Easy example first:

$f(x) = nx$

$f'(x) = (f(x+0)-f(x))/0 = (nx+0n-nx)/0 = (0n)/0 = n$

Hard one:

$f(x) = a^x$

$f'(x) = (f(x+0)-f(x))/0 = (a^{x+0}-a^x)/0 = (a^x(a^0-1))/0 = (a^x(e^{\ln(a^0)}-1))/0 = (a^x(e^{0\ln(a)}-1))/0$

Place $e^x = 1 + x + x^2/2 + ... + x^k/!k$ in $f'(x)$.

$f'(x) = (a^x(1 + 0\ln(a) + ... - 1))/0 = (a^x(0\ln(a) + ...))/0 = a^x(\ln(a) + (...)/0)$

Discard values that are $0$ that we can't rescue them with that $/0$.

$f'(x) = a^x(\ln(a)) = a^x\ln(a) $

Why is this working? Is there a name for this type of equations? (Or type of a way of solving equations) This way of solving involves expanding $x$ of $x/0$ until some $*0$ to elimnate the outside $/0$.

But as you see this is not exactly limit, but I tagged it because it's similar.

  • $\begingroup$ Why the downvote? $\endgroup$ Nov 23, 2014 at 17:08
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    $\begingroup$ As the question is tagged "nonstandard-analysis" and "infinitesimals", you can save the argument by replacing your $0$s with an infinitesimal $\epsilon$, expand suitably, cancel what can be canceled, then strike out all that is still a multiple of $\epsilon$ (i.e. round to the nearest standard real) $\endgroup$ Nov 23, 2014 at 17:53

3 Answers 3


There are a number of ways to do this formally; we really wish to talk about infinitesimals rather than $0$. I will describe a simple approach and a more sophisticated one in this post. However, before that, I think it's worth noting that there is some historical precedent for doing this sort of thing (though it, rightfully, would never live up to modern standards of proof) and interesting insights can be found - indeed, I once (but cannot seem to relocate) found a proof due to Euler in which he lets $k$ be infinite and then uses the binomial theorem $$(1+\frac{x}k)^k={k\choose 0}1+{k\choose 1}\frac{x}k+{k\choose 2}\frac{x}k^2+\ldots=1+\frac{k}{1!}\frac{x}k+\frac{k(k-1)}{2!}\frac{x^2}{k^2}+\ldots$$ and then figure that $\frac{1}k=0$ when it's convenient (i.e. when we're not going to multiply it by an infinite number later), so the above evaluates to $$1+\frac{1}{1!}x+\frac{1}{2!}x^2+\ldots$$ which is a familiar result defining the exponential function. It's not formal, but it's certainly not without merit and can guide intuition, since the above is easily formalized. For more normal people, I think it's too often frowned upon to do things like what you did with $a^x$ - I did similar things too.

The first approach to a formal concept is a very simple algebraic construction. In particular, we want to add a new infinitesimal number to our arithmetic - and to only add one "layer" of infinitesimals. So, we'll let $\varepsilon$ be some symbol to represent our new "number" and we'll think of it as being some smidgen greater than $0$. Since $\varepsilon^2$ ought to be less than $a\varepsilon$ for any real $a$, but we wish to keep our system simple, we will consider it negligible and define $\varepsilon^2=0$ (but $\varepsilon\neq 0$).

We have now constructed the dual numbers. We will use the standard rules of arithmetic when dealing with them, applying the identity $\varepsilon^2=0$ when possible. Let's just compute some things using this - and we'll start with polynomials, because we need no additional definitions for that. $$(x+\varepsilon)^3=x^3+3x^2\varepsilon + 3x\varepsilon^2 + \varepsilon^3 = x^3+3x^2\varepsilon$$ where we cancel out the latter three terms, since they are multiples of $\varepsilon^2=0$. Moreover, using the binomial identity and noticing that all but the first two terms will be multiples of $\varepsilon^2$, we can write $$(x+\varepsilon)^n=x^n+nx^{n-1}\varepsilon.$$ Hey, doesn't that $nx^{n-1}$ term look familiar? We can, using the above, quickly get to the remarkable result that, if $f$ is a polynomial, then $$f(x+\varepsilon)=f(x)+f'(x)\varepsilon$$ which makes sense. This simple states "the derivative measures how much an infinitesimal change in input affects the output". Moreover, the above expression is exactly that linear approximation for $f(x+\varepsilon)$ taken at $x$. What's remarkable here is that there is never a limit involved - everything here is purely algebraic.

However, your computations with $e^x=1+x+\frac{1}2x^2+\frac{1}6x^3+\ldots$ can easily be used to establish, for of all, that $$e^{x+\varepsilon}=e^x+e^x\varepsilon$$ where we once again get the form $f(x+\alpha \varepsilon)=f(x)+\alpha f'(x)\varepsilon$ by algebraic methods (if we expand every $(x+\alpha \varepsilon)^n=x^n+\alpha nx^{n-1}\varepsilon$). Moreover, you could easily extend any function less amenable to algebraic methods by defining $$f(x+\alpha \varepsilon)=f(x)+\alpha f'(x)\varepsilon$$ and then, when you compose, multiply, and add such functions together, the above will still hold.

Now, we do still run into an issue when we get expressions like $$\frac{f(x+\varepsilon)-f(x)}{\varepsilon}$$ since $\varepsilon$ has no multiplicative inverse (i.e. no dual number $x$ such that $x\varepsilon=1$ exists), so division can't be done simply. However, if we compute the above and receive $$\frac{\varepsilon f'(x)}{\varepsilon}$$ which can be seen as asking for the solution to $z\varepsilon = \varepsilon f'(x)$ (since division is the inverse of multiplication), so we could reasonably assign any value of the form $$f'(x)+\alpha \varepsilon$$ to the above quotient (since multiplying by $\varepsilon$ will vanish the $\alpha\varepsilon$ term) - which we can interpret as meaning "we are able to determine how quickly $f$ is changing - that is $f'$ - but we are unable to determine how quickly this quantity ($f'$) is changing (i.e. $f''$)."

The real beauty of this is that, firstly, it is algebraic, and secondly, it avoids any infinite sums of powers of $0$ (as you use) or anything like that - it always characterizes things as a sum of reals and a coefficient of $\varepsilon$. This makes it really easy to make a computer use and hence allows code that usually handles only real numbers compute derivatives too (which is useful when a system does not admit a closed form but can be simulated in a numerically stable way).

However, we could extend this further; if we put $\varepsilon^3=0$ and $\varepsilon^2\neq 0$, we would start tracking information about the second derivative and $f(x+\varepsilon)$ would then equal $f(x)+f'(x)\varepsilon+\frac{1}2f''(x)\varepsilon^2$ - and so on. Indeed, to truly take this thought to the extreme, one could say that every power of $\varepsilon$ is non-zero, and then it becomes perfectly reasonable to include $\frac{1}{\varepsilon}$ as a value too - and then we end up with quotients involving these infinitesimals being perfectly reasonable and only differing infinitesimally from the true answer. Largely, we can work intuitively with this new field, although if we define $e^x=1+x+\frac{1}{2!}x^2+\ldots$ or something like that, we do run into topological issues about what convergence means in a space like this (though algebra works just the same as it did before).

Essentially, taking this direction to the extreme we end up working in the hyperreal numbers, which are an extension of the real numbers with new infinite elements - and they are a model of the same axioms as the real numbers, which means, vaguely, if a statement is true for real numbers, it is also true for hyperreal numbers. This is the primary object of study of non-standard analysis and is an intuitive idea in nature, but it is much harder to state precisely than what I present above.

  • $\begingroup$ It is really not clear what you mean by Moreover, you could easily extend any function less amenable to algebraic methods by defining $f(x+\alpha \varepsilon)=f(x)+\alpha f'(x)\varepsilon$. You write that hyperreals are "much harder", however what you present above is an ardent hope rather than a mathematical theory. $\endgroup$ Dec 4, 2014 at 13:38
  • $\begingroup$ @user What I present is well-defined and of practical use. Defining $f(x+a\varepsilon)=f(x)+af'(x)\varepsilon$ is an admittedly artificial measure to extend their utility. It's never inconsistent to extend f to the duals like that - and it already follows for algebraic functions and, if you use the typical topology on $\mathbb R^2$ for the duals, it follows for analytic functions too. Yes, the algebraic properties of dual numbers does not cover all cases alone. The point is they cover many cases already and can be consistently extended to all cases (even if that appeals to a limit). $\endgroup$ Dec 4, 2014 at 22:28
  • $\begingroup$ If you extend the function artifially to the duals via defining them by $f(x+\alpha \varepsilon)=f(x)+\alpha f'(x)\varepsilon$, then you haven't accomplished your goal of clarifying what the derivative is at all. There have been nonarchimedean systems since Euclid book 6 where he treats the so-called cornicular angles, and there have been numerous theories of infinitesimals throughout the 19th and first half of 20th centuries. However, none of this was ever a useful tool in analysis until we had a theory of infinitesimals with the transfer principle/Los's theorem, namely the hyperreals... $\endgroup$ Dec 5, 2014 at 9:40
  • $\begingroup$ ...It is not really useful to obscure this basic point. $\endgroup$ Dec 5, 2014 at 9:40

The definition of a derivative involves a limit: $$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$ Which is why there is no division by zero.

For your first example: $$f'(x)= \lim_{h \to 0} \frac{(nx+nh)-(nx)}{h}=\lim_{h \to 0} \frac{nh}{h}=\lim_{h \to 0} n=n$$

In response to the edit: You can't find a derivative without taking the limit or using s shortcut - and all the shortcuts are based on limits!

Let me see if I can try to improve this answer (as per Meelo's suggestions).

When defining a derivative, we use the form $$\lim_{h \to 0}$$

This is the primary difference between your method and taking the derivative by its definition. This avoids the issues with forms like $\frac{0}{0}$. For example, take the function $$f(x)=x^2$$ If you used your method, we would get $$f'(x)=\frac{f(x+0)-f(x)}{0}=\frac{(x+0)^2-x^2}{0}=\frac{x^2-(2)(0)x-x^2}{0}=\frac{0}{0}$$ which is unhelpful. However, if we use the limit definition, we get $$f'(x)=\lim_{h \to x}\frac{f(x+h)-f(x)}{h}=\lim_{h \to 0}\frac{(x+h)^2-x^2}{h}=\lim_{h \to 0}\frac{x^2 + 2xh + h^2-x^2}{h}$$ $$=\lim_{h \to 0}\frac{2xh +h^2}{h}=\lim_{h \to 0}2x+h$$ When $h \to 0$, the last term disappears, and we get $$f'(x)=2x$$ The reason this is so convenient is that $h$ always disappears when we take this final limit. This goes for all higher powers of $h$.

  • $\begingroup$ I am starting to think that this is really just $\lim$ but primitive that you directly prove something. (I am looking at the hard example mostly) I will wait a bit, thanks. $\endgroup$ Nov 23, 2014 at 17:27
  • $\begingroup$ Downvoter - Is this answer incorrect? $\endgroup$
    – HDE 226868
    Nov 25, 2014 at 1:00
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    $\begingroup$ It's correct, but I don't think it is relevant to the question; the question, as it is stated, asks why symbolic manipulations involving $0$ yield the same answer as limits do. It doesn't suggest that we ought to define $f'(x)=\frac{f(x+0)-f(x)}0$, but asks why we can compute it as if we could. The answer doesn't so much as refer to the asker's method with $0$. I think it would be valuable if you revised the answer to show how the asker's method is equivalent to limits (e.g. how them canceling $\frac{0}0$ is really cancelling $\frac{h}h$), but the answer doesn't have that value (yet). $\endgroup$ Nov 25, 2014 at 1:12
  • $\begingroup$ @Meelo I tried to do that, but I suppose it didn't work too well. The reason I did use limits was because the question originally did not allude to them, except in the tag, and I thought there might be circular logic somewhere in it. I'll work to improve this. $\endgroup$
    – HDE 226868
    Nov 25, 2014 at 1:14
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    $\begingroup$ The edit is certainly better; though I suspect the OP would like manipulate $$f'(x)=\frac{f(x+0)-f(x)}0=\frac{x^2 + 2\cdot 0 \cdot x + 0^2 - x^2}0=\frac{2\cdot 0 \cdot x + 0^2}0=2\cdot x + 0$$ ignoring the indeterminate form - but I think that the connection here is that the above is exactly the same as if we replaced $h$ by $0$ in the typical notation - it's like if we wrote $$f'(x)=\lim_{0\rightarrow 0}\frac{f(x+0)-f(x)}0$$ where we use $0$ as a variable (which is bad notation, but it explains why their method gets the same answer - it's just not actually different in this view) $\endgroup$ Nov 25, 2014 at 1:33

The manipulations with $0$ which is discarded at the end are roughly the procedure pioneered by Fermat called adequality though Fermat did not have the notion of derivative.


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