Derivative at point but point itself converges Say, that $\lim_{x\to 0} f(x)=1$ anď $\lim_{x\to 0} \frac{f(x)-1}{x} = 1$ then we get $\lim_{x\to 0} f'(x)=1$
Is this an obvious conclusion?
I think there is one omitted thing for this approach.
$$\lim_{x\to 0, \tau\to 0} \frac{f(x)-f(\tau)}{x-\tau} = \lim_{x\to 0} f'(x)$$
Am I right? I don't know how to prove this since I haven't learned higher math but naive definition of derivative with graph (slope of tangent line)
 A: You're partially right. If we assume that $f$ is continuous at $0$, that is $f(0) = \lim_{x\rightarrow 0} f(x) = 1$, then $\lim_{x\rightarrow 0}\frac{f(x)-1}{x} = 1$ means only that $f'(0)=1$, but it doesn't have to mean that $f'(x)$ is continuous at $x=0$, which would imply $\lim_{x\rightarrow 0} f'(x) = 1$.
However, if $f(x)$ is continuous at $0$, then you still have
$$ \lim_{x\rightarrow 0} \Big(\lim_{\tau\rightarrow 0} \frac{f(x)-f(\tau)}{x-\tau}\Big) = \lim_{x\rightarrow 0} \frac{f(x)-f(0)}{x-0} = f'(0)$$
What you want is the limit
$$ \lim_{x\rightarrow 0} \Big(\lim_{\tau\rightarrow x} \frac{f(x)-f(\tau)}{x-\tau}\Big) = \lim_{x\rightarrow 0} f'(x)$$
ADDENDUM: An example.
Function $$f(x)=1+x+x^2\sin\frac{1}{x}$$ satisfies $\lim_{x\rightarrow 0} f(x) = \lim_{x\rightarrow 0} \frac{f(x)-1}{x} = 1$, but limits $$\lim_{x\rightarrow 0} f'(x) = \lim_{x\rightarrow 0} \big(1+2x\sin\frac{1}{x} - \cos\frac{1}{x}\big) $$
and $$\lim_{x\rightarrow 0, \tau\rightarrow 0} \frac{f(x)-f(\tau)}{x-\tau} = \lim_{x\rightarrow 0, \tau\rightarrow 0} \frac{x-\tau + x^2\sin\frac{1}{x} - \tau^2\sin\frac{1}{\tau}}{x-\tau}$$
don't exist. To see that the last limit doesn't exist, consider the sequence $(x_n,\tau_n) = (\frac{1}{(n+\frac12)\pi},\frac{1}{n\pi})$.
A: Your question implicitly assumes that $f$ is differentiable in a neighborhood $I$ of $0$ except possibly at $0$ (if it were not so then it would not make sense to talk of limit of $f'(x) $).
Let $g$ be another function defined in the same neighborhood $I$ of $0$ such that $g(x) =f(x), x\neq 0$ and $g(0)=1$. Then $g$ is continuous in its domain $I$ and further $g$ is differentiable in $I$ with $g'(x) =f'(x), x\neq 0,g'(0)=1$.
With these conditions we can't conclude that $\lim_{x\to 0}f'(x)=\lim_{x\to 0}g'(x)$ exists. However by mean value theorem one can show that if this limit exists then it must equal $g'(0)=1$.
