Expert: Paul Klarreich - 10/10/2009

Question
I can not understand why you have written
| 3x - 6 |  < e  

2 | x - 2 | < e

| x - 2 | < e/2

Am I missing something or does this factor 3 | x - 2 | < e
Answer
Questioner:   jino
Category:  Calculus

Subject:  limits
Question:  the definition of limit is
a function f is said to tend to a limit,l,as x tends to c,if to any positive number e arbitrarily assigned there corresponds a positive number d, such that for every x in (c-d,c+d)-{c},f(x) differs from l numerically by a number less than  e
modulus of f(x)-l < e
for every value of x , such that
0 < modulus x-c <d

i did not understand this definition could u quote some example putting some  values
===========================================
Hi, Jino,

Symbolically, we write:

lim  f(x) = L
x->a

if WHENEVER  x is close to a,  f(x) is close to L.  

Analytically, we define those terms to mean that we shall play the 'epsilon-delta' game.  

f(x) can be made as close as we like to  L,  JUST by making  x close enough to a.

What does 'as close as we like' mean?  We can specify a (small) number epsilon, (which I can't make and will have to write as 'e') and then we should be able to compute, from this epsilon, a (small) number delta, [which I can't make and will be 'd'].

In playing the game, the first person picks the epsilon.  It is up to the second person to compute the proper delta.  Let's play the game for:

f(x) = 3x + 5,  and  lim(x->2), which we think will be  11.  [I leave it to you to see how we guessed the 11.]

First guy picks  e = 0.00001.  That's real small, right?

Second guy says:

I need: | f(x) - 11 |  < 0.00001

So I'll do some algebra.

| 3x + 5 - 11 |  < 0.00001

| 3x - 6 |  < 0.00001  

2 | x - 2 | < 0.00001

| x - 2 | < 0.000005

Aha!  That's it.  If I use that  0.000005 as my delta, I can 'climb' back up the ladder and win the game.  I will just say that if:

| x - 2 | < 0.000005
then
2 | x - 2 | < 0.00001
and
| 3x - 6 |  < 0.00001  
and
| 3x + 5 - 11 |  < 0.00001  
which is
| f(x) - 11 |  < 0.00001  
and the  11  was my L and the 0.00001  was the epsilon.

Now how can we be sure this will always work?  0.00001 might seem small, but maybe someone thinks it's big?  Just do it symbolically:

First guy picks  e,  and we don't care how small it is. We want  

| 3x + 5 - 11 |  < e

| 3x - 6 |  < e  

2 | x - 2 | < e

| x - 2 | < e/2

So that's the answer.  Whatever the first guy picks as e, we pick d = e/2, and climb the ladder.
.......................................
OK, is that all there is to it?  No, unfortunately some examples are harder, but I'll leave those to your teacher.  

Answer
Questioner: Dan
Country: Australia
Category: Calculus
Private: No
Subject: Analysis
Question: I can not understand why you have written


| 3x - 6 |  < e  

2 | x - 2 | < e

| x - 2 | < e/2

Am I missing something or does this factor 3 | x - 2 | < e
.............................................
>>  Oops -- of course you are correct.  That should read:

| 3x - 6 |  < e  

3 | x - 2 | < e

| x - 2 | < e/3

I will try to post a correction.