Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am having a problem with this question

For what functions do we have:

$$\lim_{h \to 0} \frac{f^2(x+h)-f^2(x)}{h}=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$$

We have $$\lim_{h \to 0} \frac{(f(x+h)-f(x))(f(x+h)+f(x))}{h}=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h} \Leftrightarrow \lim_{h \to 0}(f(x+h)+f(x))=1 $$ hence $f(x)=\frac{1}{2}$ .But what if $f(x) \neq \frac{1}{2}$ Is what I did up to now correct? Please help.

share|cite|improve this question
$\frac{f^2(x+h)-f(x)}{h}\neq\frac{(f(x+h)-f(x))(f(x+h)+f(h))}{h}$ – Salech Alhasov Oct 17 '12 at 15:39
Sorry I fixed that – Carpediem Oct 17 '12 at 15:40
@user43758 That wasn't his point. $f^2(x+h)-f(x)$ is not in the form $a^2-b^2$, unless you want to introduce some square roots in there, which I doubt. Is the question supposed to be $f^2(x+h)-f^2(x)?$ – user39572 Oct 17 '12 at 15:45
Its perhaps should be, $$\lim_{h \to 0} \frac{f^2(x+h)-f^2(x)}{h}=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$$ – Salech Alhasov Oct 17 '12 at 15:45
No! $f^2(x)=(f(x))^2$ in my case – Carpediem Oct 17 '12 at 15:46
up vote 0 down vote accepted

This answers the problem as stated originally, i.e determines for which $f$ the following holds $$ \lim_{h\to 0} \frac{f^2(x+h) - f(x)}{h} = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h} $$

I assume that $f^2(x)$ is supposed to mean $(f(x))^2$. Then, for $\lim_{h\to 0}\frac{f^2(x+h) - f(x)}{h}$ to exist, it must be that $\lim_{h\to 0}f^2(x+h) = f(x)$. If $f$ is continuous, you get $f^2(x) = f(x)$ and thus $f(x) \in \{0,1\}$. If $f$ is not continous, the question makes no sense because it then certainly is not derivable, hence the right-hand side limit in your question does not exist.

Note that $f$ must not necessarily be constant globally, but it must be constant on every connected $A \subset \text{dom }f$. Thus, $f: \mathbb{R}\setminus\{0\} \to \{0,1\}$, $$ f(x) = \begin{cases} 0 &\text{if } x < 0\\ 1 &\text{if } x > 0 \end{cases} $$ is a possible solution. As are the constant functions $0$ and $1$, of course.

And now for $$ \lim_{h\to 0} \frac{f^2(x+h) - f^2(x)}{h} = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h} $$

This is quite obviously equivalent to $((f(x))^2)' = f'(x)$, which by applying the chain law yields $2f(x)f'(x) = f'(x)$. To have that, it must thus either be that $f(x) = \frac{1}{2}$ or that $f'(x) = 0$ (where the format also implies the latter). Thus, any function which is constant on all connected $A \subset \text{dom } f$ works.

share|cite|improve this answer
@user43758 This answers both your original question and your modified question. And btw, simply stating "This is wrong" is rather rude, especially if it was you who changed the question! – fgp Oct 17 '12 at 16:13
What happens if f is not constant? – Carpediem Oct 17 '12 at 16:14
@user43758 Than the two limit cannot be equal. If it isn't constant, to have $2ff'=f'$ it must be that $2f = 1$, hence $f = 1/2$, hence $f$ is constant... – fgp Oct 17 '12 at 16:15
Ok Thank you and sorry for my the way I responded earlier. – Carpediem Oct 17 '12 at 16:16
@user43758 You're welcome! – fgp Oct 17 '12 at 16:17

Let us assume that the right hand limit exists. Then, $f$ is differentiable at $x$. Hence, $f^2$ is differentibale at $x$ with derivative $2 f f'$. You can now add and subtract $f^2(x)$ in the numerator of the left hand side. You should be able to recognize the derivative of $f^2$ at $x$ and then proceed with the argument.

You should get that $f(x) = 0$ or $1$ otherwise the left hand side is infinite. Then you should conclude that $f'(x) = 0$.

EDIT: this was assuming your original question where you have $f(x)$ on the left-hand side and not $f^2(x)$.

share|cite|improve this answer
But it is $f^2(x)$ – Carpediem Oct 17 '12 at 15:51
The answer is wrong since we are considering $f^2(x)$ – Carpediem Oct 17 '12 at 16:03

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.