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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}$2012-10-17
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    Sorry I fixed that2012-10-17
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    @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)?$2012-10-17
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    Its perhaps should be, $$\lim_{h \to 0} \frac{f^2(x+h)-f^2(x)}{h}=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$$2012-10-17
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    No! $f^2(x)=(f(x))^2$ in my case2012-10-17
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    Can somebody just provide me with the solution to this question. I did 10 Problems before encoutering this question. I just want to finish at this point2012-10-17
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    @user43758 Well, I just did, but unfortunately not for the question as it appears now. You really should try to get the *basics* of the question right from the starts - answering a question only to discover that you changed it *while* I was answering is quite frustrating. Reading over your question before clicking "Post" really isn't that hard...2012-10-17
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    @user43758 You introduced the second square 13 minutes after posting, as the history quite clearly shows... Do as you want, of course, I'm just trying to tell you that this isn't a great way to motivate people to help you...2012-10-17

2 Answers 2

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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!2012-10-17
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    What happens if f is not constant?2012-10-17
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    @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...2012-10-17
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    Ok Thank you and sorry for my the way I responded earlier.2012-10-17
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    @user43758 You're welcome!2012-10-17
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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)$2012-10-17
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    The answer is wrong since we are considering $f^2(x)$2012-10-17