If $f:]0,a[\rightarrow \mathbb{R}$ is diff. with $\lim_{x\downarrow 0}f(x)=0$ and $\lim_{x\downarrow 0}f'(x)=c$ then $\lim_{x\downarrow 0}f(x)/x=c$ I am not familiar with lhospialrule and everything after that, please consider this in your answer I have tried to first look at $c=0$ with an estimate by employing the triangleequality $|f(x)/x|\leq (|f(x)-f(x_0)|/x)+f(x_0)/x$ in combination with the fact that there exists a $x_0$ such that $|f'(x)|<\epsilon$ for all $x\leq x_0$ in conjuncture with a corollary that if the derivative is bounded then the secant is a also bounded (i.e. $\forall x\in ]a,b[:|f'(x)|\leq\epsilon\Rightarrow \forall x_1\leq x_2\in [a,b]:f(x_2)-f(x_1)\leq \epsilon(x_2-x_1)$). But it didnt work out. I also dont know how to use that $\lim_{x\downarrow0}f(x)=0$. Please help.

Since you know that $f(x) \to 0$ as $x \to 0$ you can define the extended function $F: [0,1) \to \mathbb{R}$ via $$ F(x) = \begin{cases} f(x) & \text{if }x >0 \\\ 0 &\text{if }x=0. \end{cases} $$ Then $F$ is continuous on $[0,1)$ and differentiable on $(0,1)$. By the mean-value theorem, for $0 < x <1$, $$ \frac{f(x)}{x} = \frac{F(x) - F(0)}{x -0} = f'(y) $$ for some $0 < y < x$. From this and the fact that $f'(x) \to c$ as $x \to 0$ we can readily deduce that $f(x)/x \to c$ as $x \to 0$.