A great refresher. I'm actually working on those tonight, since in series and sequences the limits come back!! I was hoping they would disappear after the derivative ;)

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- April 4th 2012, 07:00 PMDKSanders123Re: For those that never learnt well the epsilon-delta proofs
A great refresher. I'm actually working on those tonight, since in series and sequences the limits come back!! I was hoping they would disappear after the derivative ;)

- July 16th 2012, 02:05 AMnhellie16Re: For those that never learnt well the epsilon-delta proofs
thanks for the file.. it helps me lot... By the way its my birthday today.. thanks guys.. God bless..

- August 20th 2012, 08:57 AMebbRe: For those that never learnt well the epsilon-delta proofs
Could I just check that my answer to this question is correct? Well, can I check that the d in my answer is correct at any rate!

Give an e d proof that the function

f(x) = (x^2 + 1)/(x + 1)

is continuous at x = 1.

I got d=min(1, 1.5e)

Thanks! - August 20th 2012, 09:26 AMebbRe: For those that never learnt well the epsilon-delta proofs
Also for f(x) = e^|x| + 1 at x=0. I got d=ln( epsilon + 1 )

- September 19th 2012, 04:23 AMmittevansRe: For those that never learnt well the epsilon-delta proofs
Delta-Epsilon proofs well-past my BC calc days when I was taking an introductory proofs class. I assume that many individuals going into or emerging of Calc just do not have evidences down sufficient yet to be able to whip one out.

- September 19th 2012, 05:32 AMProve ItRe: For those that never learnt well the epsilon-delta proofs
To show is continuous at , you need to show , i.e. that . Working on the second inequality we have

It would now help us if we had an upper bound on , to give us the maximum value of . If we bound (say), that gives

and to get an upper bound on we can use the Triangle Inequality

So that gives

Therefore choosing and reversing the process will finish the proof. - September 19th 2012, 05:46 AMProve ItRe: For those that never learnt well the epsilon-delta proofs
- October 13th 2012, 01:06 PMAntiwasserstoffRe: For those that never learnt well the epsilon-delta proofs
Hello! I hope its right to ask here, since i wanted to use a example from the pdf file, I am just using the proof of example 3 to show my problem.

I almost understand every single step to make the proof, the mainproblem is this step:

|x^2 - 25| = |x + 5||x - 5| < 11|x-5|

I know to get the 11|x-5|, but i dont get why it is less than my original function? In fact, its one of the first times i really work with inequallities, could someone explain me this little simple step? i solved |x+5| for 11, and thats why i can replace |x+5| with 11 which gives me 11|x-5|, but why is this new function < than my old function? arent both functions the same until i put my delta in? - November 19th 2012, 12:44 AMjesongRe: For those that never learnt well the epsilon-delta proofs
I was taking an introductory proofs class. I think that most people going into or coming out of Calc just don't have proofs down enough yet to be able to whip one out.Luxury Goods Store

- November 19th 2012, 08:25 AMProve ItRe: For those that never learnt well the epsilon-delta proofs
- December 10th 2012, 12:43 AMkingsureRe: For those that never learnt well the epsilon-delta proofs
I understand that a little bit but I am having a hard time with how to start my proof.

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Cheap Air Yeezy 2 - September 27th 2013, 08:47 PMmathlover10Re: For those that never learnt well the epsilon-delta proofs
thanks for the nice paper! so in example 3 do you choose delta=1 arbitrarily and M is equal to the upperbound of 11? can you explain further what the meaning of the M derivation is?

for instance if you choose in ex 3 delta=0.1 instead of 1 then would get M=10.1 and define delta as min{0.1, 10.1} and is that an equivalent proof?

do most mathematicians learn this very quickly? - September 28th 2013, 03:59 AMProve ItRe: For those that never learnt well the epsilon-delta proofs
The idea is that as long as your x is sufficiently close to the c value, then you are guaranteed for your function to be within your tolerance of . So yes, the value was chosen arbitrarily. Any number smaller than this will also work as we know this value works.

- January 9th 2014, 07:37 PMJeanGunterRe: For those that never learnt well the epsilon-delta proofs
- March 19th 2014, 06:08 AMjohnadam4Re: For those that never learnt well the epsilon-delta proofs
A proof of a formula on limits based on the epsilon-delta definition. An example is the following proof that every linear function f(x)=ax+b (a,b in R,a!=0) is continuous at every point x_0. The claim to be shown is that for every epsilon>0 there is a delta>0 such that whenever |x-x_0|<delta, then |f(x)-f(x_0)|<epsilon fap turbo 2