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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 

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
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 

- Doctor Rick, The Math Forum 
Associated Topics:
High School Calculus

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