Let $\Delta x = x_1 - x$, where $x$, $x_1$ are real numbers.

Let $f$ be a real valued function; we define the difference quotient as: $$\frac{\Delta f}{\Delta x} = \frac{f(x + \Delta x) - f(x)}{\Delta x} \text{.}$$

I've read this in a physics book:

When $\Delta x \to 0$, the denominator of the difference quotient tends to zero; in order for the limit $\lim_{\Delta x \to 0}{\frac{\Delta f}{\Delta x}}$ to exists (and be finite), also the numerator must tend to zero. That is, $\lim_{\Delta x \to 0}{f(x + \Delta x) = f(x)}.$

Is the book saying that the limit of a quotient is finite when the denominator tends to zero iff the numerator tends to zero too?

If so, why this happens?

  • 1
    $\begingroup$ Looks like a mistake in the book. I guess they didn't want to say $\lim_{\Delta x \to 0}{(f(x + \Delta x) - f(x))} = f(x)$ but one of $\lim_{\Delta x \to 0}{(f(x + \Delta x) - f(x))} = 0$ or $\lim_{\Delta x \to 0}{f(x + \Delta x)} = f(x)$. (Maybe the author wanted to change one to the other but changed it only half-way through.) $\endgroup$
    – user491874
    Commented Dec 30, 2017 at 19:24
  • $\begingroup$ It is a my mistake (I'll correct it immediately), i should have written $\lim_{\Delta x \to 0}{f(x + \Delta x)} = f(x)$, and not what I wrote. $\endgroup$
    – user457568
    Commented Dec 30, 2017 at 19:42
  • $\begingroup$ So the question now simply is: "when the limit of a quotient where the denominator approaches to zero exists and is finite? $\endgroup$
    – user457568
    Commented Dec 30, 2017 at 19:55
  • $\begingroup$ It is a necessary but not sufficient condition. $\endgroup$
    – user
    Commented Dec 31, 2017 at 8:16

2 Answers 2


To answer your question: No. The book doesn't try to say "iff". The book says that the limit of a quotient is finite when the denominator tends to zero only if the numerator tends to zero too. (Note I replaced "iff" with "only if" in your sentence.) Only one direction follows directly, the other doesn't, and for some functions and some choice of $x$ it is true, for others it is false.


It is an error in your book (or a sloppy notation):

enter image description here

The diagram shows the basic principle of a derivative. It is a triangular approximation of the variation of a function evaluated at 2 points x and x+$\Delta$x.

What he should have written in the book is $f'(x)=\tan(\alpha)=\lim_{\Delta x->0} \frac{f(x+\Delta x)-f(x)}{\Delta x}=\frac{\Delta f}{\Delta x}$. Now, he should have written is $\lim_{\Delta x->0} {f(x+\Delta x)-f(x)}={\Delta f}$, which goes to zero also but at a different rate than $\Delta x$.


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