Are these two definitions of differentiability equal?

Question

A function $f$ is differentiable in $x$ iff

1. the limit $~\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}~~\text{exists}$ ("normal" definition)
2. $|f(x+h)-f(x)|<C|h|~$ holds for small $h$ with $C>0$

Now for the sake of comparison, I called the first limit $C$:

$$\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}=C \iff \frac{|f(x+h)-f(x)|}{|h|}<C$$

but maybe that's not the right approach to understanding it intuitively. Could someone explain the connection/equivalence? I'm not seeing through.

Answer 1

Those conditions are not the same. To see this, try $f(x) = |x|$ at $0.$

Answer 2

Those two definitions of differentiability are not equivalent. Consider the famous function

$$f(x)=\begin{cases} x\sin\left(\frac 1x\right) & x\ne 0 \\ 0 & x=0 \end{cases}$$

It is easy to show that $|f(0+h)-f(0)|\le |h|$, so you can choose any value of $C$ greater than $1$ and the second definition holds. However, the first limit does not hold, since the value of the difference quotient oscillates between $-1$ and $1$, so there is no limit. There is not even a one-sided limit.