Calculus (Math 1A) Lecture 6 Vivek Shende September 5, 2017 Hello - - PowerPoint PPT Presentation

Calculus (Math 1A) Lecture 6

Vivek Shende September 5, 2017

Hello and welcome to class!

Hello and welcome to class!

Last time

Hello and welcome to class!

Last time

We introduced limits, and discussed slopes of tangent lines.

Hello and welcome to class!

Last time

We introduced limits, and discussed slopes of tangent lines.

Today

Hello and welcome to class!

Last time

We introduced limits, and discussed slopes of tangent lines.

Today

We give rules for computing limits, and use them in examples.

Hello and welcome to class!

Last time

We introduced limits, and discussed slopes of tangent lines.

Today

We give rules for computing limits, and use them in examples. These rules will be justified later, in terms of the precise ǫ − δ definition of limits.

Limits

For a function f (x), we say lim

x→x0 f (x) = y0

if, by restricting our attention to x very close but not equal to x0, we can guarantee that f (x) is very close to y0.

The limit of a constant function

The limit of a constant function

For c a constant, lim

x→a c = c

(7)

The limit of f (x) = x

lim

x→a x = a

(8)

Arithmetic of limits

If c is a constant, and limx→a f (x) and limx→a g(x) exist and are finite,

Arithmetic of limits

If c is a constant, and limx→a f (x) and limx→a g(x) exist and are finite, then: lim

x→a[f (x) + g(x)]

= [ lim

x→a f (x)] + [ lim x→a g(x)]

(1) lim

x→a[f (x) − g(x)]

= [ lim

x→a f (x)] − [ lim x→a g(x)]

(2) lim

x→a[cf (x)]

= c[ lim

x→a f (x)]

(3) lim

x→a[f (x)g(x)]

= [ lim

x→a f (x)] · [ lim x→a g(x)]

(4) lim

x→a

f (x) g(x) = limx→a f (x) limx→a g(x) if lim

x→a g(x) = 0

(5)

Arithmetic of limits

In words:

Arithmetic of limits

In words: a sum, difference, constant multiple, product, or quotient

Arithmetic of limits

In words: a sum, difference, constant multiple, product, or quotient

• f limits which exist

Arithmetic of limits

In words: a sum, difference, constant multiple, product, or quotient

• f limits which exist

is the limit of the corresponding sum, difference, constant multiple, product, or quotient.

Arithmetic of limits

If c is a constant, and limx→a f (x) and limx→a g(x) exist and are finite, then: lim

x→a[f (x) + g(x)]

= [ lim

x→a f (x)] + [ lim x→a g(x)]

(1) lim

x→a[f (x) − g(x)]

= [ lim

x→a f (x)] − [ lim x→a g(x)]

(2) lim

x→a[cf (x)]

= c[ lim

x→a f (x)]

(3) lim

x→a[f (x)g(x)]

= [ lim

x→a f (x)] · [ lim x→a g(x)]

(4) lim

x→a

f (x) g(x) = limx→a f (x) limx→a g(x) if lim

x→a g(x) = 0

(5)

Arithmetic of limits

Example: lim

x→5

x + 1 x + 2

Arithmetic of limits

Example: lim

x→5

x + 1 x + 2 By the quotient law we have lim

x→5

x + 1 x + 2 = lim

x→5(x + 1)

lim

x→5(x + 2)

Arithmetic of limits

Example: lim

x→5

x + 1 x + 2 By the quotient law we have lim

x→5

x + 1 x + 2 = lim

x→5(x + 1)

lim

x→5(x + 2)

assuming these two limits exist

Arithmetic of limits

Example: lim

x→5

x + 1 x + 2 By the quotient law we have lim

x→5

x + 1 x + 2 = lim

x→5(x + 1)

lim

x→5(x + 2)

assuming these two limits exist and the one in the denominator is nonzero.

Arithmetic of limits

We continue with lim

x→5(x + 1)

Arithmetic of limits

We continue with lim

x→5(x + 1)

By the sum law, lim

x→5(x + 1) = lim x→5 x + lim x→5 1

Arithmetic of limits

We continue with lim

x→5(x + 1)

By the sum law, lim

x→5(x + 1) = lim x→5 x + lim x→5 1

assuming these two limits exist.

Arithmetic of limits

We continue with lim

x→5(x + 1)

By the sum law, lim

x→5(x + 1) = lim x→5 x + lim x→5 1

assuming these two limits exist. They do, and we know how to compute them: lim

x→5(x + 1) = lim x→5 x + lim x→5 1 = 5 + 1 = 6

Arithmetic of limits

We have learned that lim

x→5(x + 1) = lim x→5 x + lim x→5 1 = 5 + 1 = 6

and in particular, that the limit exists.

Arithmetic of limits

We have learned that lim

x→5(x + 1) = lim x→5 x + lim x→5 1 = 5 + 1 = 6

and in particular, that the limit exists. Similarly, lim

x→5(x + 2) = lim x→5 x + lim x→5 2 = 5 + 2 = 7

Arithmetic of limits

We have learned that lim

x→5(x + 1) = lim x→5 x + lim x→5 1 = 5 + 1 = 6

and in particular, that the limit exists. Similarly, lim

x→5(x + 2) = lim x→5 x + lim x→5 2 = 5 + 2 = 7

Putting these together, we see that lim

x→5

x + 1 x + 2 = lim

x→5(x + 1)

lim

x→5(x + 2) = 6

7

Try it yourself!

Using only the laws we have seen so far — i.e., the sum, difference, scalar multiple, product, quotient rules and the rules for limits of the constant and identity functions — compute: lim

x→4(x + 1)(x + 2)(x + 3)

Try it yourself!

Try it yourself!

lim

x→4(x + 1)(x + 2)(x + 3) = ( lim x→4(x + 1))( lim x→4(x + 2))( lim x→4(x + 3))

Try it yourself!

lim

x→4(x + 1)(x + 2)(x + 3) = ( lim x→4(x + 1))( lim x→4(x + 2))( lim x→4(x + 3))

= ( lim

x→4 x + lim x→4 1)( lim x→4 x + lim x→4 2)( lim x→4 x + lim x→4 3)

Try it yourself!

lim

x→4(x + 1)(x + 2)(x + 3) = ( lim x→4(x + 1))( lim x→4(x + 2))( lim x→4(x + 3))

= ( lim

x→4 x + lim x→4 1)( lim x→4 x + lim x→4 2)( lim x→4 x + lim x→4 3)

= (4 + 1)(4 + 2)(4 + 3) = 5 · 6 · 7 = 210

Try it yourself!

lim

x→4(x + 1)(x + 2)(x + 3) = ( lim x→4(x + 1))( lim x→4(x + 2))( lim x→4(x + 3))

= ( lim

x→4 x + lim x→4 1)( lim x→4 x + lim x→4 2)( lim x→4 x + lim x→4 3)

= (4 + 1)(4 + 2)(4 + 3) = 5 · 6 · 7 = 210 Technical note: the first equalities are justified only because we learn from subsequent further calculations that the limits on the right hand side of each actually exist.

Roots and powers

Roots and powers

Repeatedly applying the product law, we learn lim

x→a[f (x)]n = [ lim x→a f (x)]n

(6)

Roots and powers

Repeatedly applying the product law, we learn lim

x→a[f (x)]n = [ lim x→a f (x)]n

(6) In particular, taking f (x) = x: lim

x→a xn = [ lim x→a x]n = an

(9)

Roots and powers

Roots and powers

Applying this to f (x) = n g(x) for g(x) ≥ 0 lim

x→a g(x) = lim x→a[n

g(x)]n = [ lim

x→a n

g(x)]n Taking n’th roots on both sides gives

n

lim

x→a g(x) = lim x→a n

g(x) (11)

Roots and powers

Applying this to f (x) = n g(x) for g(x) ≥ 0 lim

x→a g(x) = lim x→a[n

g(x)]n = [ lim

x→a n

g(x)]n Taking n’th roots on both sides gives

n

lim

x→a g(x) = lim x→a n

g(x) (11) In particular, taking g(x) = x: lim

x→a n√x = n√a

(10)

Continuity

Continuity

If f is any polynomial, rational, or algebraic function,

Continuity

If f is any polynomial, rational, or algebraic function, and a is in its domain,

Continuity

If f is any polynomial, rational, or algebraic function, and a is in its domain, then lim

x→a f (x) = f (a)

Continuity

If f is any polynomial, rational, or algebraic function, and a is in its domain, then lim

x→a f (x) = f (a)

Why this is true: such functions are built from arithmetic

• perations and taking roots; so repeatedly use the limit laws we

just learned.

Continuity

If f is any polynomial, rational, or algebraic function, and a is in its domain, then lim

x→a f (x) = f (a)

Why this is true: such functions are built from arithmetic

• perations and taking roots; so repeatedly use the limit laws we

just learned. In section 2.3, your book calls this the direct substitution property.

Continuity

If f is any polynomial, rational, or algebraic function, and a is in its domain, then lim

x→a f (x) = f (a)

Why this is true: such functions are built from arithmetic

• perations and taking roots; so repeatedly use the limit laws we

just learned. In section 2.3, your book calls this the direct substitution property. However from section 2.5 onwards, this will become the definition

• f continuity:

Continuity

If f is any polynomial, rational, or algebraic function, and a is in its domain, then lim

x→a f (x) = f (a)

Why this is true: such functions are built from arithmetic

• perations and taking roots; so repeatedly use the limit laws we

just learned. In section 2.3, your book calls this the direct substitution property. However from section 2.5 onwards, this will become the definition

• f continuity: a function with the above property is continuous.

Continuity

• Example. Compute lim

x→5

x + √x − 1 x2 + 2x .

Continuity

• Example. Compute lim

x→5

x + √x − 1 x2 + 2x . After checking that 5 is in the domain of the expression, we just plug it in: lim

x→5

5 + √5 − 1 52 + 2 · 5 = 7 35

Try it yourself!

Compute lim

x→2

3x + 3√x − 1 2x − 2 .

Try it yourself!

Compute lim

x→2

3x + 3√x − 1 2x − 2 . After checking that 2 is in the domain of the expression, we just plug it in: lim

x→2

6 + 3√2 − 1 4 − 2 = 7 2

Continuity

• Example. Compute lim

x→5

x2 − 5x x − 5 .

Continuity

• Example. Compute lim

x→5

x2 − 5x x − 5 . After checking that 5 is in the domain of the expression,

Continuity

• Example. Compute lim

x→5

x2 − 5x x − 5 . After checking that 5 is in the domain of the expression, We discover that it’s not!

Continuity

• Example. Compute lim

x→5

x2 − 5x x − 5 . After checking that 5 is in the domain of the expression, We discover that it’s not! So we have to do something other than just plugging it in.

Independence of the value

Independence of the value

If f (x) = g(x) except possibly at x = a

Independence of the value

If f (x) = g(x) except possibly at x = a then lim

x→a f (x) = lim x→a g(x)

Independence of the value

If f (x) = g(x) except possibly at x = a then lim

x→a f (x) = lim x→a g(x)

More precisely, if one limit is defined, so is the other, and in this case they are equal.

Independence of the value

If f (x) = g(x) except possibly at x = a then lim

x→a f (x) = lim x→a g(x)

More precisely, if one limit is defined, so is the other, and in this case they are equal. This is true because the limit only takes into account values near but not at a.

Independence of the value

If f (x) = g(x) except possibly at x = a then lim

x→a f (x) = lim x→a g(x)

More precisely, if one limit is defined, so is the other, and in this case they are equal. This is true because the limit only takes into account values near but not at a. Note: f nor g need not have the same domain, and need not be defined at a.

Independence of the value

If f (x) = g(x) except possibly at x = a then lim

x→a f (x) = lim x→a g(x)

More precisely, if one limit is defined, so is the other, and in this case they are equal. This is true because the limit only takes into account values near but not at a. Note: f nor g need not have the same domain, and need not be defined at a. One needs only f (x) = g(x) over some [b, a) ∪ (a, c].

Independence of the value

• Example. Compute lim

x→5

x2 − 5x x − 5 .

Independence of the value

• Example. Compute lim

x→5

x2 − 5x x − 5 . lim

x→5

x2 − 5x x − 5

Independence of the value

• Example. Compute lim

x→5

x2 − 5x x − 5 . lim

x→5

x2 − 5x x − 5 = lim

x→5

x(x − 5) x − 5

Independence of the value

• Example. Compute lim

x→5

x2 − 5x x − 5 . lim

x→5

x2 − 5x x − 5 = lim

x→5

x(x − 5) x − 5 = lim

x→5 x

Independence of the value

• Example. Compute lim

x→5

x2 − 5x x − 5 . lim

x→5

x2 − 5x x − 5 = lim

x→5

x(x − 5) x − 5 = lim

x→5 x = 5

Independence of the value

• Example. Compute lim

x→5

x2 − 5x x − 5 . lim

x→5

x2 − 5x x − 5 = lim

x→5

x(x − 5) x − 5 = lim

x→5 x = 5

Here in the second equality, we used that x(x−5)

x−5

= x for any value

• f x other than 5.

Independence of the value

• Example. Compute lim

x→2

x − 2 √ x2 − 3 − 1 .

Independence of the value

• Example. Compute lim

x→2

x − 2 √ x2 − 3 − 1 . Note we cannot just plug in x = 2,

Independence of the value

• Example. Compute lim

x→2

x − 2 √ x2 − 3 − 1 . Note we cannot just plug in x = 2, because even though the limit

• f the numerator and denominator both exist,

Independence of the value

• Example. Compute lim

x→2

x − 2 √ x2 − 3 − 1 . Note we cannot just plug in x = 2, because even though the limit

• f the numerator and denominator both exist, the limit of the

denominator is 0.

Independence of the value

• Example. Compute lim

x→2

x − 2 √ x2 − 3 − 1 . Note we cannot just plug in x = 2, because even though the limit

• f the numerator and denominator both exist, the limit of the

denominator is 0. Remark: If the limit of the numerator were not zero,

Independence of the value

• Example. Compute lim

x→2

x − 2 √ x2 − 3 − 1 . Note we cannot just plug in x = 2, because even though the limit

• f the numerator and denominator both exist, the limit of the

denominator is 0. Remark: If the limit of the numerator were not zero, the limit of the quotient would be infinite or undefined. But it is zero.

We want to cancel some zeroes from the numerator and denominator.

We want to cancel some zeroes from the numerator and

• denominator. We start by removing the square-root from the

denominator: lim

x→2

x − 2 √ x2 − 3 − 1 = lim

x→2

x − 2 √ x2 − 3 − 1 · √ x2 − 3 + 1 √ x2 − 3 + 1 = lim

x→2

(x − 2)( √ x2 − 3 + 1) (x2 − 3) − 1 = lim

x→2

(x − 2)( √ x2 − 3 + 1) x2 − 4 = lim

x→2

(x − 2)( √ x2 − 3 + 1) (x − 2)(x + 2)

Now we are able to cancel:

Now we are able to cancel: lim

x→2

(x − 2)( √ x2 − 3 + 1) (x − 2)(x + 2) = lim

x→2

√ x2 − 3 + 1 x + 2

Now we are able to cancel: lim

x→2

(x − 2)( √ x2 − 3 + 1) (x − 2)(x + 2) = lim

x→2

√ x2 − 3 + 1 x + 2 Finally, we can plug in 2.

Now we are able to cancel: lim

x→2

(x − 2)( √ x2 − 3 + 1) (x − 2)(x + 2) = lim

x→2

√ x2 − 3 + 1 x + 2 Finally, we can plug in 2. lim

x→2

√ x2 − 3 + 1 x + 2 = √ 22 − 3 + 1 2 + 2 = 1 2

Try it yourself!

Compute: lim

x→3

√ x2 − 5 − 2 x − 3

Limits from one-sided limits

Limits from one-sided limits

If limx→x−

0 f (x) and limx→x+ 0 f (x) exist

Limits from one-sided limits

If limx→x−

0 f (x) and limx→x+ 0 f (x) exist and are equal

Limits from one-sided limits

If limx→x−

0 f (x) and limx→x+ 0 f (x) exist and are equal

then also limx→x0 f (x) exists

Limits from one-sided limits

If limx→x−

0 f (x) and limx→x+ 0 f (x) exist and are equal

then also limx→x0 f (x) exists and lim

x→x−

f (x) = lim

x→x0 f (x) = lim x→x+

f (x)

Limits from one-sided limits

If limx→x−

0 f (x) and limx→x+ 0 f (x) exist and are equal

then also limx→x0 f (x) exists and lim

x→x−

f (x) = lim

x→x0 f (x) = lim x→x+

f (x) All the limit laws hold for one-sided limits as well.

Limits from one-sided limits

We find the one-sided limits lim

x→0+ f (x) = 0 and lim x→0− f (x) = 0 by

direct substitution.

Limits from one-sided limits

We find the one-sided limits lim

x→0+ f (x) = 0 and lim x→0− f (x) = 0 by

direct substitution. Thus lim

x→0 f (x) = 0.

Try it yourself

Find lim

x→1 f (x), where

f (x) =

• x2−1

x−1

x > 1 x2 + 1 x < 1

Squeeze theorem

Squeeze theorem

If f (x) ≤ g(x) ≤ h(x) on some (a, b) ∪ (b, c)

Squeeze theorem

If f (x) ≤ g(x) ≤ h(x) on some (a, b) ∪ (b, c) and lim

x→b f (x) = lim x→b h(x)

Squeeze theorem

If f (x) ≤ g(x) ≤ h(x) on some (a, b) ∪ (b, c) and lim

x→b f (x) = lim x→b h(x)

Then lim

x→b g(x) exists and

lim

x→b f (x) = lim x→b g(x) = lim x→b h(x)

Also sometimes called the sandwich theorem

Squeeze theorem

If f (x) ≤ g(x) ≤ h(x) on some (a, b) ∪ (b, c) and lim

x→b f (x) = lim x→b h(x)

Then lim

x→b g(x) exists and

lim

x→b f (x) = lim x→b g(x) = lim x→b h(x)

Also sometimes called the sandwich theorem or the two policemen and one drunk theorem.

If a police officer stands on each side of a drunk, and both police go to the jail, then no matter how the drunk wobbles about, he will also end up in jail.

sin(1/x)

Squeeze theorem

lim

x→0 x2 sin(1/x) = ?

x2 sin(1/x)

Squeeze theorem

Squeeze theorem

−1 ≤ sin(1/x) ≤ 1

Squeeze theorem

−1 ≤ sin(1/x) ≤ 1 −x2 ≤ x2 sin(1/x) ≤ x2

Squeeze theorem

−1 ≤ sin(1/x) ≤ 1 −x2 ≤ x2 sin(1/x) ≤ x2 lim

x→0 −x2 = 0 = lim x→0 x2

Squeeze theorem

−1 ≤ sin(1/x) ≤ 1 −x2 ≤ x2 sin(1/x) ≤ x2 lim

x→0 −x2 = 0 = lim x→0 x2

So by the squeeze theorem lim

x→0 x2 sin(1/x) = 0

sin(x)/x

Graphs of sin(x)/x, x, sin(x)

Some trigonometric limits from the squeeze theorem

Some trigonometric limits from the squeeze theorem

Some trigonometric limits from the squeeze theorem

Area(Triangle ADB) ≤ Area(Green sector) ≤ Area(Triangle ADF)

Some trigonometric limits from the squeeze theorem

Area(Triangle ADB) ≤ Area(Green sector) ≤ Area(Triangle ADF) 1 2 sin(x) ≤ x 2π · π ≤ 1 2 tan(x)

Some trigonometric limits from the squeeze theorem

1 2 sin(x) ≤ x 2π · π ≤ 1 2 tan(x)

Some trigonometric limits from the squeeze theorem

1 2 sin(x) ≤ x 2π · π ≤ 1 2 tan(x) sin(x) ≤ x ≤ sin(x) cos(x)

Some trigonometric limits from the squeeze theorem

1 2 sin(x) ≤ x 2π · π ≤ 1 2 tan(x) sin(x) ≤ x ≤ sin(x) cos(x) x cos(x) ≤ sin(x) ≤ x

Some trigonometric limits from the squeeze theorem

1 2 sin(x) ≤ x 2π · π ≤ 1 2 tan(x) sin(x) ≤ x ≤ sin(x) cos(x) x cos(x) ≤ sin(x) ≤ x cos(x) ≤ sin(x) x ≤ 1

Some trigonometric limits from the squeeze theorem

1 2 sin(x) ≤ x 2π · π ≤ 1 2 tan(x) sin(x) ≤ x ≤ sin(x) cos(x) x cos(x) ≤ sin(x) ≤ x cos(x) ≤ sin(x) x ≤ 1 By the squeeze theorem, lim

x→0

sin(x) x = 1