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So, what about δ? How to include it in the answer?
Why δ≤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.
Compare |f(x)-L|=|1/(2x+1)| to |x−c|=|x−1/2|, like those in my common practice to elementary exercise?
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.
Do you mean because they are inequality?
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...
Is it B?
|[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.
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.
Is it B?
|[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.
I agree my fundamental problem is that I don't really understand what I'm trying to do with these epsilons and deltas, but I don't really know what can I do with that. I just learn those things online, with only telling me to substitute numbers into the definition written. Are there any suggestions for what can I do other than asking question here? Thanks.
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).
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.I now know how to get δ=0.5
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