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Let function $f:\mathbb{R}\to \mathbb{R}$ be such that

$$ \lim_{\Large{(y,z)\rightarrow (x,x) \atop y\neq z}} \frac{f(y)-f(z)}{y-z}=0. $$ Is it then $f'(x)=0$ ?

Pedro
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user101
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    And what do you think about it? – Did Apr 01 '13 at 13:34
  • I would be curious to see you expand the haiku-like accepted answer into a detailed proof. For example, "to take $z=x$ in the double limit", as you suggest in a comment, is a pretty sure way to get a wrong solution. – Did Apr 01 '13 at 14:29
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    For the record, the limiting notion you've asked about is called the strong derivative, and you can find more than you'd ever want to know about it in my answer here. – Dave L. Renfro Apr 02 '13 at 14:46
  • @Did:I added a detailed proof. Can you please explain to me where I am wrong? – P.. Apr 25 '13 at 21:11
  • @P.. The detailed proof works. – Did Apr 25 '13 at 21:56

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What is $f'(x)=\displaystyle\lim_{y\to x}\dfrac{f(y)-f(x)}{y-x}$?

Let $\epsilon>0$. From the hypothesis it follows that exists a $\delta>0$ such that if $y\neq z$ and $0<\|(y,z)-(x,x)\|<\delta$ then $\left|\dfrac{f(y)-f(z)}{y-z}\right|<\epsilon$.
Now if $0<|y-x|<\delta$ then $0<\|(y,x)-(x,x)\|=|y-x|<\delta$ and $y\neq x$.

Therefore $\left|\dfrac{f(y)-f(x)}{y-x}\right|<\epsilon$.

It follows that $f'(x)=\displaystyle\lim_{y\to x}\dfrac{f(y)-f(x)}{y-x}=0.$

P..
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