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If $f$ is differentiable at $x=0$ and $\lim_{x\to0}{f(x)\over x}=3$ then $f(0)=0$ and $f'(0)=3 $.

so after a few failed counter examples I decided to prove this since it also kinda seems to be true... this is where I got to:

$f$ is differentiable at $x=0\Rightarrow$ $f'(0)$ exists.

$f'(0)=\lim_{x\to0}{f(x)-f(0)\over x-0}=\lim_{x\to0}{{f(x)\over x}-{f(0)\over x}}$

$f'(0)-3=\lim_{x\to0}({{f(x)\over x}-{f(0)\over x})-\lim_{x\to0}{f(x)\over x}}=\lim_{x\to0}{-{f(0)\over x}}$

$\Rightarrow f'(0)=3-\lim_{x\to0}{{f(0)\over x}}$

now comes the part where i'm not sure about...

$\Rightarrow f'(0)={3\lim_{x\to0}{x\over x}}-\lim_{x\to0}{{f(0)\over x}}={\lim_{x\to0}{3x\over x}}-\lim_{x\to0}{{f(0)\over x}}=\lim_{x\to 0}{{3x-f(0)}\over{x}-0}$

which is by definition the derivative of the function $3x$ at $x=0$ which satisfies both $f'(0)=3$ and $f(0)=0$....

is this true? it seems a little stinky since I forced in a specific function....

cheers

Rubenz
  • 421

2 Answers2

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The statement is true. $f$ is differentiable at 0 implies that $$ f(x) = f'(0)x + f(0) + r(x), $$ where $r(x)$ is a remainder term such that $\lim_{x\rightarrow 0}\frac{r(x)}{x} = 0.$ Thus $$ \lim_{x\rightarrow 0}\frac{f(x)}{x} = \lim_{x\rightarrow 0}\left(f'(0) + \frac{f(0)}{x}+ \frac{r(x)}{x}\right). $$

On LHS we get 3. On the RHS we have $$ f'(0) + \lim_{x\rightarrow 0}\frac{f(0)}{x}. $$ If $f(0) \neq 0$, then this term would be infinite, giving a contradiction.

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In a neighborhood of $x=0$ we have $f(x)=3x+o(x)$, $f$ is differentiable and then continuous at $x=0$ and then $f(x)\to 0$ for $x\to 0$, i.e. $f(0)=0$ and $f'(0)=\lim_{x\to0}{f(x)\over x}=3$.

alexjo
  • 14,976