- #1

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- Thread starter ChloeYip
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- #1

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- #2

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- #3

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Why

For

Thanks.

- #4

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δ? How to include it in the answer?

Whyδ≤ϵ? Is it part of the definition of limit? As I know, it only mentioned "for each ϵ>0 there exists δ>0", according to "S. L. Salas, G. J. Etgen, E. Hille,Calculus: One and Several Variables".

For(2x+1)>1 or x>0, are they just an assumption?

Thanks.

There are several ways to tackle this problem. But, I think your first difficulty is that you don't really understand the epsilon-delta definition.

In any case, you are

(A) Let ##|x - 1/2| < \epsilon/100 \ \dots##

What can you do with that?

- #5

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I really have no idea.

After watching this youtube video: , I tried calculate...

Substitute 1/2-∈ & 1/2+∈ to f(x)=1/(2x+1) → ∂=min{1/[2(1-∈)] or 1/[2(1+∈)]} ?

That's totally different to the given options!

What's wrong with that?

Thanks.

- #6

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I really have no idea.

After watching this youtube video: , I tried calculate...

Substitute 1/2-∈ & 1/2+∈ to f(x)=1/(2x+1) → ∂=min{1/[2(1-∈)] or 1/[2(1+∈)]} ?

That's totally different to the given options!

What's wrong with that?

Thanks.

That may be a solution (I haven't checked). There are lots of possible solutions for how to choose delta (depending on epsilon) in these cases. You have to choose between the four you are given. There will be other ways to tackle this limit.

Let me ask a more general question. Suppose, for another problem, you came up with:

##\delta = \epsilon/2##

Can you see why

##\delta = \epsilon/4##

is also a solution?

Or, do you think those can't both be correct?

- #7

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So do you think my steps are correct?

To guess with a more reasonable answer, B or D should be chosen?

I have never learned to put an unknown from denominator to numerator...

- #8

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So do you think my steps are correct?

To guess with a more reasonable answer, B or D should be chosen?

I have never learned to put an unknown from denominator to numerator...

I think you should go back to post #4 and follow my suggestion. Try with B) or D) first if you like. In any case, you need to do some algebra.

- #9

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|[1/(2x+1)]-1/2|=1/2<ϵ → ϵ can be ≥ 1/4 as the only answer.

Am I right?

And how to get 5/2ϵ?

Thanks.

- #10

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|[1/(2x+1)]-1/2|=1/2<ϵ → ϵ can be ≥ 1/4 as the only answer.

Am I right?

And how to get 5/2ϵ?

Thanks.

Other than Q1,

May I ask one more question(Q32)?

I can't even find a constant in Q32.

The following are my steps in Q32:

lx^(2)-x-xl=l(x-2)(x+1)l<ϵ

Let M=x+1,Mδ<ϵ → δ<ϵ/M

l(x-2)l<1 → -1<x-2<1 → 1<x<3

2<x+1<4 → M=4 → δ<ϵ/4 → NO THIS OPTION!

What's wrong with that?

Thanks.

- #11

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So B) doesn't work , because according to B) you take ##\delta\leq\frac{\epsilon}{3}## and this choice doesn't necessarily satisfies ##\delta<\frac{\epsilon}{4}##Other than Q1,

May I ask one more question(Q32)?

View attachment 102111

I can't even find a constant in Q32.

The following are my steps in Q32:

lx^(2)-x-xl=l(x-2)(x+1)l<ϵ

Let M=x+1,Mδ<ϵ → δ<ϵ/M

l(x-2)l<1 → -1<x-2<1 → 1<x<3

2<x+1<4 → M=4 → δ<ϵ/4 → NO THIS OPTION!

What's wrong with that?

Thanks.

Check C) which I believe is the right answer. If all go well you 'll find that delta has to satisfy ##\delta<\frac{\epsilon}{5}##. But C) tells you that you take ##\delta\leq\frac{\epsilon}{6}## which for sure satisfies ##\delta<\frac{\epsilon}{5}##.

- #12

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|[1/(2x+1)]-1/2|=1/2<ϵ → ϵ can be ≥ 1/4 as the only answer.

Am I right?

And how to get 5/2ϵ?

Thanks.

Well, unfortunately, as I said before, your fundamental problem is that you don't really understand what you're trying to do with these epsilons and deltas. You've been given some strong hints and suggestions in this thread, but you don't seem able to follow them up.

My only suggestion is to try to understand what that definition really means. Questions like these should help you, but at the moment you seem to be struggling in the dark and it's difficult to help.

- #13

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- #14

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It's not easy, that's for sure. Almost everyone takes some time to get to grips with this. Also, there is no formulaic approach to these questions. There's no technique, like there might be for solving a certain differential equation, for example, that you can simply learn. In fact, to do real analysis you need to be good with a whole range of algebraic techniques. You will start to see common themes and ideas, but you need to be generally good with algebra and numbers.

First, let me try to describe the idea behind epsilon-delta limits. If we take a point, ##x_0## and we have a function ##f(x)## what we are trying to do is to show that as ##x \rightarrow x_0## we have ##f(x) \rightarrow L##. That means that as ##x## gets close to ##x_0## we must have ##f(x)## getting close to ##L## (and staying close to ##L##).

The epsilon-delta definition you quoted in post #1 captures this idea formally and rigorously. At the centre of the definition is an implication: a certain condition on ##x## must imply a certain condition on ##f(x)##. The key points are:

You must be able to prove the implication for any ##\epsilon##.

The ##\delta## you choose in each case depends on ##\epsilon##.

You must show that ##0 < |x - x_0| < \delta## implies ##|f(x) - L| < \epsilon##.

So, in general, you would start by saying:

1) Let ##\epsilon > 0##.

2) Choose ##\delta = \dots## (this is the first tricky bit: you have to find your delta for each epsilon).

3) If ##0 < |x - x_0| < \delta## then ##\dots## (this is the second tricky bit. By algebraic, logic, and any means at your disposal, you must have a rock-solid series of logical steps that leads to ##\dots \ |f(x) - L| < \epsilon##

This is as far as you can go in general. In step 3) almost any algebraic trick you could think of might be needed. For step 3) you really need to think on your feet.

In these questions, they have given you some options for ##\delta##. So, you need to focus on step 3).

The first thing I would do is to work on the expression ##|f(x) - L|##. You need to find how this expression depends on ##|x-x_0|##. Let's take your first question as an example:

##|\frac{1}{2x+1} - \frac{1}{2}| = |\frac{2 - (2x+1)}{2(2x+1)}| = |\frac{1-2x}{2(2x+1)}| = |\frac{2x - 1}{2(2x+1)}| = |\frac{x - 1/2}{2x+1}|##

Do you see what I did in those last two steps? We're now getting somewhere. But, we need to do something about the ##x## in the denominator. Any ideas?

- #15

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rationalize them?

- #16

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the following are the steps:

If δ<1, then lx-0.5l<δ<1 ⇒ -1<x-0.5<1 ⇒-0.5<x-0.5<1.5 ⇒ ϵ = 0.5

Am I right? Thanks

- #17

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In fact, it is a multiple choice question and I think we need to focus on answering it first (i.e. to find δ), then learn ways to solve it...In these questions, they have given you some options for ##\delta##. So, you need to focus on step 3).

And as I know, finding delta is the essential to learn to solve it. Hope you may help me learn it first because I still don't understand how to get it done.

Thank you very much for helping me~

- #18

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This is not correct. To get back on track you need to go back to the last expression in post #14 and deal with the denominator.

the following are the steps:

If δ<1, then lx-0.5l<δ<1 ⇒ -1<x-0.5<1 ⇒-0.5<x-0.5<1.5 ⇒ ϵ = 0.5

Am I right? Thanks

hint: What happens if ##x > 0##? Would that help?

I'll be offline for a bit, but hopefully someone else can help.

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