Definition of the LimitDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/