MAT137 | Calculus! | Lecture 8

Today:

Squeeze Theorem (v. 2.11-2.12)Trigonometric Limit (v. 2.17)

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Watch videos on:Continuity (v. 2.13 - 2.16)

official website http://uoft.me/MAT137Beatriz Navarro-Lameda L0601 MAT137 03 October 2017

Bad theorem

Bad Theorem

Let f and g be functions with domain R, except perhaps a.IF lim

x→af (x) = 0,

THEN limx→a

f (x)g(x) = 0.

Bad Proof

limx→a

f (x)g(x) =[

limx→a

f (x)]·[

limx→a

g(x)]

= 0 ·[

limx→a

g(x)]

= 0, because 0

times anything is 0.

Find the error in the proof.

Show the theorem is false with a counterexample.

Beatriz Navarro-Lameda L0601 MAT137 03 October 2017

Limit: Bounded × something that goes to 0

Theorem

Let a ∈ R. Let f and g be functions with domain R, except perhaps a.

IF

limx→a

f (x) = 0

g is bounded.

THEN limx→a

f (x)g(x) = 0.

Definition (Bounded function)

Let g : R→ R. We say that g is bounded if

∃M > 0 s.t. ∀x ∈ R |g(x)| ≤ M.

Beatriz Navarro-Lameda L0601 MAT137 03 October 2017

The Squeeze Theorem

Theorem

Let p > 0. Suppose that for all x with 0 < |x − a| < p,

f (x) ≤ g(x) ≤ h(x).

IFlimx→a

f (x) = L and limx→a

h(x) = L,

THENlimx→a

g(x) = L.

Beatriz Navarro-Lameda L0601 MAT137 03 October 2017

The Squeeze Theorem

Exercise

Use the Squeeze Theorem to prove that:IF f (x) is bounded and lim

x→ag(x) = 0, THEN lim

x→af (x)g(x) = 0.

Exercise

IF limx→a|f (x)| = 0, THEN lim

x→af (x) = 0.

Hint: First show that −|f (x)| ≤ f (x) ≤ |f (x)|.

Beatriz Navarro-Lameda L0601 MAT137 03 October 2017

The Squeeze Theorem

Exercise

Use the Squeeze Theorem to prove that:IF f (x) is bounded and lim

x→ag(x) = 0, THEN lim

x→af (x)g(x) = 0.

Exercise

IF limx→a|f (x)| = 0, THEN lim

x→af (x) = 0.

Hint: First show that −|f (x)| ≤ f (x) ≤ |f (x)|.

Beatriz Navarro-Lameda L0601 MAT137 03 October 2017

Limits involving sin(1/x) Part I

The reason that limx→0

sin(1/x) does not exist is:

1 because no matter how close x gets to 0, there are x ’s near 0 forwhich sin(1/x) = 1, and some for which sin(1/x) = −1

2 because the function values oscillate around 0

3 because 1/0 is undefined

4 all of the above

-1.5 -1.0 -0.5 0.5 1.0 1.5

-1.0

-0.5

0.5

1.0

Beatriz Navarro-Lameda L0601 MAT137 03 October 2017

Limits involving sin(1/x) Part I

The reason that limx→0

sin(1/x) does not exist is:

1 because no matter how close x gets to 0, there are x ’s near 0 forwhich sin(1/x) = 1, and some for which sin(1/x) = −1

2 because the function values oscillate around 0

3 because 1/0 is undefined

4 all of the above

-1.5 -1.0 -0.5 0.5 1.0 1.5

-1.0

-0.5

0.5

1.0

Beatriz Navarro-Lameda L0601 MAT137 03 October 2017

Limits involving sin(1/x) Part II

The limit limx→0

x2 sin(1/x)

1 does not exist because no matter how close x gets to 0, there are x ’snear 0 for which sin(1/x) = 1, and some for which sin(1/x) = −1

2 does not exist because the function values oscillate around 0

3 does not exist because 1/0 is undefined

4 equals 0

5 equals 1

Beatriz Navarro-Lameda L0601 MAT137 03 October 2017

Limits involving sin(1/x) Part II

The limit limx→0

x2 sin(1/x)

1 does not exist because no matter how close x gets to 0, there are x ’snear 0 for which sin(1/x) = 1, and some for which sin(1/x) = −1

2 does not exist because the function values oscillate around 0

3 does not exist because 1/0 is undefined

4 equals 0

5 equals 1

Beatriz Navarro-Lameda L0601 MAT137 03 October 2017

Some Trigonometric Limits

Trig. Limit 1

You already know that limx→0

sin(x)

x= 1.

Use to evaluate limx→0

sin(3x)

x.

Answer:

1 1

2 3

3 13

-1.5 -1.0 -0.5 0.5 1.0 1.5

1

2

3

Beatriz Navarro-Lameda L0601 MAT137 03 October 2017

Some Trigonometric Limits

Trig. Limit 1

You already know that limx→0

sin(x)

x= 1.

Use to evaluate limx→0

sin(3x)

x.

Answer:

1 1

2 3

3 13

-1.5 -1.0 -0.5 0.5 1.0 1.5

1

2

3

Beatriz Navarro-Lameda L0601 MAT137 03 October 2017

Some Trigonometric Limits

Trig. Limit 2

The limit limx→0

1− cos(x)

xis

Answer:

1 1

2 0

3 12

Hint

Use the trigonometric identity sin2 x + cos2 x = 1.

Beatriz Navarro-Lameda L0601 MAT137 03 October 2017