@PeterTamaroff Yup, but that is not enough for differentiability, it has to be continuous in some open interval around the point where you are checking the differentiability.
In the second case, say $a+h=m/u$, then I get that $$\frac{{f\left( {a + h} \right) - f\left( a \right)}}{h} = \frac{{\frac{1}{u} - 0}}{{\frac{m}{u}}} = \frac{1}{m}$$
Once this is done, then, you can fix any integer $p$ and then show that for for $h < qp$
The ratio is greater than q.
Hence, for any epsilon,delta combination, you can make your difference of ratios always greater than $q-1$
and hence, the for every epsilon > 0 there exists delta>0 condition is broken, since the choice of q is upto you.
Wait I guess I am being totallt obscure here.
Let me formulate my arguments again.
You want to prove for every $\epsilon > 0$ there exists $\delta > 0$ such that $|x-a| < \delta \implies \left|\frac{f(x+h) - f(x)}{h} - L\right| <\epsilon$
We have now simplified it to. You want to prove for every $\epsilon > 0$ there exists $\delta > 0$ such that $|x-a| < \delta \implies \left|f\left(\frac{m}{u}\right)\frac{1}{h} - L\right| <\epsilon$
Now consider the interval $(a,a+\delta)$ for any delta.
Can you make $f\left(\frac{m}{u}\right)\frac{1}{h}$ vary wildly in this interval?
