Expert: Paul Klarreich - 1/18/2006

Question
The formal definition of a limit is as follows:  Let f(x) be defined on an open interval about x0, except possibly at x0 itself. We say that if(x) approaches the limit L as x approaches x0, and write :

lim of f(x) as x approaches x0 = L

if, for every number (episilon = E) E>0, there exists a corresponding number delta (d) > 0 such that for all x

0 < |x-x0| < d   ===>     |f(x) - L| < E  

THE QUESTION IS This:

Show BY EXAMPLE that the following statement is wrong:  

"The number L is the limit of f(x) as x approaches x0 if, given any E> 0, there exists a value of x for which |f(x) - L| < E"

Explain why the function in your example does not have the value of L as a limit as x approaches x0.  

the problem im having in coming up with an example is this..  if i draw any continous function  say for example f(x) = x^2 , it seems that the above statement Would be true  cause i Can find a value for x for every E >0 for which |f(x) - L| < E.  ( and yes that value would have to be between x + or - a delta) but, im having trouble coming up with an example that proves this statment wrong, although it is intuitively obvious to me through the precise definition of a limit why x must be with a range of plus or minus delta.

In other words, despite having thought about this for hours, I cant seem to come up with a graphical example or function that proves the statement wrong...   thanks for your help

Im 48 years old and trying to relearn calculus  

Answer
Hi, Jerry,

You wrote:

The formal definition of a limit is as follows: Let f(x) be defined on an open interval about x0, except possibly at x0 itself. We say that f(x) approaches the limit L as x approaches x0, and write :

lim  f(x) = L
x->x0
if, for every number (epsilon = E) E>0, there exists a corresponding number delta (d) > 0 such that for all x { you mean x not= x0 }

0 < |x-x0| < d   ===>    |f(x) - L| < E

THE QUESTION IS This:

Show BY EXAMPLE that the following statement is wrong:

"The number L is the limit of f(x) as x approaches x0 if, given any E> 0, there exists a value of x for which |f(x) - L| < E"

Explain why the function in your example does not have the value of L as a limit as x approaches x0.

The problem I'm having in coming up with an example is this..

If I draw a continous function say, f(x) = x^2 , it seems that the above statement Would be true cause i Can find a value for x for every E >0 for which |f(x) - L| < E. ( and yes that value would have to be between x + or - a delta) but, im having trouble coming up with an example that proves this statment wrong, although it is intuitively obvious to me through the precise definition of a limit why x must be with a range of plus or minus delta.

In other words, despite having thought about this for hours, I cant seem to come up with a graphical example or function that proves the statement wrong...   thanks for your help

Im 48 years old and trying to relearn calculus
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Good for you. So am I.  That's why I volunteered.

OK, I think you left something out of your 'wrong statement' -- something about a neighborhood of x0.  You probably wanted to say why THIS statement is wrong :

Lim  f(x) = L  if:
x->x0

given any E>0 there exists d (delta) such that the interval |x - x0| < d contains a point x1 such that |f(x) - L| < E.
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Clearly if you had an example like  f(x) = x^2 and wanted to 'prove' that:

lim  f(x) = 36,  for example,
x->0

you could show that given any E, you can find an x (such as 6) where
|f(6) - 36| < E.
Of course, the value x1 = 6 is not very close to x0 = 0.

So you want to say (repeating it) that:

Lim  f(x) = L  if:
x->x0

given any E>0 there exists d (delta) such that the interval |x - x0| < d contains a point x1 such that |f(x) - L| < E.

In words, you are saying here that no matter how close to f(x) = L you want to get, you can find a point close to x0 whose f(x) is close enough.  You are supposed to say that no matter how close to f(x) = L you want to get, ALL f(x)'s will be close enough to L if you are close to x = x0.

(I like to use the word 'whenever' in this context.  As in: Whenever x is close to x0, f(x) is close to L.)
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I think you will get nowhere looking at a continuous function, because the definition of a continuous function at x = x0 requires that the limit exists.  (It also has to be equal to f(x0), for that matter.)  And if the limit exists, both your right and wrong definitions work.

So you have to look for DIScontinuous functions, and forget about drawing nice graphs of them.  You can't even get by with a function that is continuous at x = x0, just because it has some discontinuities elsewhere.  

For example, a simple step function such as:

f(x) = GIF(x), meaning the largest integer not greater than x,

will be discontinuous at the integers, but continuous at other points, such as x = 2.5, and would satisfy your 'wrong' definition.

No, you need a function with lots and lots of discontinuities, and forget about writing them with the usual algebraic forms.  Here is an example:

Let f(x) = 1, whenever x is a rational number.
          2, otherwise.  (i.e. irrational)

Now take x0 = 0, for example.  What is  lim f(x)?  Is it L = 1?

Given any E, i.e. E = 0.01 perhaps, there will be points arbitrarily close to x0 = 0 where f(x) is close to L = 1, because the rational numbers are dense.  But there will also be points arbitrarily close to x0 = 0 where f(x) is NOT close to L = 1, because the irrationals are also dense, and they give values of 2, which is 'far away' from L = 1.

If this doesn't make it clear, send back some issues and we will have a conversation.  Limits and continuity is a fascinating subject.

Paul