# Condition functions and limits

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.

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$\frac{f^2(x+h)-f(x)}{h}\neq\frac{(f(x+h)-f(x))(f(x+h)+f(h))}{h}$ – 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

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.

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@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 $f^2(x)$.

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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