# Constant on [epsilon, 2epsilon] implies f'(0)=K

1. Dec 7, 2015

### RBG

1. The problem statement, all variables and given/known data
Warning: I realize the title is misleading... the function itself isn't what's constant.
Mod note: Edited to fix the LaTeX

If $f$ is a continuous at 0 such that $\lim_{x \to 0}\frac{f(x)-f(g(x))}{g(x)}=M$, where $g(x)\to 0$ as $x\to 0$ does this generally mean that $f'(0)=M$? I have been working on the case that $g(x)=x/2$. It seems like this should be the case since you are saying the secant line on any $[\epsilon, 2\epsilon]$ is constant, so as $\epsilon$ tends to zero, the slope will approach the slope of the tangent line. I've been trying to fudge the definition of continuity to get this and show $|f'(0)-h(0)|<\epsilon$ where $h(t) = \lim_{x\to t} \frac{f(x)-f(g(t))}{x-g(t)}$

2. Relevant equations
Definition of continuity - \epsilon-delta and limit
Definition of differentiability
Mean Value Theorem
Sequences of functions?

Last edited: Dec 7, 2015
2. Dec 7, 2015

### SammyS

Staff Emeritus
Hello RBG. Welcome to PF !

Code (Text):
Use  (Double dollar sign.) to set off LaTeX on an individual line.
Use $to set off LaTeX for in-line style. As follows: If$f$is a continuous at 0 such that \underset{x\to 0}{\lim}\frac{f(x)-g(x)}{g(x)}=M\,, does this generally mean that$f'(0)=M$? 3. Dec 7, 2015 ### RBG Thank you! Also, I edited my question to what I meant. I am really more just interested in the$g(x)=\frac{x}{k}$where$k\in\mathbb{N}$case. 4. Dec 7, 2015 ### PeroK \underset{x\to 0}{\lim}\frac{f(x)-x/k}{x/k}=M \ \Rightarrow \ \underset{x\to 0}{\lim}\frac{f(x)}{x} = \frac{M+1}{k} 5. Dec 7, 2015 ### Ray Vickson Your condition does not imply$f'(0) = M$. For example, take$f(x) = (M+1) g(x) \, \forall \, x$, which certainly satisfies your limit condition. If you take$g(x) = \sqrt{|x|}$you will see that$f'(0)$need not even exist. For the case of interest, namely,$g(x) = kx$you can compute$f'(0)$and see what you get. 6. Dec 7, 2015 ### RBG Woops! Man, for a first post I am messing up a lot. I meant what my edits say, that is$f(g(x))$not just$g(x)$in the numerator! 7. Dec 8, 2015 ### PeroK$f'(0) = \underset{x\to 0}{\lim}\frac{f(x/k)-f(0)}{x/k} \ne \underset{x\to 0}{\lim}\frac{f(x) - f(x/k)}{x/k}$For example. Try$f(x) = x$:$\underset{x\to 0}{\lim}\frac{f(x) - f(x/k)}{x/k} = k - 1 \ne f'(0)$(unless$k = 2$) 8. Dec 8, 2015 ### RBG So if we're in the$k=2$case and the only thing I know about$f$is that it's continuous at$0$, how would I show that$f'(0)=M$? 9. Dec 9, 2015 ### PeroK See Samy_A's post below.$k = 2$is a special case. Last edited: Dec 9, 2015 10. Dec 9, 2015 ### Samy_A Now I'm confused too. Lets assume that$f$is differentiable at$x=0$, and$M=\underset{x\to 0}{\lim}\frac{f(x) - f(x/2)}{x/2}\frac{f(x)-f(0)}{x}=\frac{f(x)-f(x/2)+f(x/2)-f(0)}{x}=\frac{f(x)-f(x/2)}{x}+\frac{f(x/2)-f(0)}{x}=\frac{1}{2}\frac{f(x)-f(x/2)}{x/2}+\frac{1}{2}\frac{f(x/2)-f(0)}{x/2}$Taking the limit for$x \to 0$, we get$f'(0)=\frac{1}{2}M+\frac{1}{2}f'(0)$, so$f'(0)=M$. 11. Dec 9, 2015 ### Samy_A But I assumed that$f$is differentiable at$x=0$. What he wants to prove is that if$M=\underset{x\to 0}{\lim}\frac{f(x) - f(x/2)}{x/2}$and$f$is continuous at$x=0$, then$f## is also differentiable in 0.

For now, I find neither a proof, nor a counterexample.

Last edited: Dec 9, 2015