# How should I define the limit definition of a derivative using negative numbers?

Typically the derivative is defined at a point $x$, assuming it is differentiable at it, by

$$\lim_{n \rightarrow \infty} \frac{f(x + \frac{1}{n}) - f(x)}{\frac{1}{n}}$$

But I want to define it using $f(x - \frac{1}{n})$. Should I use

$$\lim_{n \rightarrow \infty} \frac{f(x - \frac{1}{n}) - f(x)}{|-\frac{1}{n}|}$$

or

$$\lim_{n \rightarrow \infty} \frac{f(x) - f(x - \frac{1}{n}) }{|-\frac{1}{n}|}$$

• The latter $\lim_{n \rightarrow \infty} \frac{f(x)-f(x - \frac{1}{n}) }{\frac{1}{n}}$ – draks ... Dec 13 '13 at 12:22
• This is not the typical definition of derivative, rather $\lim_{h\rightarrow 0} \frac{f(x+h) - f(x)}{h}$, which works to assure limits exist on both sides of zero. – hardmath Dec 13 '13 at 12:27
• Remember what this definition means geometrically; you're essentially seeing how the rate of change, the slope of the tangent line at x, behaves as you make the change approach zero. The definition $lim_{n \to \infty} \frac{f(x+\frac{1}{n}) - f(x-\frac{1}{n})}{\frac{2}{n}}$ is equivalent to those two. – Lost Dec 13 '13 at 12:36
• Also, if $n$ is supposed to be an integer, so the limit is over the integers, as it would seem from using a variable name like $n$, then there's a more subtle problem in that this definition will "differentiate" things that shouldn't really be differentiable. Consider a function that's $0$ at $x = 0$ and also at $x = \frac{1}{n}$ for every integer $n \ne 0$, but $1$ everywhere else. This definition will differentiate it at $0$ and say the derivative is $0$, but the correct definition, taking the limit over the reals, says it is not differentiable at $0$. – The_Sympathizer Dec 13 '13 at 12:39
• @user1876508 : your "typical" definition is not typical or even correct. What you wrote is equivalent to $\lim_{h\to 0^+}(f(x+h)-f(x))/h$, which is not equivalent to the correct definition given by Sami below, because the limit is one-sided. – Stefan Smith Dec 13 '13 at 21:39

The definition of the derivative of $f$ at $x$ is $$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$ and by the change of variable you can find different version of the definition, for example if $k=-h$ we find $$f'(x)=\lim_{k\to0}\frac{f(x)-f(x-k)}{k}$$