Show that $\frac{f(x+h)-f(x)}{h} \leq \limsup_{h \to 0+}\frac{f(x+h)-f(x)}{h}?$. > If $f$ is a continuous function on $[a,b]$ such that $$\frac{f(x+h)-f(x)}{h} < \frac{f(y)-f(x)}{y-x}\tag{$\ast$}$$ for any $y \in (x,x+h)$, then $$\frac{f(x+h)-f(x)}{h} \leq \limsup_{p \to 0+}\frac{f(x+p)-f(x)}{p}?$$ Actually, I'm trying to prove that if $$\limsup_{p \to 0+}\frac{f(x+p)-f(x)}{p} \leq k$$ so, $$\frac{f(x+h)-f(x)}{h} \leq k$$ for $a\leq x<x+h\leq b$. This is only part of an question. I (think that I) showed it without $(\ast)$, but it was very different from the rest of the question. Just for preciosity, I'd like to prove it using $(\ast)$. Can someone help me?

Suppose that for $y \in (x,x+h)$,

$$k < \frac{f(x+h)-f(x)}{h} < \frac{f(y)-f(x)}{y-x}$$

We then have for $\delta > 0$,

$$\frac{f(x+h)-f(x)}{h} \leqslant \sup_{y \in (x,x+\delta)}\frac{f(y)-f(x)}{y-x} \\\ \implies\frac{f(x+h)-f(x)}{h} \leqslant \lim_{\delta \to 0+}\sup_{y \in (x,x+\delta)}\frac{f(y)-f(x)}{y-x} = \limsup_{y \to x+} \frac{f(y)-f(x)}{y-x}, $$

leading to a contradiction

$$k < \limsup_{y \to x+}\frac{f(y)-f(x)}{y-x}= D^+f(x) \leqslant k$$