I need to decide whether it is true or false and show why. I think that it is true (and after spending a lot of time thinking about counter examples I probably understand why), but I don't know how to prove it.
If $\lim \limits_{x \to x_0}\frac{f(x)}{|x-x_0|} = 1$, then $f$ is not differentiable at $x_0$.
The equation implies that $f(x) \simeq |x-x_{0}|$ when $x$ is close to $x_{0}$. Intuitively, this implies that the graph of $f(x)$ looks like that of $g(x) = |x-x_{0}|$ near $x_{0}$. Then $f(x)$ may not be differentiable since left and right limit of $f(x)/(x-x_{0})$ might be different as $-1$ and $1$.
More rigorously, it can be done in the following way: Let's assume that $f$ is differentiable at $x_{0}$. Then $f$ is continuous at $x_{0}$ and $f(x_{0}) = \lim_{x\to x_{0}}f(x) = \lim_{x\to x_{0}}f(x)/|x - x_{0}| \cdot |x-x_{0}| = 0$. Now we have
$$ \lim_{x\to x_{0} + 0} \frac{f(x) - f(x_{0})}{x-x_{0}} = \lim_{x\to x_{0} + 0} \frac{f(x)}{|x-x_{0}|} = 1 $$ since $x > x_{0}$ ($x ->x_{0} + 0$ means the limit is right limit). Similarly, for left limit, we have $$ \lim_{x\to x_{0} - 0} \frac{f(x) - f(x_{0})}{x-x_{0}} = \lim_{x\to x_{0} - 0} - \frac{f(x)}{|x-x_{0}|} = -1 $$ so the limit of $(f(x)-f(x_{0}))/(x-x_{0})$ does not exist, and $f$ is not differentiable at $x = x_{0}$.