Fifth lecture for MAT137 - LEC5201 Topics for today: 1. More limits - - PowerPoint PPT Presentation

Fifth lecture for MAT137 - LEC5201

Topics for today:

• 1. More limits
• 2. Squeeze theorems
• 3. Continuity

Required videos for next week: 2.21-2.22, 3.1-3.8 We will do some review for the test next lecture.

Jeffrey Im MAT137 October 10, 2018 1 / 16

The course’s convention for when limits don’t exist

Suppose f is a real-valued function with domain R and that a ∈ R.

Nonexistence of limits

We say that lim

x→a f (x) does not exist

when there is not an L ∈ R for which limx→a f (x) = L. Limits which tend to infinity will be treated as limits not existing in a particular way.

Jeffrey Im MAT137 October 10, 2018 2 / 16

Indeterminate form

Let a ∈ R and let f and g be positive functions defined near a. Assume lim

x→a f (x) = lim x→a g(x) = 0.

What can be concluded about lim

x→a

f (x) g(x)?

• 1. The limit is 1.
• 2. The limit is 0.
• 3. The limit is ∞.
• 4. The limit does not exist.
• 5. We do not have enough information to decide.

Jeffrey Im MAT137 October 10, 2018 3 / 16

A theorem about limits

Let f be a function with domain R such that lim

x→0 f (x) = 3.

Prove that lim

x→0[5f (2x)] = 15

directly from the definition of the limit (No appeals to limit laws!).

• 1. Write down what needs to be proved.
• 2. Do some scratch work.
• 3. Write up a proof.

Jeffrey Im MAT137 October 10, 2018 4 / 16

Proof feedback

• 1. Is the structure of the proof correct?

(First fix ε, then choose δ, then ...)

• 2. Did you say exactly what δ is?
• 3. Is the proof self-contained?

(I do not need to read the rough work)

• 4. Are all variables defined? In the right order?
• 5. Do all steps follow logically from what comes before?

Do you start from what you know and prove what you have to prove?

• 6. Did you remember not to assume your conclusion?

Jeffrey Im MAT137 October 10, 2018 5 / 16

Infinite version of the squeeze theorem

The usual squeeze theorem

Let a, L ∈ R and suppose f , g, h are functions defined near a except possibly at a. If f (x) ≤ g(x) ≤ h(x) for x close to a, but not a, limx→a f (x) = L, limx→a h(x) = L, then limx→a g(x) = L. Come up with a version of the theorem where the conclusion is lim

x→a g(x) = ∞.

Jeffrey Im MAT137 October 10, 2018 6 / 16

Infinite version of the squeeze theorem

Theorem

Let a ∈ R and let f and g be functions defined near a, except possibly at

• a. If

for x close to a but not a, f (x) ≤ g(x), and limx→a f (x) = ∞, then lim

x→a g(x) = ∞.

Now let’s prove it.

• 1. Make the hypothesis more precise.
• 2. Write down what you want to prove.
• 3. Do some scratch work.
• 4. Write down a proof.

Jeffrey Im MAT137 October 10, 2018 7 / 16

Infinite version of the squeeze theorem

Proof.

Let M ∈ R be given. From our hypotheses, we know that

• 1. there exists a δ > 0 such that 0 < |x − a| < δ implies f (x) > M,
• 2. and f (x) ≤ g(x).

Using these hypotheses together, we have that g(x) ≥ f (x) > M whenever 0 < |x − a| < δ. This shows that limx→a g(x) = ∞, as desired.

Jeffrey Im MAT137 October 10, 2018 8 / 16

A composition theorem

Write a formal proof for

Theorem

Let a, L ∈ R. Let f and g be functions. IF lim

x→a f (x) = L

g is continuous at L THEN lim

x→a g(f (x)) = g(L)

Jeffrey Im MAT137 October 10, 2018 9 / 16

A composition theorem

Proof.

Let ε > 0 be given. Our hypotheses guarantee these two points:

• 1. (continuity of g at L) there exists an η > 0 such that |y − L| < η

implies |g(y) − g(L)| < ε.

• 2. (limx→a f (x) = L) there exists a δ > 0 such that 0 < |x − a| < δ

implies |f (x) − L| < η. In point (2), we chose δ so that 0 < |x − a| < δ implies |f (x) − L| < η and in point (1), we chose η so that this guarantees |g(f (x)) − g(L)| < ε. This shows that limx→a g(f (x)) = g(L).

Jeffrey Im MAT137 October 10, 2018 10 / 16

Intermission

Exercise

Show that lim

x→0(x3 + x2) = 0.

First proof we came up with today

Let ε > 0 be given. Then for |x| < 1/2 we have that |x2 + x| < 1. Pick δ := min{ε, 1/2}. Then 0 < |x| < δ implies |x3 + x2| = |x||x2 + x| < |x| · 1 < ε.

2nd one

Let ε > 0 be given and pick δ := min{

• ε/2,

3

• ε/2}. Then 0 < |x| < δ

implies |x3 + x2| ≤ |x|3 + |x|2 < ε/2 + ε/2 = ε.

Jeffrey Im MAT137 October 10, 2018 11 / 16

New continuous functions

We want to prove the following theorem

Theorem

IF f and g are continuous functions THEN h(x) = max{f (x), g(x)} is also a continuous function. You are allowed to use all results that we already know. What is the fastest way to prove this? Hint: What is the number a + b + |a − b| 2 ? There is a way to prove this quickly without writing any epsilons.

Jeffrey Im MAT137 October 10, 2018 12 / 16

New continuous functions

Proof.

• 1. Show that a+b+|a−b|

2

= max{a, b}. (Cases... a ≥ b and a < b).

• 2. So max{f (x), g(x)} = f (x)+g(x)+|f (x)−g(x)|

2

.

• 3. (f (x) + g(x)) is continuous.
• 4. (f (x) − g(x)) is continuous and so |f (x) − g(x)| is also continuous.
• 5. From the previous two points, f (x) + g(x) + |f (x) − g(x)| is

continuous.

• 6. Lastly,

f (x) + g(x) + |f (x) − g(x)| 2 = max{f (x), g(x)} is continuous, as needed.

Jeffrey Im MAT137 October 10, 2018 13 / 16

A question from last year’s test

The only thing we know about the function g is that lim

x→0

g(x) x2 = 2. Use it to compute the following limits (or explain that they do not exist):

• 1. lim

x→0

g(x) x

• 2. lim

x→0

g(x) x4

• 3. lim

x→0

g(3x) x2

Jeffrey Im MAT137 October 10, 2018 14 / 16

Computations!

Using that lim

x→0

sin x x = 1, compute the following limits:

• 1. lim

x→2

sin x x

• 2. lim

x→0

sin(5x) x

• 3. lim

x→0

tan2(2x2) x4

• 4. lim

x→0

sin ex ex

• 5. lim

x→0

1 − cos x x

• 6. lim

x→0

tan10(2x20) sin200(3x)

Jeffrey Im MAT137 October 10, 2018 15 / 16

Limits at infinity Compute:

• 1. lim

x→∞

• x7 − 2x5 + 11
• 2. lim

x→∞

• x2 −
• x5 + 1
• 3. lim

x→∞

x2 + 11 x + 1

• 4. lim

x→∞

x2 + 2x + 3 3x2 + 4x + 5

• 5. lim

x→∞

x3 + √ 2x6 + 1 2x3 + √ x5 + 1

Jeffrey Im MAT137 October 10, 2018 16 / 16