Expert: Paul Klarreich - 3/6/2010

Question
Hi I don't understand epsilon delta definition of the limit I just continued with the course thinking I will understand it by time but even when we reached integration I still don't understand it I understand all normal limit problems but when someone mentions epsilon delta I just like freeze and stop getting things So can you give me a way to understand it better and an example of a problem?

Answer
Questioner: hamad
Country: Kuwait
Category: Calculus
Private: No
Subject: epsilon delta limit
Question: Hi I don't understand epsilon delta definition of the limit I just
continued with the course thinking I will understand it by time but even
when we reached integration I still don't understand it I understand all
normal limit problems but when someone mentions epsilon delta I just like
freeze and stop getting things So can you give me a way to understand it
better and an example of a problem?
..............................................................

Below are two answers I have sent in the past.  But here is a simple example:

Prove:  lim[x->3] (2x - 1) = 5

Claim:  given e, there exists d such that for all |x - 3| < d,

| (2x - 1) - 5 | < e

Do algebra:

| 2x - 6 | < e

2 | x - 3 | < e

| x - 3 | < e/2

Now take  d = e/2 and reverse the steps.

But study the explanations below.
------------------------------------------
Questioner:   Len
Category:  Calculus
Private:  No
 
Subject:  calculus: limit proofs
Question:  The two problems are the following:

Give the proof of a limit for a linear function,and the proof of the sum and product of special functions.

I'm guessing you have to use the delta-epsilon definition, but am not sure how to go about it.  Thanks.
........................................................
Hi, Len,

I am afraid your question is not at all clear.

I suspect your first one says:

Prove that  

lim   ax + b = ar + b
x->r

but I am not sure what you mean by 'special functions.'  In the meantime, we can give the first one a try.  You have to prove that:

given  e > 0, there exists  d > 0  [I cannot make epsilons and deltas, so I will use e and d.]  such that:

whenever | x - r | < d,  |ax + b - (ar + b) | < e

Now |ax + b - (ar + b) | =

|ax + b - ar - b | =

|ax - ar | =

|a(x - r)| =

|a| |x - r|

So if |x - r| < d then

|a||x - r | < d|a|

So all we need is to take  d = e/|a| and we have our proof.
--------------------------------------------
Questioner:  sammy
Category:  Calculus
Private:  no
 
Subject:  limit question
Question:  Prove that
lim as x->a (sqrt x)= sqrt a
I just cant figure out the right epsilon delta combination.

Also can you prove that lim x->a of x^n=a^n
I think this one involves the bernoulli inequality.
Thank you.
...............................
Hi, Sammy,

It has been a while since I did things like this, but maybe the following will work out:
[I will write  'e' and 'd' in this reply -- can't make epsilons and deltas.]

You want to prove this limit:
 
lim  sqrt(x) = sqrt(a)
x->a

OR, given e > 0,  find  d > 0 such that whenever |x - a| < d,

|sqrt(x) - sqrt(a)| < e.

Now if we note that a bit of rationalizing might help:  Multiply both sides by (sqrt(x) + sqrt(a)):

| x - a | < e (sqrt(x) + sqrt(a))

That might give us the clue.  

| x - a | = | sqrt(x) - sqrt(a) | (sqrt(x) + sqrt(a))

Now when x is near x, sqrt(x) is near sqrt(a), and the second factor is near 2 sqrt(a).  If it is near  2 sqrt(a), we can require that it be greater than 1 sqrt(a) or sqrt(a)

So if we simply choose  d = sqrt(a) e, we will have:

| x - a | < d

| x - a | < e sqrt(a)

| x - a | < e (sqrt(x) + sqrt(a))

   | x - a |         
--------------------  < e
(sqrt(x) + sqrt(a))


|sqrt(x) - sqrt(a)| (sqrt(x) + sqrt(a))
---------------------------------------  < e
        (sqrt(x) + sqrt(a))


|sqrt(x) - sqrt(a)| < e

Which is what we wanted to prove.
...........................................
About:

lim  x^n = a^n
x->a

Also can you prove that lim x->a of x^n=a^n
I think this one involves the bernoulli inequality.

Try this approach:

x^n - a^n  has  (x-a) as a factor. [Remember your Factor Theorem and Remainder Theorem from precalculus?]

x^n - a^n = (x - a)(x^n-1 + x^n-2 a + x^n-3 a^2 + ... + a^n-1)

[I'm being a bit sloppy with the parenthesizing.]

Now again, suppose we make a restriction on x.  It's supposed to be near a, right?  Then assume it is less than 2a.  Then we have:

x^n - a^n < (x - a)(2^n-1 a^n-1 + 2^n-2 a^n-2 a + 2^n-3 a^n-3 a^2 + ... + a^n-1)

x^n - a^n < (x - a)(2^n-1 a^n-1 + 2^n-2 a^n-1   + 2^n-3 a^n-1     + ... + a^n-1)

x^n - a^n < (x - a)(2^n-1 + 2^n-2  + 2^n-3  + ... + 1)a^n-1

x^n - a^n < (x - a)(n 2^n-1)a^n-1

Now let  d = e/(n 2^n-1)a^n-1, and then

|x - a| < e/(n 2^n-1)a^n-1 and so:

x^n - a^n < (n 2^n-1)a^n-1[e/(n 2^n-1)a^n-1] = e