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

[RBG]

Homework Statement

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 $h(t) = \lim_{x\to t} \frac{f(x)-f(g(t))}{x-g(t)}$

Homework Equations

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

 

[SammyS]

Homework Statement

Warning: I realize the title is misleading... the function itself isn't what's constant.

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 $f'(0)=M$? 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 $h(t) = \underset{x\to t}{\lim}\frac{f(x)-f(t)}{x-t}$

Homework Equations

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

Hello RBG. Welcome to PF !

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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 $f'(0)=M$?

 

[RBG]

Hello RBG. Welcome to PF !

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 $f'(0)=M$?

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.

 

[PeroK]

$$\underset{x\to 0}{\lim}\frac{f(x)-x/k}{x/k}=M \ \Rightarrow \ \underset{x\to 0}{\lim}\frac{f(x)}{x} = \frac{M+1}{k}$$

 

[Ray Vickson]

Homework Statement

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 $\lim_{x \to 0}\frac{f(x)-g(x)}{g(x)}=M$, 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 $h(t) = \lim_{x\to t} \frac{f(x)-f(t)}{x-t}$

Homework Equations

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

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.

 

[RBG]

$$\underset{x\to 0}{\lim}\frac{f(x)-x/k}{x/k}=M \ \Rightarrow \ \underset{x\to 0}{\lim}\frac{f(x)}{x} = \frac{M+1}{k}$$

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!

 

[PeroK]

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!

$f'(0) = \underset{x\to 0}{\lim}\frac{f(x/k)-f(0)}{x/k} \ne \underset{x\to 0}{\lim}\frac{f(x) - f(x/k)}{x/k}$

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 $k = 2$)

 

[RBG]

$f'(0) = \underset{x\to 0}{\lim}\frac{f(x/k)-f(0)}{x/k} \ne \underset{x\to 0}{\lim}\frac{f(x) - f(x/k)}{x/k}$

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 $k = 2$)

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 $f'(0)=M$?

 

[PeroK]

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 $f'(0)=M$?

See Samy_A's post below. $k = 2$ is a special case.

 

[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 $f'(0)=M$.

 

[Samy_A]

See Samy_A's post below. $k = 2$ is a special case.

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.