Derivative Definition and Division by Zero
Date: 03/31/2007 at 16:43:52 From: Adrian Subject: First Principles and Dividing by Zero Hi, I'm having a bit of trouble with the concept behind first principles. Even though one aims to cancel out the denominator, which is technically zero, proofs that show that 1 = 2 may perform the same act (canceling out a denominator that was 0) and these proofs are considered invalid. Why is First Principles considered valid then? My thoughts are that perhaps the solution lies in the fact that in First Principles, the denominator approaches 0 as opposed to actually being 0, although I figured these two things are more or less equivalent since usually at the end of a First Principles problem one substitutes in 'h' as 0, even though it is stated that h is only approaching 0. Any thoughts?
Date: 03/31/2007 at 20:23:05 From: Doctor Rick Subject: Re: First Principles and Dividing by Zero Hi, Adrian. In order to make sense of what you've written, I have to add some words. You're talking about CALCULATING A DERIVATIVE from first principles (that is, the definition of a derivative as a limit), aren't you? I just used a word that makes all the difference: the derivative is defined as a LIMIT. We never actually divide by zero; rather, we divide by a very small number, and we find the limit of that quotient as h approaches zero. For example, let's find the derivative of f(x) = 2x^2 at x=3: 2(x+h)^2 - 2x^2 4hx + 2h^2 lim --------------- = lim ---------- = lim 4x + 2h = 4x h->0 h h->0 h h->0 The quantity whose limit we are taking is undefined at h = 0, for just the reason you say. But the LIMIT at h = 0 exists. Maybe it will help you if you go back to the first-principles definition of a limit: (this is from my old calculus textbook) Suppose f is a function defined for values of x near a. (The domain of f need not include a, although it may.) We say that L is the limit of f(x) as x approaches a, provided that, for any epsilon > 0 there corresponds a deleted neighborhood N of a such that L - epsilon < f(x) < L + epsilon whenever x is in N and in the domain of f. Notice that the domain of f need not include the point at which the limit is to be taken, and that the condition only needs to hold when x is in the domain of f. This shows that a function, such as our function of h, can have a limit at a point (h=0 in our case) where the function itself is undefined. There is no contradiction there. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/
Date: 03/31/2007 at 21:03:26 From: Adrian Subject: First Principles and Dividing by Zero Thanks Doctor Rick, it is a bit clearer though I still have one small problem. Sticking with the example you gave, you eventually arrive at lim 4x + 2h = 4x. What is confusing is that before reaching h->0 this step, we were unable to simply sub in h as 0 and I was under the impression that this was because h didn't really equal zero, but rather a number extremely close to zero. However, once we have eliminated the problem of getting 0/0 if we were to sub in h as zero, it seems like it's perfectly alright to treat h as if it is exactly zero, as we can sub it into the equation as zero in order to arrive at 4x. Is it possible to think of 'h' as being BOTH extremely close to zero and exactly zero, and therefore limits to be kind of a way of sidestepping the rule of not being able to divide by zero, or am I totally missing the concept?
Date: 03/31/2007 at 22:06:14 From: Doctor Rick Subject: Re: First Principles and Dividing by Zero Hi, Adrian. What's going on here is that we have replaced a function f(h) that has no value at h=0, with a CONTINUOUS function g(h) = 4x + 2h. The two functions f(h) and g(h) are equal for every h EXCEPT h=0; therefore they have the same limit at h=0. The limit at point a of a function g(x) that is CONTINUOUS at a is the value g(a). (That's the definition of a continuous function.) Therefore we easily find the limit of 4x + 2h as h approaches 0, namely 4x; and this must also be the limit of f(h) as h approaches 0. Does that help? We are not sidestepping the "rule" (that division by zero is undefined) in the sense of violating it in any way. We are finding the LIMIT of a function with h in the denominator as h approaches 0; the limit, by definition, never invokes division by EXACTLY zero. We evaluate the limit by finding a function that has the same values for every x NEAR zero (but not EXACTLY zero); this function does not involve division by h, so there is nothing wrong with evaluating it at exactly zero. You could say we are "getting around" division by zero in the literal sense that we are working in the immediate neighborhood of zero without going exactly there. It may be tricky, but it's perfectly legal. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/
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