Question

see attached question
Prove the following limits using only the ǫ, δ-definition.

Answer 1

\[\forall \epsilon >0\;\exists \delta\; \forall x:\quad
x>\delta\implies \left|\frac{x^2+2x}{x^2+1}-1\right| <\epsilon
\]

Answer 2

\[
\left|\frac{x^2+2x}{x^2+1}-1\right|
= \left|\frac{x^2+2x}{x^2+1}-\frac{x^2+1}{x^2+1}\right|
= \left|\frac{2x-1}{x^2+1}\right|
\]

Answer 3

i got choose δ = 2/ϵ. do u get this answer

Answer 4

How did you get that answer?

Answer 5

Doesn't really explain \(2/\epsilon\)

Answer 6

is that right? what do u think ?

Answer 7

the lesser the denominator, the greater the fraction; the greater the numerator, the greater the fraction.

Answer 8

Wait, don't you want it to be smaller rather than bigger?

Answer 9

what do you mean?

Answer 10

You don't want to over estimate their \(\epsilon\), you want to underestimate it.

Answer 11

If they give you \(\epsilon=1\) then it's not safe to think \(\epsilon = 2\) while it is safe to think \(\epsilon=1/2\)

Answer 12

okay. what is the answer then? Help me

Answer 13

However, I think you were on the right course. Clearly we have something similar to \(2/x\)

Answer 14

I didn't really over estimate ϵ actually. we treate ϵ is as a constant..... otherwise no point to write δ in terms of ϵ

Answer 15

The point is, it is not safe to have the fraction getting larger, you want it to be getting smaller.

Answer 16

That way it remains in the epsilon range.

Answer 17

the open statement is given ϵ >0, there exists δ >0 such that x >δ for all x, then the limit.
we have treated ϵ as a constant in the beginning. we need to find δ

Answer 18

I know what you mean. I'll double check. if you have the answer. Please let me know .thanks

Answer 19

Let's say \(\delta = 1\implies x>1\):\[
\left|\frac{2x-1}{x^2+1}\right| > \left|\frac{2x-x}{x^2+x}\right|=\left|\frac{x}{x(x+1)}\right|=\left|\frac{1}{x+1}\right|>\left|\frac{1}{2x}\right|
\]

Answer 20

\[
\frac{1}{2x} < \epsilon
\]

Answer 21

So we could say: \[
\frac{1}{2\epsilon} < \delta
\]When we know \(\delta >1\)

Answer 22

Then I guess the result is \[\delta = f(\epsilon) =\min\left\{1,\frac{1}{2\epsilon}\right\}\]

Answer 23

so u choose a random number. what about δ=100

Answer 24

What about it?

Answer 25

well. not really like that

Answer 26

I was wrong, I should have said \(\max\)

Answer 27

please explain this step to me

Answer 28

It should be: \[\delta = f(\epsilon) =\max\left\{1,\frac{1}{2\epsilon}\right\}\]

Answer 29

yes, maybe u should say max.

Answer 30

\[
\frac{1}{2x}<\epsilon \implies \frac{1}{2}<\epsilon x\implies \frac{1}{2\epsilon}<x
\]