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Definition of the Limit

Date: 11/05/2002 at 22:13:06
From: Stephen Adams
Subject: Definition of the Limit

I need some help with the definition of the limit, particularly 
choosing delta for a given epsilon. For example, consider the limit 
as x approaches 2 of (x^2 + 3) = 7 with epsilon = .05. Here is my 
work:

|x^2 + 3 - 7| < .05
|x^2 - 4| < .05
-.05 < x^2 - 4 < .05
3.95 < x^2 < 4.05
1.98746... < x < 2.01246...
-.01253... < x - 2 < .01246

The lower and upper values for delta are different, so how do I choose 
which one should be delta? I seem to remember reading that the smaller 
value is usually chosen, so delta would equal .01246. However, I can 
find no explanation as to why the smaller value is chosen. Is there 
anything wrong with saying that delta equals .01253? Is one value 
considered more correct than the other, or are they both considered 
correct?  

Moreover, on a test should .01253 be considered a correct answer?  
Any help is greatly appreciated.


Date: 11/05/2002 at 23:26:13
From: Doctor Peterson
Subject: Re: Definition of the Limit

Hi, Stephen.

Any number is acceptable if you can PROVE that it fits the definition; 
you can't just look at an inequality and say "I seem to recall that 
this one will work."

So let's see whether it is true that, whenever |x-2| < 0.01253, we 
will have |x^2-4| < 0.05. For a quick test, we can just try making 
|x-2| = 0.01252 and see if it works. In one direction, we have x = 
2.01252, x^2 = 4.0502367504, and x^2-4 = 0.0502367504. This is NOT 
less than 0.05, so we see that it doesn't work.

Now, how could we have thought through this to see why we should 
choose delta as 0.01246? Look back at what you found:

    -.01253 < x - 2 < .01246

You showed that this is equivalent to |x^2 - 4| < 0.05 (given that 
x > 0). You want to choose a delta so that this will be true whenever 
|x-2| < delta. Notice that the latter is the same as

    -delta < x - 2 < delta

If you pick delta = 0.01246, then this interval is inside the interval 
from -0.01253<x-2<0.01246, so that any x within delta of 2 does what 
you want. If you choose delta = 0.01253, then

    -0.01253 < x - 2 < 0.01253

does NOT imply

    -0.01253 < x - 2 < 0.01246

because the former includes some numbers that are not in the latter.

Do you see how the reasoning goes? Essentially we want a symmetrical 
interval that is INSIDE the interval equivalent to the epsilon goal, 
so that x being within delta IMPLIES that f(x) is within epsilon. So 
we can choose any delta LESS THAN OR EQUAL TO 0.01246.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
College Analysis
College Calculus
High School Analysis
High School Calculus

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