Lipschitz function differentiable Let $f:R\to R$ be a Lipschitz function.
Suppose $\lim_{n\to \infty}n[f(x+\frac{1}{n})-f(x)]=\lim_{n\to \infty}n[f(x-\frac{1}{n})-f(x)]=0$.
Show that $f$ is differentiable on $R$.
I think this problem is too simple so I want to ask if I missed something. $\lim_{n\to \infty}n[f(x+\frac{1}{n})-f(x)]=0$ would imply $\lim_{y\to x^{+}}\frac{f(y)-f(x)}{y-x}=0$ because f is continuous, right? Then similarly we can conclude $f'(x)=0, \forall x\in R$? 
 A: No, your argument is not complete. You have to show that $\frac {f(x+h)-f(x)} h \to 0$ in whatever way $h \to 0$, not just along the sequence $h =\frac 1 n$.   For this use the bound on $f(x+h)-f(x+\frac 1 n)$ given  by Lipschitz condition . I will give more details if you cannot complete the argument. 
Some hints: let $\frac 1 {n+1} <h <\frac 1 n$. Then $$|f(x+h)-f(x)| \leq |f(x+\frac 1 n)-f(x)|+|f(x+h)-f(x+\frac 1 n)|$$ $$\leq |f(x+\frac 1 n)-f(x)|+M|h-\frac 1 n|.$$ Divide by $h$ and try to show that the ratio tends to $0$.
A: Let's take the variable $h\to 0^{+}$ and set $h=1/t$ so that $t\to \infty $ and we can write $t=[t] +\{t\} $ where $[t], \{t\} $ represent integral and fractional part of $t$ respectively. Then $$h=-\frac{\{t\}}{t[t]}+\frac{1}{[t]}$$ and $$f(x+h) - f(x) =f(x+1/[t])-f(x)+f(x+h)-f(x+1/[t])$$ and by Lipschitz condition on $f$ we can write $$|f(x+h) - f(x) |\leq |f(x+1/[t])-f(x)|+M\cdot \frac{\{t\}} {t[t]} $$ and hence $$\left |\frac{f(x+h) - f(x)} {h} \right |\leq \frac{t} {[t]} \cdot|[t] (f(x+1/[t]) - f(x)) |+M\cdot \frac{\{t\}}{[t]} $$ and the RHS clearly tends to $0$ as $h\to 0^{+} $. A similar argument can be given for $h\to 0^{-}$ and hence $f'(x) =0$ for all $x$. 
