2.1 The Tangent and Velocity Problems

Chapter 2 Notes, Stewart 7e 2.1 Chalmeta The Tangent and Velocity Problems Suppose we want to find the slope of the line between (1, 2) and (-2, -...
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Chapter 2 Notes, Stewart 7e

2.1

Chalmeta

The Tangent and Velocity Problems

Suppose we want to find the slope of the line between (1, 2) and (-2, -4). We could do this easily with the slope formula: y2 − y 1 −4 − 2 m= = = 2. x2 − x1 −2 − 1

Now consider the function y = x2 − 4. Suppose we want to find the slope of the secant line shown below. y 4 3 2 1 x −5 −4 −3 −2 −1 −1 −2 −3 −4 −5

1

2

3

4

We know two points on the line so we can find the slope as we before.

What about the slope of the tangent line at (2, 0)? We don’t know two points so we can’t calculate it. What we can do is approximate the slope by taking a second point which is close to (2, 0) and calculating the slope of the secant line. The closer we get to (2, 0) the closer our approximation will be to the true slope. For example we could try find the slope between the points (x, x2 − 4) and (2, 0). The slope of the secant line between those two points would be m=

x2 − 4 − 0 . x−2

What if we try some values of x close to 2? x m=

1

1.5

1.9

2.1

x2 − 4 x−2

1

2.5

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Chapter 2 Notes, Stewart 7e

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Example 2.1.1. The point P(0.5, 2) lies on the curve y = 1/x. If Q is the point (x, 1/x), use your calculator to find the slop of the secant line PQ for the following values of x. a) .8 b) .6 c) .51 d) .49 e) .45 Guess the value of the slope of the tangent line at P(0.5,2). Average Velocity

Average Velocity =

change in distance change in time

Distance is usually given as a function of time s = s(t). This function tells us that if we know how long the object has been moving, t, we know where it is in space s(t). Example 2.1.2. Suppose that the position of an object moving in a straight line is given by the equation s = t3 /6 where t is in seconds, s in meters. Find the average velocity over the given time periods then find the instantaneous velocity when t = 1. 1. [1, 3] 2. [1, 2] 3. [1, 1.5] 4. [1, 1.1]

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Chapter 2 Notes, Stewart 7e

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Chalmeta

The Limit of a Function

The Question: What happens to f (x) as x approaches a? The Mathematical notation: lim f (x) = L x→a

In words: ”The limit of f (x) as x approaches a, equals L” Consider the following functions y

y

6

6

4

4

2

2

h

x −3 −2 −1

1

2

3

x

4

−3 −2 −1

f (x) = x + 2

g(x) =

f (2) = 4

1

2

3

4

x2 − 4 x−2

g(2) is undefined

As x gets close to 2,

As x gets close to 2,

y = f (x) gets close to 4.

y = g(x) gets close to 4.

lim f (x) = lim g(x) = 4

x→2

x→2

Translation: So when we write lim f (x) = L what we mean in English is ”As x gets CLOSE to a then f (x) gets x→a CLOSE to L”. In Words: So when we write lim f (x) = L what we say out loud is ”The limit as x gets approaches a of f (x) is L”. The key word here is

x→a

CLOSE.

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Chapter 2 Notes, Stewart 7e

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Consider the following graph: y 4 3 2 1 x −4

−3

−2

−1 −1

1

2

3

4

5

6

7

−2 −3 −4 Evaluate the following: (a) lim f (x) =

(b) lim f (x) =

(c) lim f (x) =

(d) f (−3) =

(e) f (1) =

(f) f (3) =

x→−3

x→1

x→3

One Sided Limits The limit as x approaches a from the left: lim f (x) = L x→a−

The limit as x approaches a from the right: lim f (x) = L x→a+

The limit exists if and only if the limit from the left is equal to the limit from the right. lim f (x) = L ⇔ lim f (x) = lim f (x) = L

x→a

x→a−

x→a+

Infinite Limits lim f (x) = ±∞

x→a

For example f (x) = tan x or f (x) =

1 (x + 2)2

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Chapter 2 Notes, Stewart 7e

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Example 2.2.1. Sketch the graph of the following function and use it to determine the values of a for which lim f (x) exists: x→a

f (x) =

   2−x  

if x < −1

x    (x − 1)2

if −1 ≤ x < 1 if x ≥ 1

4 3 2 1 −3 −2 −1 −1

1

2

3

4

−2 −3 Example 2.2.2. Sketch the graph of a function f that satisfies these conditions: lim f (x) = 4,

x→3+

lim f (x) = 2,

x→3−

lim f (x) = 2,

x→−2

5

f (3) = 3,

f (−2) = 1.

Chapter 2 Notes, Stewart 7e

Chalmeta

Example 2.2.3. Determine the infinite limits: a) lim

6 x−5

b) lim

6 x−5

x→5+

x→5−

c) lim

x−1 + 2)

x→0 x2 (x

Example 2.2.4. Determine the limits from the graph of f (x). y 4 3 2 1 x −5 −4 −3 −2 −1 −1 −2 −3 −4 −5

1

2

3

4

5

a) lim f (x) = x→3+

b) lim f (x) = x→3−

c) d)

lim f (x) =

x→−3+

lim f (x) =

x→−3−

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Chapter 2 Notes, Stewart 7e

2.3

Chalmeta

Calculating Limits Using the Limit Laws

Recall: Notation lim f (x) = L means as x gets close to a, f (x) gets close to L. x→a

Limits have certain properties which make them easy to work with. Properties: Suppose lim f (x) = L and lim g(x) = K,

x→a

x→a

then: 1) lim b · f (x) = b · L x→a

2) lim [f (x) ± g(x)] = L ± K x→a

3) lim [f (x) × g(x)] = L × K x→a



 f (x) L 4) lim when K 6= 0 = x→a g(x) K 5) lim [f (x)]n = Ln x→a

We can use these properties to help us solve limits analytically (using algebra). Example 2.3.1. lim x2 + 7x + 2 x→2

EVERY time you get to a limit problem the first thing you always want to do is to plug in the x value that you are given. STEP 1: Always plug in the x value. f (2) = 4 + 14 + 2 = 20 If you get a number, as in this case, then the problem is done. Here the answer is 20. As x gets close to 2, x2 + 7x + 2 gets close to 20 2x2 − x − 3 x→−1 x+1

Example 2.3.2. lim

STEP 1: Plug in for x:

2(−1)2 − (−1) − 3 0 = (−1) + 1 0

0 Here we didn’t get a number but rather an undefined expression of the form . This means that we can 0 do some algebra and cancel some common factors.

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Chapter 2 Notes, Stewart 7e

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Step 2: Do some algebra and cancel.

√ 2+x− 2 Example 2.3.3. lim x→0 x √ √ 2+0− 2 0 = STEP 1: Plug in for x: 0 0 √

0 Here we didn’t get a number but rather an undefined expression of the form . This means that we can 0 do some algebra and cancel some common factors. Step 2: Do some algebra and cancel. In this case we can’t factor so we will multiply by the conjugate of the numerator to get rid of the square roots. This is known as rationalizing the numerator.

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Chapter 2 Notes, Stewart 7e

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9−t √ x→3 3 − t

Example 2.3.4. lim

Example 2.3.5. lim

x→0

1 x+4

− x

1 4

 3x + 9 if x < −2 Example 2.3.6. lim f (x) if f (x) = x→−2 x2 − 1 if x > −2

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Chapter 2 Notes, Stewart 7e

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(x + h)2 − x2 h→0 h

Example 2.3.7. lim

STEP 1: Plug in for h = 0: 0 Here we didn’t get a number but rather an undefined expression of the form . This means that we 0 can do some algebra and cancel some common factors. Step 2: Do some algebra and cancel. In this case we will multiply out the numerator.

f (x + h) − f (x) where f (x) = x2 − 4x. h→0 h

Example 2.3.8. Find lim

f (x)

f (x+h)

}| { z }| { z (x + h)2 − 4(x + h) − (x2 − 4x) f (x + h) − f (x) = lim lim h→0 h→0 h h

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Chapter 2 Notes, Stewart 7e

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The Squeeze Theorem If f (x) ≤ g(x) ≤ h(x) when x is near a (except possibly a) and lim f (x) = lim h(x) = L

x→a

x→a

then lim g(x) = L

x→a

  1 Example 2.3.9. Evaluate lim x sin x→0 x 2

y

f (x) = x2

0.15 0.10 0.05 x −1.0

−0.5 −0.05

0.5

1.0

−0.10

−0.15 h(x) = −x2 y f (x) = x2

0.05

x −0.2 −0.1

−0.05

0.1

0.2

0.3

h(x) = −x2

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Chapter 2 Notes, Stewart 7e

2.4

Chalmeta

The Precise Definition of a Limit

Recall: A. Informal Definition of Limit Let f (x) be defined on an open interval about a except possibly at a itself. If f (x) gets arbitrarily close to L (as close to L as we like) for all x sufficiently close to a, we say that f approaches the limit L as x approaches a and we write: lim f (x) = L (“the limit of f (x), as x approaches a, equals L”). x→a

Notes: (1) x → a means that you approach x = a from both sides of a. (2) f (a) does not have to be defined. B. Solving an inequality 1. Fill in the blanks: To say that |x − 3| < 1 means that x is less than

units from

.

2. Solve |x − 3| < 1

Definition 2.1. The ǫ − δ Definition of Limit Let f (x) be defined on an open interval about a except possibly at a itself. Then we say that f approaches the limit L as x approaches a and we write: lim f (x) = L

x→a

if for all ǫ > 0 there exists a δ > 0 such that |f (x) − L| < ǫ whenever 0 < |x − a| < δ. OR if 0 < |x − a| < δ then |f (x) − L| < ǫ.

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Chapter 2 Notes, Stewart 7e

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Example 2.4.1. Consider the function f (x) = −2x + 5. How close to a = 1, must we hold x to be sure that f (x) lies within 1.5 units of f (a) = 3? In terms of the definition f (x) = −2x + 5, a = 1, L = 3 and ǫ = 1.5 the distance between x1 and 1 OR the distance between x2 and 1 is the desired value of δ. y 6 5

L+ǫ

4 3

L

2

L−ǫ −2

−1

1 x1 −1

x2

1

x 2

3

4

−2 We can also use algebra. We want to know when is |f (x) − L| < 1.5? We solve the inequality: |f (x) − 3| < 1.5 |(−2x + 5) − 3| < 1.5

√ Example 2.4.2. Use the graph of f (x) = 19 − x, a = 10, L = 3 and ǫ = 1 to determine the value of δ √ using the definition of lim 19 − x = 3. To do this we will find a number δ such that if 0 < |x − 10| < δ x→10 √ then | 19 − x − 3| < 1. y 5 L+ǫ 4 L

3

L−ǫ 2 1

−1

a−δ

10

x

a+δ

20

−2 13

Chapter 2 Notes, Stewart 7e

Chalmeta

If the value of ǫ is not specified then you solve for δ in terms of ǫ. This usually involves using algebra on the inequality |f (x) − L| < ǫ until you can get it to look like |x − a| < something. At that point something = δ. Example 2.4.3. Prove that lim 3x + 5 = 8. x→1

Given any ǫ > 0 we need to find δ > 0 such that 0 < |x − a| < δ =⇒ 0 < |x − 1| < δ =⇒ |(3x + 5) − 8| < ǫ =⇒ |3x − 3| < ǫ This suggests that we choose δ = =⇒ 3|x − 1| < ǫ ǫ =⇒ |x − 1| < 3 ǫ ǫ Let’s show that δ = works in the definition. If 0 < |x − 1| < δ = then 3 3 ǫ |(3x + 5) − 8| = |3x − 3| = 3|x − 1| < 3δ = 3 = ǫ. 3

|f (x) − L| < ǫ. ǫ . 3

Example 2.4.4. The interior of a typical 1-L measuring cup is a right circular cylinder of radius 6cm. How closely must we measure the height, h, in order to measure out 1 L (1000 cm3 ) with an error of no more than 1% (i.e. 10 cm3 )? (Use: V = πr2 h)

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Chapter 2 Notes, Stewart 7e

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Example 2.4.5. Use the Graph of f (x) = x2 to find a number δ such that |x2 − 4| < 0.5 whenever |x − 2| < δ. y 6 5 L+ǫ L L−ǫ

4 3 2 1

−1.0

−0.5

0.5

1.0

1.5

x1

−1

15

2.0

x2

x 2.5

3.0