(5 + h)2 − 52

= 10 + h when h 6= 0

We write:limh→0

(10 + h) = 10

We write:limh→0

“The limit as h goes to 0 of (10 + h) is 10.”

As h gets extremely close to 0, (10 + h) gets extremely close to 10.

Limit Notation

s(5 + h)− s(5)

=(5 + h)2 − 52

= 10 + h when h 6= 0

We write:limh→0︸︷︷︸

limit as h goes to 0

(10 + h) = 10

We write:limh→0

Limit Notation

s(5 + h)− s(5)

=(5 + h)2 − 52

= 10 + h when h 6= 0

We write:limh→0

Limit Notation

s(5 + h)− s(5)

=(5 + h)2 − 52

= 10 + h when h 6= 0

We write:limh→0

Limits 1.3 The Limit of a Function

Definition

limx→a

f (x) = L

where a and L are real numbersWe read the above as “the limit as x goes to a of f (x) is L.”Its meaning is: as x gets very close to (but not equal to) a, f (x) gets very close to L.

We DESPERATELY NEED limits to find slopes of tangent lines.

Slope of secant line:∆y

∆x, ∆x 6= 0.

Slope of tangent line: can’t do the same way.

If the position of an object at time t is given by s(t), then its instantaneous velocity is

given by limh→0

s(t + h)− s(t)

Definition

limx→a

f (x) = L

∆x, ∆x 6= 0.

given by limh→0

s(t + h)− s(t)

Definition

limx→a

f (x) = L

∆x, ∆x 6= 0.

given by limh→0

s(t + h)− s(t)

Definition

limx→a

f (x) = L

∆x, ∆x 6= 0.

given by limh→0

s(t + h)− s(t)

Definition

limx→a

f (x) = L

∆x, ∆x 6= 0.

given by limh→0

s(t + h)− s(t)

Definition

limx→a

f (x) = L

∆x, ∆x 6= 0.

given by limh→0

s(t + h)− s(t)

Definition

limx→a

f (x) = L

∆x, ∆x 6= 0.

given by limh→0

s(t + h)− s(t)

Definition

limx→a

f (x) = L

∆x, ∆x 6= 0.

given by limh→0

s(t + h)− s(t)

Definition

limx→a

f (x) = L

∆x, ∆x 6= 0.

given by limh→0

s(t + h)− s(t)

Definition

limx→a

f (x) = L

∆x, ∆x 6= 0.

given by limh→0

s(t + h)− s(t)

Definition

limx→a

f (x) = L

∆x, ∆x 6= 0.

given by limh→0

s(t + h)− s(t)

Definition

limx→a

f (x) = L

∆x, ∆x 6= 0.

given by limh→0

s(t + h)− s(t)

Definition

limx→a

f (x) = L

∆x, ∆x 6= 0.

given by limh→0

s(t + h)− s(t)

Definition

limx→a

f (x) = L

∆x, ∆x 6= 0.

given by limh→0

s(t + h)− s(t)

Let f (x) =x3 + x2 − x − 1

x − 1.

We want to evaluate limx→1

f (x).

DNE (can’t divide by zero)

limx→1

f (x) = 4

Let f (x) =x3 + x2 − x − 1

x − 1.

f (x).

What is f (1)?

DNE (can’t divide by zero)

limx→1

f (x) = 4

Let f (x) =x3 + x2 − x − 1

x − 1.

f (x).

What is f (1)? DNE (can’t divide by zero)

limx→1

f (x) = 4

Let f (x) =x3 + x2 − x − 1

x − 1.

f (x).

What is f (1)? DNE (can’t divide by zero)Use the table below to guess lim

x→1f (x)

x f (x)0.9 3.610.99 3.96010.999 3.996000.9999 3.999601.1 4.411.01 4.04011.001 4.004001.0001 4.00040

limx→1

f (x) = 4

Let f (x) =x3 + x2 − x − 1

x − 1.

f (x).

What is f (1)? DNE (can’t divide by zero)Use the table below to guess lim

x→1f (x)

x f (x)0.9 3.610.99 3.96010.999 3.996000.9999 3.999601.1 4.411.01 4.04011.001 4.004001.0001 4.00040

limx→1

f (x) = 4

f (x) =

x if x < 31 if x = 3

x − 1 if x > 3

y = f (x)

0 1 2 3 4 5 60

What do you think should be the value of limx→3

f (x)?

Evaluate: limx→3−

f (x)︸︷︷︸from the left

limx→3+

f (x)︸︷︷︸from the right

f (x) =

x if x < 31 if x = 3

x − 1 if x > 3

y = f (x)

0 1 2 3 4 5 60

f (x)? DNE

limx→3+

f (x) =

x if x < 31 if x = 3

x − 1 if x > 3

y = f (x)

0 1 2 3 4 5 60

f (x)? DNE

limx→3+

f (x) =

x if x < 31 if x = 3

x − 1 if x > 3

y = f (x)

0 1 2 3 4 5 60

f (x)? DNE

= 3 limx→3+

f (x) =

x if x < 31 if x = 3

x − 1 if x > 3

y = f (x)

0 1 2 3 4 5 60

f (x)? DNE

= 3 limx→3+

Definition

The limit as x goes to a from the left of f (x) is written

limx→a−

We only consider values of x that are less than a.

Definition

The limit as x goes to a from the right of f (x) is written

limx→a+

We only consider values of x that are greater than a.

Theorem

In order for limx→a

f (x) to exist, both one-sided limits must exist and be equal.

Definition

The limit as x goes to a from the left of f (x) is written

limx→a−

We only consider values of x that are less than a.

Definition

The limit as x goes to a from the right of f (x) is written

limx→a+

We only consider values of x that are greater than a.

Theorem

In order for limx→a

f (x) to exist, both one-sided limits must exist and be equal.

Consider the function f (x) =1

(x − 1)2.

For what value of x is f (x) not defined?

(x − 1)2.

Based on the graph below, what would you like to write for:

limx→1

f (x) =

yy = f (x)

−2 −1 0 1 2 3 4 5 6

(x − 1)2.

limx→1

f (x) =∞

yy = f (x)

−2 −1 0 1 2 3 4 5 6

(x − 1)2.

limx→1

f (x) =∞

yy = f (x)

−2 −1 0 1 2 3 4 5 6

THIS LIMIT DOES NOT EXIST

Let f (x) = sin

). Graphed by Google:

Let f (x) = sin

To understand why it looks this way, first evaluate: limx→0+

Let f (x) = sin

To understand why it looks this way, first evaluate: limx→0+

x= ∞

Now, what should the graph of f (x) look like when x is near 0?

Let f (x) = sin

What is limx→0

f (x)?

DNE. This is sometimes called “infinite wiggling”

Let f (x) = sin

What is limx→0

f (x)? DNE. This is sometimes called “infinite wiggling”

Let f (x) = sin

What is limx→0

f (x)? DNE. This is sometimes called “infinite wiggling”

Does a limit exist for other points?

Suppose f (x) = log(x).

−2 0 2 4 6 8

Where is f (x) defined, and where is it not defined?

What can you say about the limit of f (x) near 0?

limx→0+

log(x) = −∞

Suppose f (x) = log(x).

−2 0 2 4 6 8

Where is f (x) defined, and where is it not defined?

What can you say about the limit of f (x) near 0? limx→0+

log(x) = −∞

f (x) =

{x2 x 6= 12 x = 1

y = f (x)

−2 −1 1 2

What is limx→1

f (x)?

A. limx→1

f (x) = 2

B. limx→1

f (x) = 1

C. limx→1

f (x) DNE

D. none of the above

f (x) =

{x2 x 6= 12 x = 1

y = f (x)

−2 −1 1 2

What is limx→1

f (x)?

A. limx→1

f (x) = 2

B. limx→1

f (x) = 1

C. limx→1

f (x) DNE

f (x) =

{x2 x 6= 12 x = 1

y = f (x)

−2 −1 1 2

What is limx→1

f (x)?

A. limx→1

f (x) = 2

B. limx→1

f (x) = 1

C. limx→1

f (x) DNE

f (x) =

{4 x ≤ 0x2 x > 0

y = f (x)

−2 −1 1 2

What is limx→0

f (x)?

A. limx→0

f (x) = 4

B. limx→0

f (x) = 0

C. limx→0

f (x) =

{4 x ≤ 00 x > 0

limx→0

f (x) DNE

limx→0−

f (x) = 4

“the limit as x approaches 0from the left is 4”

limx→0+

f (x) = 0

“the limit as x approaches 0from the right is 0”

f (x) =

{4 x ≤ 0x2 x > 0

y = f (x)

−2 −1 1 2

What is limx→0

f (x)?

A. limx→0

f (x) = 4

B. limx→0

f (x) = 0

C. limx→0

f (x) =

{4 x ≤ 00 x > 0

limx→0

f (x) DNE

limx→0−

f (x) = 4

limx→0+

f (x) = 0

f (x) =

{4 x ≤ 0x2 x > 0

y = f (x)

−2 −1 1 2

What is limx→0

f (x)?

A. limx→0

f (x) = 4

B. limx→0

f (x) = 0

C. limx→0

f (x) =

{4 x ≤ 00 x > 0

limx→0

f (x) DNE

limx→0−

f (x) = 4

limx→0+

f (x) = 0

f (x) =

{4 x ≤ 0x2 x > 0

y = f (x)

−2 −1 1 2

What is limx→0

f (x)?

A. limx→0

f (x) = 4

B. limx→0

f (x) = 0

C. limx→0

f (x) =

{4 x ≤ 00 x > 0

D. none of the abovelimx→0

f (x) DNE

limx→0−

f (x) = 4

limx→0+

f (x) = 0

f (x) =

{4 x ≤ 0x2 x > 0

y = f (x)

−2 −1 1 2

What is limx→0

f (x)?

A. limx→0

f (x) = 4

B. limx→0

f (x) = 0

C. limx→0

f (x) =

{4 x ≤ 00 x > 0

f (x) DNE

limx→0−

f (x) = 4

limx→0+

f (x) = 0

f (x) =

{4 x ≤ 0x2 x > 0

y = f (x)

−2 −1 1 2

What is limx→0

f (x)?

A. limx→0

f (x) = 4

B. limx→0

f (x) = 0

C. limx→0

f (x) =

{4 x ≤ 00 x > 0

f (x) DNE

limx→0−

f (x) = 4

limx→0+

f (x) = 0

Suppose limx→3−

f (x) = 1 and limx→3+

f (x) = 1.5. Does limx→3

f (x) exist?

A. Yes, certainly, because the limits from both sides exist.

B. No, never, because the limit from the left is not the same as the limit from the right.

C. Can’t tell. For some functions is might exist, for others not.

Suppose limx→3−

f (x) = 22 = limx→3+

f (x). Does limx→3

f (x) exist?

A. Yes, certainly, because the limits from both sides exist and are equal to each other.

B. No, never, because we only talk about one-sided limits when the actual limit doesn’texist.

C. Can’t tell. We need to know the value of the function at x = 3.

Suppose limx→3−

f (x) exist?

Suppose limx→3−

f (x) = 22 = limx→3+

f (x) exist?

Suppose limx→3−

f (x) exist?

Suppose limx→3−

f (x) = 22 = limx→3+

f (x) exist?

Suppose limx→3−

f (x) exist?

Suppose limx→3−

f (x) = 22 = limx→3+

f (x) exist?

Suppose limx→3−

f (x) exist?

Suppose limx→3−

f (x) = 22 = limx→3+

f (x) exist?

Suppose limx→3−

f (x) exist?

Suppose limx→3−

f (x) = 22 = limx→3+

f (x) exist?

Limits 1.4 Calculating Limits with Limit Laws

Calculating Limits in Simple Situations

Direct Substitution

If f (x) is a polynomial or rational function, and a is in the domain of f , then:

limx→a

f (x) = f (a).

Direct Substitution

limx→a

f (x) = f (a).

Calculate: limx→3

(x2 − 9

(32 − 9

Calculate: limx→3

(x2 − 9

x − 3

Can’t do it the same way: 3 not in domain

Direct Substitution

limx→a

f (x) = f (a).

Calculate: limx→3

(x2 − 9

(32 − 9

Calculate: limx→3

(x2 − 9

x − 3

Direct Substitution

limx→a

f (x) = f (a).

Calculate: limx→3

(x2 − 9

(32 − 9

Calculate: limx→3

(x2 − 9

x − 3

Direct Substitution

limx→a

f (x) = f (a).

Calculate: limx→3

(x2 − 9

(32 − 9

Calculate: limx→3

(x2 − 9

x − 3

)Can’t do it the same way: 3 not in domain

Direct Substitution

limx→a

f (x) = f (a).

Algebra with Limits

Suppose limx→a

f (x) = F and limx→a

g(x) = G , where F and G are both real numbers. Then:

- limx→a

(f (x) + g(x)) = F + G

- limx→a

(f (x)− g(x)) = F − G

- limx→a

(f (x)g(x)) = FG

- limx→a

(f (x)/g(x)) = F/G PROVIDED G 6= 0

Direct Substitution

Algebra with Limits

Suppose limx→a

f (x) = F and limx→a

g(x) = G , where F and G are both real numbers. Then:

- limx→a

(f (x) + g(x)) = F + G

- limx→a

(f (x)− g(x)) = F − G

- limx→a

(f (x)g(x)) = FG

- limx→a

(f (x)/g(x)) = F/G PROVIDED G 6= 0

Calculate: limx→1

[2x + 4

x + 2+ 13

(x + 5

2x − 1

Direct Substitution

Algebra with Limits

Calculate: limx→1

[2x + 4

x + 2+ 13

(x + 5

2x − 1

= limx→1

(2x + 4

2x + 2

limx→1

2x − 1

(2(1) + 4

2(1) + 2

)+ (13)

((1) + 5

2(1)− 1

= (2) + 13(2)(1) = 28

Direct Substitution

Algebra with Limits

Calculate: limx→1

[2x + 4

x + 2+ 13

(x + 5

2x − 1

= limx→1

(2x + 4

2x + 2

limx→1

2x − 1

(2(1) + 4

2(1) + 2

)+ (13)

((1) + 5

2(1)− 1

= (2) + 13(2)(1) = 28

Direct Substitution

Algebra with Limits

Calculate: limx→1

[2x + 4

x + 2+ 13

(x + 5

2x − 1

= limx→1

(2x + 4

2x + 2

limx→1

2x − 1

(2(1) + 4

2(1) + 2

)+ (13)

((1) + 5

2(1)− 1

= (2) + 13(2)(1) = 28

Powers and Roots of Limits

Which of the following gives a real number?

A. 41/2

B. (−4)1/2

=√−4

C. 4−1/2

D. (−4)−1/2

= 1√−4

Powers of Limits

If n is a positive integer, and limx→a

f (x) = F (where F is a real number), then:

limx→a

(f (x))n = F n.

Furthermore,limx→a

(f (x))1/n = F 1/n

UNLESS n is even and F is negative.

A. 41/2 = 2 B. (−4)1/2

=√−4

C. 4−1/2

D. (−4)−1/2

= 1√−4

Powers of Limits

limx→a

(f (x))n = F n.

(f (x))1/n = F 1/n

A. 41/2 = 2 B. (−4)1/2=√−4 C. 4−1/2

D. (−4)−1/2

= 1√−4

Powers of Limits

limx→a

(f (x))n = F n.

(f (x))1/n = F 1/n

A. 41/2 = 2 B. (−4)1/2=√−4 C. 4−1/2 = 1

2D. (−4)−1/2

= 1√−4

Powers of Limits

limx→a

(f (x))n = F n.

(f (x))1/n = F 1/n

A. 41/2 = 2 B. (−4)1/2=√−4 C. 4−1/2 = 1

2D. (−4)−1/2 = 1√

Powers of Limits

limx→a

(f (x))n = F n.

(f (x))1/n = F 1/n

A. 41/2 = 2 B. (−4)1/2=√−4 C. 4−1/2 = 1

2D. (−4)−1/2 = 1√

A. 81/3

B. (−8)1/3

= −2

C. 8−1/3

D. (−8)−1/3

= 1−2

Powers of Limits

limx→a

(f (x))n = F n.

(f (x))1/n = F 1/n

A. 41/2 = 2 B. (−4)1/2=√−4 C. 4−1/2 = 1

2D. (−4)−1/2 = 1√

A. 81/3 = 2 B. (−8)1/3

= −2

C. 8−1/3

D. (−8)−1/3

= 1−2

Powers of Limits

limx→a

(f (x))n = F n.

(f (x))1/n = F 1/n

A. 41/2 = 2 B. (−4)1/2=√−4 C. 4−1/2 = 1

2D. (−4)−1/2 = 1√

A. 81/3 = 2 B. (−8)1/3= −2 C. 8−1/3

D. (−8)−1/3

= 1−2

Powers of Limits

limx→a

(f (x))n = F n.

(f (x))1/n = F 1/n

A. 41/2 = 2 B. (−4)1/2=√−4 C. 4−1/2 = 1

2D. (−4)−1/2 = 1√

A. 81/3 = 2 B. (−8)1/3= −2 C. 8−1/3 = 12

D. (−8)−1/3

= 1−2

Powers of Limits

limx→a

(f (x))n = F n.

(f (x))1/n = F 1/n

A. 41/2 = 2 B. (−4)1/2=√−4 C. 4−1/2 = 1

2D. (−4)−1/2 = 1√

A. 81/3 = 2 B. (−8)1/3= −2 C. 8−1/3 = 12

D. (−8)−1/3 = 1−2

Powers of Limits

limx→a

(f (x))n = F n.

(f (x))1/n = F 1/n

A. 41/2 = 2 B. (−4)1/2=√−4 C. 4−1/2 = 1

2D. (−4)−1/2 = 1√

A. 81/3 = 2 B. (−8)1/3= −2 C. 8−1/3 = 12

D. (−8)−1/3 = 1−2

Powers of Limits

limx→a

(f (x))n = F n.

(f (x))1/n = F 1/n

Powers of Limits

limx→a

(f (x))n = F n.

(f (x))1/n = F 1/n

limx→4

(x + 5)1/2

limx→4

(x + 5)]1/2

= 91/2 = 3

Powers of Limits

limx→a

(f (x))n = F n.

(f (x))1/n = F 1/n

limx→4

(x + 5)1/2 =[

limx→4

(x + 5)]1/2

= 91/2 = 3

Powers of Limits

limx→a

(f (x))n = F n.

(f (x))1/n = F 1/n

limx→4

(x + 5)1/2 =[

limx→4

(x + 5)]1/2

= 91/2

Powers of Limits

limx→a

(f (x))n = F n.

(f (x))1/n = F 1/n

limx→4

(x + 5)1/2 =[

limx→4

(x + 5)]1/2

= 91/2 = 3

Cautionary Tales

limx→0

(5 + x)2 − 25

0; NEED ANOTHER WAY

limx→3

(x − 6

→ 8√−1; DANGER DANGER

limx→0

x→ 32

0; THIS EXPRESSION IS MEANINGLESS

limx→5

(x2 + 2

= (52 + 2)1/3 = 3√

27 = 3

Cautionary Tales

limx→0

(5 + x)2 − 25

x→ 0

0; NEED ANOTHER WAY

limx→3

(x − 6

limx→0

x→ 32

limx→5

(x2 + 2

= (52 + 2)1/3 = 3√

27 = 3

Cautionary Tales

limx→0

(5 + x)2 − 25

x→ 0

0; NEED ANOTHER WAY

limx→3

(x − 6

limx→0

x→ 32

limx→5

(x2 + 2

= (52 + 2)1/3 = 3√

27 = 3

Cautionary Tales

limx→0

(5 + x)2 − 25

x→ 0

0; NEED ANOTHER WAY

limx→3

(x − 6

limx→0

x→ 32

limx→5

(x2 + 2

= (52 + 2)1/3 = 3√

27 = 3

Cautionary Tales

limx→0

(5 + x)2 − 25

x→ 0

0; NEED ANOTHER WAY

limx→3

(x − 6

limx→0

→ 32

limx→5

(x2 + 2

= (52 + 2)1/3 = 3√

27 = 3

Cautionary Tales

limx→0

(5 + x)2 − 25

x→ 0

0; NEED ANOTHER WAY

limx→3

(x − 6

limx→0

x→ 32

limx→5

(x2 + 2

= (52 + 2)1/3 = 3√

27 = 3

Cautionary Tales

limx→0

(5 + x)2 − 25

x→ 0

0; NEED ANOTHER WAY

limx→3

(x − 6

limx→0

x→ 32

limx→5

(x2 + 2

= (52 + 2)1/3 = 3√

27 = 3

Cautionary Tales

limx→0

(5 + x)2 − 25

x→ 0

0; NEED ANOTHER WAY

limx→3

(x − 6

limx→0

x→ 32

limx→5

(x2 + 2

= (52 + 2)1/3 = 3√

27 = 3

Functions that Differ at a Single Point

Suppose limx→a

g(x) exists, and f (x) = g(x)

when x is close to a (but not necessarily equal to a).

Then limx→a

f (x) = limx→a

Evaluate limx→1

x3 + x2 − x − 1

x − 1.

x3 + x2 − x − 1

x − 1=

(x + 1)2(x − 1)

x − 1

= (x + 1)2 whenever x 6= 1

So, limx→1

x3 + x2 − x − 1

x − 1= lim

x→1(x + 1)2 = 4

Suppose limx→a

Then limx→a

f (x) = limx→a

Evaluate limx→1

x3 + x2 − x − 1

x − 1.

x3 + x2 − x − 1

x − 1=

(x + 1)2(x − 1)

x − 1

= (x + 1)2 whenever x 6= 1

So, limx→1

x3 + x2 − x − 1

x − 1= lim

x→1(x + 1)2 = 4

Suppose limx→a

Then limx→a

f (x) = limx→a

Evaluate limx→1

x3 + x2 − x − 1

x − 1.

x3 + x2 − x − 1

x − 1=

(x + 1)2(x − 1)

x − 1

= (x + 1)2 whenever x 6= 1

So, limx→1

x3 + x2 − x − 1

x − 1= lim

x→1(x + 1)2 = 4

Suppose limx→a

Then limx→a

f (x) = limx→a

Evaluate limx→1

x3 + x2 − x − 1

x − 1.

x3 + x2 − x − 1

x − 1=

(x + 1)2(x − 1)

x − 1

= (x + 1)2 whenever x 6= 1

So, limx→1

x3 + x2 − x − 1

x − 1= lim

x→1(x + 1)2 = 4

Evaluate limx→5

√x + 20−

√4x + 5

x − 5

√x + 20−

√4x + 5

x − 5=

√x + 20−

√4x + 5

x − 5

(√x + 20 +

√4x + 5√

x + 20 +√

4x + 5

(x + 20)− (4x + 5)

(x − 5)(√x + 20 +

√4x + 5)

=−3x + 25

(x − 5)(√x + 20 +

√4x + 5)

=−3√

x + 20 +√

4x + 5

limx→5

√x + 20−

√4x + 5

x − 5= lim

−3√x + 20 +

√4x + 5

√5 + 20 +

√4(5) + 5

Evaluate limx→5

√x + 20−

√4x + 5

x − 5

√x + 20−

√4x + 5

x − 5=

√x + 20−

√4x + 5

x − 5

(√x + 20 +

√4x + 5√

x + 20 +√

4x + 5

(x + 20)− (4x + 5)

(x − 5)(√x + 20 +

√4x + 5)

=−3x + 25

(x − 5)(√x + 20 +

√4x + 5)

=−3√

x + 20 +√

4x + 5

limx→5

√x + 20−

√4x + 5

x − 5= lim

−3√x + 20 +

√4x + 5

√5 + 20 +

√4(5) + 5

A Few Strategies for Calculating Limits

First, hope that you can directly substitute (plug in). If your function is made up of thesum, difference, product, quotient, or power of ploynomials, you can do this PROVIDEDthe function exists where you’re taking the limit.

limx→1

(√35 + x5 +

x − 3

(√35 + 15 +

1− 3

If you have a function as described above and the point where you’re taking the limit isNOT in its domain, if you think the limit exists, try to simplify and cancel.

limx→0

x + 71x− 1

= limx→0

x + 72

2x− 1

= limx→0

x + 71

= limx→0

2x(x + 7) = 0

Otherwise, you can try graphing the function, or making a table of values, to get a betterpicture of what is going on.

limx→1

(√35 + x5 +

x − 3

(√35 + 15 +

1− 3

limx→0

x + 71x− 1

= limx→0

x + 72

2x− 1

= limx→0

x + 71

= limx→0

2x(x + 7) = 0

limx→1

(√35 + x5 +

x − 3

(√35 + 15 +

1− 3

limx→0

x + 71x− 1

= limx→0

x + 72

2x− 1

= limx→0

x + 71

= limx→0

2x(x + 7) = 0

limx→1

(√35 + x5 +

x − 3

(√35 + 15 +

1− 3

limx→0

x + 71x− 1

= limx→0

x + 72

2x− 1

= limx→0

x + 71

= limx→0

2x(x + 7) = 0

For the limit of a fraction where the numerator goes to a non-zero number, andthe denominator goes to zero, think about what division does, and figure out thesign.

limx→0

x − 1

As x gets close to zero, the denominator gets very tiny, and the numberator is close to 1.Since a tiny, tiny , tiny number “goes into” 1 many, many, many times, the value ofx − 1

xwill get very large–but might be positive or negative depending on the sign of x .

limx→0−

x − 1

x=∞ and lim

x→0+

x − 1

x= −∞

limx→0

x − 1

−∞

limx→−4−

−3√x2 − 4

−∞

limx→0

x − 1

limx→0−

x − 1

x=∞ and lim

x→0+

x − 1

x= −∞

limx→0

x − 1

−∞

limx→−4−

−3√x2 − 4

−∞

limx→0

x − 1

limx→0−

x − 1

x=∞ and lim

x→0+

x − 1

x= −∞

limx→0

x − 1

−∞

limx→−4−

−3√x2 − 4

−∞

limx→0

x − 1

limx→0−

x − 1

and limx→0+

x − 1

x= −∞

limx→0

x − 1

−∞

limx→−4−

−3√x2 − 4

−∞

limx→0

x − 1

limx→0−

x − 1

x=∞ and lim

x→0+

x − 1

x= −∞

limx→0

x − 1

−∞

limx→−4−

−3√x2 − 4

−∞

limx→0

x − 1

x= DNE

limx→0−

x − 1

x=∞ and lim

x→0+

x − 1

x= −∞

limx→0

x − 1

−∞

limx→−4−

−3√x2 − 4

−∞

limx→0

x − 1

x= DNE

limx→0−

x − 1

x=∞ and lim

x→0+

x − 1

x= −∞

limx→0

x − 1

−∞

limx→−4−

−3√x2 − 4

−∞

limx→0

x − 1

x= DNE

limx→0−

x − 1

x=∞ and lim

x→0+

x − 1

x= −∞

limx→0

x − 1

x2= −∞

limx→−4−

−3√x2 − 4

−∞

limx→0

x − 1

x= DNE

limx→0−

x − 1

x=∞ and lim

x→0+

x − 1

x= −∞

limx→0

x − 1

x2= −∞

limx→−4−

−3√x2 − 4

= −∞

Suppose a lemonade stand diversifies to sell lemonade, raspberry juice, and raspberrylemonade. The prices of ingredients fluctuate, but the sale price of raspberry lemonade isalways between the sale prices of lemonade and raspberry juice. One day, as a promotion,the stand sells lemonade and raspberry juice each for $1 a cup. How much do they sellraspberry lemonade for on that day?

Always: lemonade ≤ raspberry lemonade ≤ raspberry juice

Today: lemonade = $1 = raspberry juice

So, today also raspberry lemonade = $1

Squeeze Theorem

Suppose, when x is near (but not necessarily equal to) a, we have functions f (x), g(x),and h(x) so that

f (x) ≤ g(x) ≤ h(x)

and limx→a

f (x) = limx→a

h(x). Then limx→a

g(x) = limx→a

f (x).

limx→0

x2 sin

Squeeze Theorem

f (x) ≤ g(x) ≤ h(x)

and limx→a

f (x) = limx→a

h(x). Then limx→a

g(x) = limx→a

f (x).

limx→0

x2 sin

Evaluate:

limx→0

x2 sin

y = x2

Evaluate:

limx→0

x2 sin

y = x2

Evaluate:

limx→0

x2 sin

y = x2

Evaluate:

limx→0

x2 sin

y = x2

limx→0

x2 sin

−1 ≤ sin

)≤ 1

so −x2 ≤ x2 sin

)≤ x2

and also limx→0−x2 = 0 = lim

x→0x2

Therefore, by the Squeeze Theorem, limx→0

x2 sin

Squeeze Theorem

f (x) ≤ g(x) ≤ h(x)

and limx→a

f (x) = limx→a

h(x). Then limx→a

g(x) = limx→a

f (x).

limx→0

x2 sin

−1 ≤ sin

)≤ 1

so −x2 ≤ x2 sin

)≤ x2

x→0x2

x2 sin

Squeeze Theorem

f (x) ≤ g(x) ≤ h(x)

and limx→a

f (x) = limx→a

h(x). Then limx→a

g(x) = limx→a

f (x).

limx→0

x2 sin

−1 ≤ sin

)≤ 1

so −x2 ≤ x2 sin

)≤ x2

x→0x2

x2 sin

Squeeze Theorem

f (x) ≤ g(x) ≤ h(x)

and limx→a

f (x) = limx→a

h(x). Then limx→a

g(x) = limx→a

f (x).

limx→0

x2 sin

−1 ≤ sin

)≤ 1

so −x2 ≤ x2 sin

)≤ x2

x→0x2

x2 sin

Squeeze Theorem

f (x) ≤ g(x) ≤ h(x)

and limx→a

f (x) = limx→a

h(x). Then limx→a

g(x) = limx→a

f (x).

Limits 1.5 Limits at Infinity

Limits at Infinity

End Behavior

We write:lim

x→∞f (x) = L

to express that, as x grows larger and larger, f (x) approaches L.Similarly, we write:

limx→−∞

f (x) = L

to express that, as x grows more and more strongly negative, f (x) approaches L.If L is a number, we call y = L a horizontal asymptote of f (x).

Common Limits at Infinity

limx→∞

limx→−∞

limx→∞

limx→−∞

limx→∞

limx→−∞

limx→∞

limx→−∞

−∞

limx→−∞

x5/3 =

−∞

limx→−∞

x2/3 =

limx→∞

13 = 13

limx→−∞

13 = 13

limx→∞

limx→−∞

limx→∞

limx→−∞

limx→∞

limx→−∞

−∞

limx→−∞

x5/3 =

−∞

limx→−∞

x2/3 =

limx→∞

13 = 13

limx→−∞

13 = 13

limx→∞

limx→−∞

limx→∞

limx→−∞

limx→∞

limx→−∞

−∞

limx→−∞

x5/3 =

−∞

limx→−∞

x2/3 =

limx→∞

13 = 13

limx→−∞

13 = 13

limx→∞

limx→−∞

limx→∞

x2 =∞

limx→−∞

limx→∞

x3 =∞

limx→−∞

−∞

limx→−∞

x5/3 =

−∞

limx→−∞

x2/3 =

limx→∞

13 = 13

limx→−∞

13 = 13

limx→∞

limx→−∞

limx→∞

x2 =∞

limx→−∞

x2 =∞

limx→∞

x3 =∞

limx→−∞

x3 = −∞

limx→−∞

x5/3 =

−∞

limx→−∞

x2/3 =

limx→∞

13 = 13

limx→−∞

13 = 13

limx→∞

limx→−∞

limx→∞

x2 =∞

limx→−∞

x2 =∞

limx→∞

x3 =∞

limx→−∞

x3 = −∞

limx→−∞

x5/3 = −∞

limx→−∞

x2/3 =

limx→∞

13 = 13

limx→−∞

13 = 13

limx→∞

limx→−∞

limx→∞

x2 =∞

limx→−∞

x2 =∞

limx→∞

x3 =∞

limx→−∞

x3 = −∞

limx→−∞

x5/3 = −∞

limx→−∞

x2/3 =∞

Arithmetic with Limits at Infinity

limx→∞

(x − x2

limx→∞

1− x

)= −∞

limx→−∞

(x2 + x3 + x4

limx→−∞

(x + 13)(x2 + 13

−∞

limx→∞

(x − x2

limx→∞

1− x

)= −∞

limx→−∞

(x2 + x3 + x4

limx→−∞

(x + 13)(x2 + 13

−∞

limx→∞

(x − x2

)= lim

x→∞x(

1− x

= −∞

limx→−∞

(x2 + x3 + x4

limx→−∞

(x + 13)(x2 + 13

−∞

limx→∞

(x − x2

)= lim

x→∞x(

1− x

)= −∞

limx→−∞

(x2 + x3 + x4

limx→−∞

(x + 13)(x2 + 13

−∞

limx→∞

(x − x2

)= lim

x→∞x(

1− x

)= −∞

limx→−∞

(x2 + x3 + x4

)= lim

x→−∞x4

limx→−∞

(x + 13)(x2 + 13

−∞

limx→∞

(x − x2

)= lim

x→∞x(

1− x

)= −∞

limx→−∞

(x2 + x3 + x4

)= lim

x→−∞x4

limx→−∞

(x + 13)(x2 + 13

−∞

limx→∞

(x − x2

)= lim

x→∞x(

1− x

)= −∞

limx→−∞

(x2 + x3 + x4

)= lim

x→−∞x4

limx→−∞

(x + 13)(x2 + 13

= −∞

Calculating Limits at Infinity

limx→∞

x2 + 2x + 1

Trick: factor out largest power of denominator.

limx→∞

x2 + 2x + 1

x3= lim

x→∞

x2 + 2x + 1

= limx→∞

+ 2x2 + 1

Now, you can do algebra

x→∞

x+ lim

x→∞

x2+ lim

x→∞

limx→∞

=0 + 0 + 0

limx→∞

x2 + 2x + 1

limx→∞

x2 + 2x + 1

x3= lim

x→∞

x2 + 2x + 1

= limx→∞

+ 2x2 + 1

x→∞

x+ lim

x→∞

x2+ lim

x→∞

limx→∞

=0 + 0 + 0

limx→∞

x2 + 2x + 1

limx→∞

x2 + 2x + 1

x3= lim

x→∞

x2 + 2x + 1

= limx→∞

+ 2x2 + 1

x→∞

x+ lim

x→∞

x2+ lim

x→∞

limx→∞

=0 + 0 + 0

limx→−∞

(x7/3 − x5/3)

Again: factor out largest power of x .

(x7/3 − x5/3) = x7/3

(1− 1

x→−∞x7/3 = −∞

limx→−∞

(1− 1

So, limx→−∞

(x7/3 − x5/3) = −∞

limx→−∞

(x7/3 − x5/3)

(x7/3 − x5/3) = x7/3

(1− 1

x→−∞x7/3 = −∞

limx→−∞

(1− 1

So, limx→−∞

(x7/3 − x5/3) = −∞

limx→−∞

(x7/3 − x5/3)

(x7/3 − x5/3) = x7/3

(1− 1

x→−∞x7/3 = −∞

limx→−∞

(1− 1

So, limx→−∞

(x7/3 − x5/3) = −∞

limx→−∞

(x7/3 − x5/3)

(x7/3 − x5/3) = x7/3

(1− 1

x→−∞x7/3 = −∞

limx→−∞

(1− 1

So, limx→−∞

(x7/3 − x5/3) = −∞

Suppose the height of a bouncing ball is given by h(t) = sin(t)+1t

, for t ≥ 1. Whathappens to the height over a long period of time?

0 ≤ sin(t) + 1

t≤ 2

limt→∞

0 = 0 = limt→∞

So, by the Squeeze Theorem,

limt→∞

sin(t) + 1

0 ≤ sin(t) + 1

t≤ 2

limt→∞

0 = 0 = limt→∞

limt→∞

sin(t) + 1

0 ≤ sin(t) + 1

t≤ 2

limt→∞

0 = 0 = limt→∞

limt→∞

sin(t) + 1

Tough One

limx→∞

√x4 + x2 + 1−

√x4 + 3x2

Multiply function by conjugate:

(√x4 + x2 + 1−

√x4 + 3x2

)(√x4 + x2 + 1 +√x4 + 3x2

√x4 + x2 + 1 +

√x4 + 3x2

−2x2 + 1√x4 + x2 + 1 +

√x4 + 3x2

Factor out highest power: x2 (same as√x4)

−2x2 + 1√x4 + x2 + 1 +

√x4 + 3x2

1/√x4

=−2 + 1

x2√1 + 1

x2 + 1x4 +

√1 + 3

limx→∞

−2 + 1x2√

1 + 1x2 + 1

x4 +√

1 + 3x2

=−2 + 0

√1 + 0 + 0 +

√1 + 0

2= −1

Tough One

limx→∞

√x4 + x2 + 1−

√x4 + 3x2

(√x4 + x2 + 1−

√x4 + 3x2

)(√x4 + x2 + 1 +√x4 + 3x2

√x4 + x2 + 1 +

√x4 + 3x2

−2x2 + 1√x4 + x2 + 1 +

√x4 + 3x2

−2x2 + 1√x4 + x2 + 1 +

√x4 + 3x2

1/√x4

=−2 + 1

x2√1 + 1

x2 + 1x4 +

√1 + 3

limx→∞

−2 + 1x2√

1 + 1x2 + 1

x4 +√

1 + 3x2

=−2 + 0

√1 + 0 + 0 +

√1 + 0

2= −1

Tough One

limx→∞

√x4 + x2 + 1−

√x4 + 3x2

(√x4 + x2 + 1−

√x4 + 3x2

)(√x4 + x2 + 1 +√x4 + 3x2

√x4 + x2 + 1 +

√x4 + 3x2

−2x2 + 1√x4 + x2 + 1 +

√x4 + 3x2

−2x2 + 1√x4 + x2 + 1 +

√x4 + 3x2

1/√x4

=−2 + 1

x2√1 + 1

x2 + 1x4 +

√1 + 3

limx→∞

−2 + 1x2√

1 + 1x2 + 1

x4 +√

1 + 3x2

=−2 + 0

√1 + 0 + 0 +

√1 + 0

2= −1

Tough One

limx→∞

√x4 + x2 + 1−

√x4 + 3x2

(√x4 + x2 + 1−

√x4 + 3x2

)(√x4 + x2 + 1 +√x4 + 3x2

√x4 + x2 + 1 +

√x4 + 3x2

−2x2 + 1√x4 + x2 + 1 +

√x4 + 3x2

−2x2 + 1√x4 + x2 + 1 +

√x4 + 3x2

1/√x4

=−2 + 1

x2√1 + 1

x2 + 1x4 +

√1 + 3

limx→∞

−2 + 1x2√

1 + 1x2 + 1

x4 +√

1 + 3x2

=−2 + 0

√1 + 0 + 0 +

√1 + 0

2= −1

Evaluate limx→−∞

√3 + x2

We factor out the largest power of the denominator, which is is x .

limx→−∞

√3 + x2

)= lim

x→−∞

√3+x2

When x < 0,√x2 = |x | = −x

= limx→−∞

√3 + x2

−√x2

= limx→−∞

√3 + x2

= limx→−∞

= −1

Evaluate limx→−∞

√3 + x2

3xWe factor out the largest power of the denominator, which is is x .

limx→−∞

√3 + x2

)= lim

x→−∞

√3+x2

When x < 0,√x2 = |x | = −x

= limx→−∞

√3 + x2

−√x2

= limx→−∞

√3 + x2

= limx→−∞

= −1

Limits 1.6 Continuity

Continuity

Definition

A function f (x) is continuous at a point a if limx→a

f (x) exists AND is equal to f (a).

A function f (x) is continuous from the left at a point a iflim

x→a−f (x) exists AND is equal to f (a).

f (x) =

{x2 sin

), x 6= 0

0 , x = 0

Is f (x) continuous at 0?

Continuity

Definition

y = f (x)

−2 −1 1 2

4 Is f (x) continuous at 0?

This kind of discontinuity is calledremovable.

f (x) =

{x2 sin

), x 6= 0

0 , x = 0

Continuity

Definition

y = f (x)

−2 −1 1 2

4 Is f (x) continuous at 0? Yes.Is f (x) continuous at 1?

f (x) =

{x2 sin

), x 6= 0

0 , x = 0

Continuity

Definition

y = f (x)

−2 −1 1 2

4 Is f (x) continuous at 0? Yes.Is f (x) continuous at 1? No.

f (x) =

{x2 sin

), x 6= 0

0 , x = 0

Continuity

Definition

y = f (x)

−2 −1 1 2

4 Is f (x) continuous at 0? Yes.Is f (x) continuous at 1? No.

f (x) =

{x2 sin

), x 6= 0

0 , x = 0

Continuity

Definition

y y = f (x)

6 Is f (x) continuous at 3?

No.Is f (x) continuous from the left at 3?

Is f (x) continuous from the right at 3?

This kind of discontinuity is called ajump.

f (x) =

{x2 sin

), x 6= 0

0 , x = 0

Continuity

Definition

y y = f (x)

6 Is f (x) continuous at 3? No.

Is f (x) continuous from the left at 3?

f (x) =

{x2 sin

), x 6= 0

0 , x = 0

Continuity

Definition

y y = f (x)

6 Is f (x) continuous at 3? No.

Is f (x) continuous from the left at 3?

f (x) =

{x2 sin

), x 6= 0

0 , x = 0

Continuity

Definition

y y = f (x)

6 Is f (x) continuous at 3? No.Is f (x) continuous from the left at 3?

f (x) =

{x2 sin

), x 6= 0

0 , x = 0

Continuity

Definition

y y = f (x)

6 Is f (x) continuous at 3? No.Is f (x) continuous from the left at 3?Yes.Is f (x) continuous from the right at 3?

f (x) =

{x2 sin

), x 6= 0

0 , x = 0

Continuity

Definition

y y = f (x)

6 Is f (x) continuous at 3? No.Is f (x) continuous from the left at 3?Yes.Is f (x) continuous from the right at 3?No.

f (x) =

{x2 sin

), x 6= 0

0 , x = 0

Continuity

Definition

y y = f (x)

−2 −1 0 1 2 3 4 5 6

Since no one-sided limits exist at 1,there’s no hope for continuity.

This is called an infinite discontinuity.

f (x) =

{x2 sin

), x 6= 0

0 , x = 0

Continuity

Definition

y y = f (x)

−2 −1 0 1 2 3 4 5 6

Since no one-sided limits exist at 1,there’s no hope for continuity.

This is called an infinite discontinuity.

f (x) =

{x2 sin

), x 6= 0

0 , x = 0

Continuity

Definition

f (x) =

{x2 sin

), x 6= 0

0 , x = 0

Continuity

Definition

A function f (x) is continuous at a point a if

limx→a

f (x) = f (a)

Functions made by adding, subtracting, multiplying, dividing, and taking appropriatepowers of polynomials are continuous for every point in their domain.

Continuity

Definition

limx→a

f (x) = f (a)

Functions made by adding, subtracting, multiplying, dividing, and taking appropriatepowers of polynomials are continuous for every point in their domain.Example:

f (x) =x2

2x − 10−

(x2 + 2x − 1

x − 1+

25− x − 1x

Continuity

Definition

limx→a

f (x) = f (a)

f (x) =x2

2x − 10−

(x2 + 2x − 1

x − 1+

25− x − 1x

f (x) is continuous at every point except 5, 1, 0, and −2.

Continuity

Definition

limx→a

f (x) = f (a)

f (x) =x2

2x − 10−

(x2 + 2x − 1

x − 1+

25− x − 1x

f (x) is continuous at every point except 5, 1, 0, and −2.A continuous function is continuous for every point in R.

Continuity

Definition

limx→a

f (x) = f (a)

f (x) =x2

2x − 10−

(x2 + 2x − 1

x − 1+

25− x − 1x

f (x) is continuous at every point except 5, 1, 0, and −2.A continuous function is continuous for every point in R.We say f (x) is continuous over (a, b) if it is continuous at every point in (a, b). So, f (x)is continuous over its domain, (−∞,−2) ∪ (−2, 0) ∪ (0, 1) ∪ (1, 5) ∪ (5,∞).

Continuity

Definition

limx→a

f (x) = f (a)

Common Functions

Functions of the following types are continuous over their domains:

- polynomials and rationals

- roots and powers

- trig functions and their inverses

- exponential and logarithm

- The products, sums, differences, quotients, powers, and compositions of continuousfunctions

Where is the following function continuous?

f (x) =

(sin x

(x − 2)(x + 3)+ e√

Over its domain: [0, 2) ∪ (2,∞).

Lots of examples in notes.

f (x) =

(sin x

(x − 2)(x + 3)+ e√

f (x) =

(sin x

(x − 2)(x + 3)+ e√

Continuity in Nature

Baby weight chart, 1909Source: http://gallery.nen.gov.uk/asset668105-.html

Continuity in Nature

Graph of luminosity of a star over time, after star explodes.Data from 1987. Source:http://abyss.uoregon.edu/ ∼js/ast122/lectures/ lec18.html

A Technical Point

Definition

A function f (x) is continuous on the closed interval [a, b] if:

f (x) is continuous over (a, b), and

f (x) is continuous form the left at

f (x) is continuous form the right at

A Technical Point

Definition

f (x) is continuous form the left at

A Technical Point

Definition

f (x) is continuous form the left at b, and

A Technical Point

Definition

f (x) is continuous form the left at b, and

f (x) is continuous form the right at a

Intermediate Value Theorem (IVT)

Theorem:

Let a < b and let f (x) be continuous over [a, b]. If y is any number between f (a) andf (b), then there exists c in (a, b) such that f (c) = y .

Theorem:

Suppose your favorite number is 45.54. At noon, your car is parked, and at 1pm you’redriving 100kph. By the Intermediate Value Theorem, at some point between noon and1pm you were going exactly 45.54 kph.

Using IVT to Find Roots: “Bisection Method”

Let f (x) = x5 − 2x4 + 2. Find any value x for which f (x) = 0.

Let’s find some points:

f (0) = 2 f (1) = 1 f (−1) = −1f (−.5) = 1.84375 f (−0.75) ≈ 1.1298 f (−.9) = 0.09731

f (−.95) ≈ −0.4

So, there has to be a root between and .We can say there is a root at approximately x = −0.9

Let f (x) = x5 − 2x4 + 2. Find any value x for which f (x) = 0.Let’s find some points:

f (0) = 2 f (1) = 1 f (−1) = −1f (−.5) = 1.84375 f (−0.75) ≈ 1.1298 f (−.9) = 0.09731

f (−.95) ≈ −0.4

1−1 −.5−.75

−.9So, there has to be a root between and .

We can say there is a root at approximately x = −0.9

f (0) = 2

f (1) = 1 f (−1) = −1f (−.5) = 1.84375 f (−0.75) ≈ 1.1298 f (−.9) = 0.09731

f (−.95) ≈ −0.4

1−1 −.5−.75

f (0) = 2 f (1) = 1

f (−1) = −1f (−.5) = 1.84375 f (−0.75) ≈ 1.1298 f (−.9) = 0.09731

f (−.95) ≈ −0.4

−1 −.5−.75

f (0) = 2 f (1) = 1 f (−1) = −1

f (−.5) = 1.84375 f (−0.75) ≈ 1.1298 f (−.9) = 0.09731f (−.95) ≈ −0.4

−.5−.75

f (0) = 2 f (1) = 1 f (−1) = −1

f (−.5) = 1.84375 f (−0.75) ≈ 1.1298 f (−.9) = 0.09731f (−.95) ≈ −0.4

−.5−.75

So, there has to be a root between x = −1 and x = 0 .

f (0) = 2 f (1) = 1 f (−1) = −1f (−.5) = 1.84375

f (−0.75) ≈ 1.1298 f (−.9) = 0.09731f (−.95) ≈ −0.4

1−1 −.5

−.75

So, there has to be a root between x = −1 and x = 0 .

f (0) = 2 f (1) = 1 f (−1) = −1f (−.5) = 1.84375

f (−0.75) ≈ 1.1298 f (−.9) = 0.09731f (−.95) ≈ −0.4

1−1 −.5

−.75

So, there has to be a root between x = −1 and x = −0.5 .

f (0) = 2 f (1) = 1 f (−1) = −1f (−.5) = 1.84375 f (−0.75) ≈ 1.1298

f (−.9) = 0.09731f (−.95) ≈ −0.4

1−1 −.5−.75

f (0) = 2 f (1) = 1 f (−1) = −1f (−.5) = 1.84375 f (−0.75) ≈ 1.1298

f (−.9) = 0.09731f (−.95) ≈ −0.4

1−1 −.5−.75

f (0) = 2 f (1) = 1 f (−1) = −1f (−.5) = 1.84375 f (−0.75) ≈ 1.1298 f (−.9) = 0.09731

f (−.95) ≈ −0.4

1−1 −.5−.75

−.9So, there has to be a root between x = −1 and x = −0.75 .

f (0) = 2 f (1) = 1 f (−1) = −1f (−.5) = 1.84375 f (−0.75) ≈ 1.1298 f (−.9) = 0.09731

f (−.95) ≈ −0.4

1−1 −.5−.75

−.9So, there has to be a root between x = −1 and x = −0.9 .

f (0) = 2 f (1) = 1 f (−1) = −1f (−.5) = 1.84375 f (−0.75) ≈ 1.1298 f (−.9) = 0.09731

f (−.95) ≈ −0.4

−0.92−0.94−0.96−0.98−1 −0.9

f (0) = 2 f (1) = 1 f (−1) = −1f (−.5) = 1.84375 f (−0.75) ≈ 1.1298 f (−.9) = 0.09731

f (−.95) ≈ −0.4

−0.92−0.94−0.96−0.98−1 −0.9

So, there has to be a root between x = −0.95 and x = −0.9 .

f (0) = 2 f (1) = 1 f (−1) = −1f (−.5) = 1.84375 f (−0.75) ≈ 1.1298 f (−.9) = 0.09731

f (−.95) ≈ −0.4

−0.92−0.94−0.96−0.98−1 −0.9

So, there has to be a root between x = −0.95 and x = −0.9 .We can say there is a root at approximately x = −0.9

Don’t use a calculator for these problems: use values that you can easily calculate.Use the Intermediate Value Theorem to show that there exists some solution to theequation ln x · ex = 4 and give a reasonable interval where that solution might occur.

- The function f (x) = ln x · ex is continuous over its domain, which is (0,∞). Inparticular, then, it is continuous over the interval (1, e).

- f (1) = ln(1)e = 0 · e = 0 and f (e) = ln(e) · ee = ee . Since e > 2, we knowf (e) = ee > 22 = 4.

- Then 4 is between f (1) and f (e).- By the Intermediate Value Theorem, f (c) = 4 for some c in (1, e).

Use the Intermediate Value Theorem to give a reasonable interval where the following istrue: ex = sin(x)

We can rearrange this: let f (x) = ex − sin(x), and note f (x) has roots exactly when theabove equation is true.

- The function f (x) = ex − sin x is continuous over its domain, which is all realnumbers. In particular, then, it is continuous over the interval

(− 3π

2, e).

- f (0) = e0 − sin 0 = 1− 0 = 0 and

f(− 3π

)= e−

3π2 − sin

(−3π2

)= e−

3π2 − 1 < e0 − 1 = 1− 1 = 0.

- Then 0 is between f (0) and f(− 3π

- By the Intermediate Value Theorem, f (c) = 0 for some c in(− 3π

2, 0).

- Therefore, ec = sin c for some c in(− 3π

2, 0).

Is there any value of x so that sin x = cos(2x) + 14?

Yes, somewhere between 0 and π2

Use the Intermediate Value Theorem to give a reasonable interval where the following istrue: ex = sin(x)

We can rearrange this: let f (x) = ex − sin(x), and note f (x) has roots exactly when theabove equation is true.

(− 3π

2, e).

- f (0) = e0 − sin 0 = 1− 0 = 0 and

f(− 3π

)= e−

3π2 − sin

(−3π2

)= e−

3π2 − 1 < e0 − 1 = 1− 1 = 0.

2, 0).

Use the Intermediate Value Theorem to give a reasonable interval where the following istrue: ex = sin(x)We can rearrange this: let f (x) = ex − sin(x), and note f (x) has roots exactly when theabove equation is true.

(− 3π

2, e).

- f (0) = e0 − sin 0 = 1− 0 = 0 and

f(− 3π

)= e−

3π2 − sin

(−3π2

)= e−

3π2 − 1 < e0 − 1 = 1− 1 = 0.

2, 0).

Use the Intermediate Value Theorem to give a reasonable interval where the following istrue: ex = sin(x)We can rearrange this: let f (x) = ex − sin(x), and note f (x) has roots exactly when theabove equation is true.

(− 3π

2, e).

- f (0) = e0 − sin 0 = 1− 0 = 0 and

f(− 3π

)= e−

3π2 − sin

(−3π2

)= e−

3π2 − 1 < e0 − 1 = 1− 1 = 0.

2, 0).

Is the following reasoning correct?

- f (x) = tan x is continuous over its domain, because it is a trigonometric function.

- In particular, f (x) is continuous over the interval[π4, 3π

- f(π4

)= 1, and f

)= −1.

- Since f(

)< 0 < f

), by the Intermediate Value Theorem, there exists some

number c in the interval(π4, 3π

)such that f (c) = 0.

y = tan x

- f(π4

)= 1, and f

)= −1.

- Since f(

)< 0 < f

y = tan x

]. FALSE

- f(π4

)= 1, and f

)= −1.

- Since f(

)< 0 < f

y = tan x

Let’s Review

Suppose f (x) is continuous at x = 1. Does f (x) have to be defined at x = 1?

Yes. Since f (x) is continuous at x = 1, limx→1

f (x) = f (1), so f (1) must exist.

Suppose f (x) is continuous at x = 1 and limx→1−

f (x) = 30.

True or false: limx→1+

f (x) = 30.

True. Since f (x) is continuous at x = 1, limx→1

f (x) = f (1), so limx→1

f (x) must exist. That

means both one-sided limits exist, and are equal to each other.

Suppose f (x) is continuous at x = 1 and f (1) = 22. What is limx→1

f (x)?

22 = f (1) = limx→1

f (x).

Suppose limx→1

f (x) = 2. Must it be true that f (1) = 2?

No. In order to determine the limit as x goes to 1, we ignore f (1). Perhaps every f (x) isnot defined at 1.

Suppose f (x) is continuous at x = 1. Does f (x) have to be defined at x = 1?Yes. Since f (x) is continuous at x = 1, lim

x→1f (x) = f (1), so f (1) must exist.

f (x) = 30.

f (x) = f (1), so limx→1

f (x)?

22 = f (1) = limx→1

f (x).

Suppose limx→1

f (x) = 30.

f (x) = f (1), so limx→1

f (x)?

22 = f (1) = limx→1

f (x).

Suppose limx→1

f (x) = 30.

f (x) = f (1), so limx→1

f (x)?

22 = f (1) = limx→1

f (x).

Suppose limx→1

f (x) = 30.

f (x) = f (1), so limx→1

f (x)?

22 = f (1) = limx→1

f (x).

Suppose limx→1

f (x) = 30.

f (x) = f (1), so limx→1

f (x)?

22 = f (1) = limx→1

f (x).

Suppose limx→1

f (x) = 30.

f (x) = f (1), so limx→1

f (x)?

22 = f (1) = limx→1

f (x).

Suppose limx→1

f (x) = 30.

f (x) = f (1), so limx→1

f (x)?

22 = f (1) = limx→1

f (x).

Suppose limx→1

f (x) =

{ax2 x ≥ 13x x < 1

For which value(s) of a is f (x) continuous?

We need ax2 = 3x when x = 1, so a = 3.

f (x) =

{ax2 x ≥ 13x x < 1

For which value(s) of a is f (x) continuous?

We need ax2 = 3x when x = 1, so a = 3.

f (x) =

{ √3x+3

x2−3x 6= ±

a x = ±√

For which value(s) of a is f (x) continuous at x = −√

3?For which value(s) of a is f (x) continuous at x =

f (x) =

{ √3x+3

x2−3x 6= ±

a x = ±√

3?By the definition of continuity, if f (x) is continuous at x = −

√3, then

f (−√

3) = limx→−

√3f (x). Note f (−

√3) = a, and when x is close to (but not equal to)

−√

3, then f (x) =√

3x+3x2−3

f (−√

3) = limx→−

√3f (x)

a = limx→−

√3x + 3

x2 − 3= lim

x→−√

√3(x +

(x +√

3)(x −√

= limx→−

x −√

−√

3−√

3= −1

So we can use a = − 12

to make f (x) continuous at x = −√

For which value(s) of a is f (x) continuous at x =√

f (x) =

{ √3x+3

x2−3x 6= ±

a x = ±√

3?For which value(s) of a is f (x) continuous at x =

By the definition of continuity, if f (x) is continuous at x =√

3, then f (√

3) = limx→√

3f (x).

When x is close to (but not equal to)√

3, then f (x) =√

3x+3x2−3

. However, as x approaches√3, the denominator of this expression gets closer and closer to zero, while the top gets

closer and closer to 6. So, this limit does not exist. Therefore, no value of a will makef (x) continuous at x =

Derivatives

For the next batch of slides, seehttp://www.math.ubc.ca/~elyse/100/2016/Deriv_Conceptual.pdf

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