Bounding $(x^n-y^n)/(x-y)$? Let $x,y \in [0,1]$ it follows then from the binomial theorem that for integers $n \ge 1:$ 
$$\sup_{x,y} \left\lvert \frac{x^n-y^n}{x-y} \right\rvert \le n.$$
Is this also true if $q \in [1,\infty)$:
$$\sup_{x,y} \left\lvert \frac{x^q-y^q}{x-y} \right\rvert \le q?$$
 A: Lagrange mean value theorem implies $$|x^q-y^q|=|q\xi^{q-1}(x-y)|,$$
where $\xi$ is between $x$ and $y$.
A: For a positive integer $n$, we have
$$\left|\frac{x^n-y^n}{x-y}\right|=\left|\sum_{r=0}^n\,x^r\,y^{n-r}\right|\leq \sum_{r=0}^n\,|x|^r\,|y|^{n-r}\leq n\,.$$
The equality does not hold (as $x\neq y$), but the inequality is sharp (by taking $y=1$ and $x\to 1^-$).  Thus, 
$$\sup_{\substack{{x,y\in[0,1]}\\{x\neq y}}}\,\left|\frac{x^n-y^n}{x-y}\right|=n\,.$$

Now, for a real number $q\geq 1$, we have by Bernoulli's Inequality that 
$$(1+t)^q\geq 1+qt\text{ for all }t\geq-1\,.\tag{*}$$
For (*) to be an equality, either $q=1$ or $t=0$.  The case $q=1$ is obvious.  From now on, we assume that $q>1$.
Without loss of generality, suppose that $x>y$.  Set $t:=\dfrac{x-y}{y}\in(0,\infty)$ in (*).  Then, the inequality becomes
$$\frac{x^q}{y^q}=\left(1+\frac{x-y}{y}\right)^q> 1+q\,\left(\dfrac{x-y}{y}\right) \,.$$
Thus,
$$\frac{x^q-y^q}{x-y}>q\,y^{q-1}\,.$$
Similarly, taking $t:=\dfrac{y-x}{x}\in(-1,0)$ in (*) yields
$$\frac{y^q}{x^q}=\left(1+\frac{y-x}{x}\right)^q> 1+q\,\left(\dfrac{y-x}{x}\right)\,,$$
so that
$$\frac{x^q-y^q}{x-y}< q\,x^{q-1}\,.$$
Ergo,
$$q\,y^{q-1}<\frac{x^q-y^q}{x-y}< qx^{q-1}\,.$$
Since $x\leq 1$, we have $$\frac{x^q-y^q}{x-y}<q\,.$$  However, this inequality is sharp by taking $x:=1$ and $y\to 1^-$.  That is,
$$\sup_{\substack{{x,y\in[0,1]}\\{x\neq y}}}\,\left|\frac{x^q-y^q}{x-y}\right|=q\text{ for all }q\geq 1\,.$$
