Why is $\lim\limits_{h \rightarrow 0} \frac{1}{h} \int_{z}^{z+he_1} P(z_1 + h, z_2) \,\mathrm dh = P(z_1, z_2)$ for continuous $P$? During proving some theorem I encountered the following difficulty. Why: 
$$\lim_{h \rightarrow 0} \frac{1}{h} \int_{z}^{z+he_1} P(z_1 + h, z_2) \,\mathrm dh = P(z_1, z_2)$$
where $P$ is a continuous function?
Can somebody explain it to me?
 A: The continuity of $P$ (at $z = (z_1,\,z_2)$) says that given an arbitrary $\varepsilon > 0$, there is a $\delta > 0$ such that $\lVert w-z\rVert < \delta \Rightarrow \lvert P(w) - P(z)\rvert <\varepsilon$.
$$\lim_{h \to 0} \frac{1}{h}\int\limits_0^h P(z_1 + t,\, z_2)\,dt = P(z_1,\,z_2)$$
means that given an arbitrary $\varepsilon > 0$, there is a $\delta > 0$ such that
$$h < \delta \Rightarrow \biggl\lvert \frac{1}{h}\int\limits_0^h P(z_1 + t,\, z_2)\,dt - P(z_1,\,z_2)\biggr\rvert < \varepsilon.$$
Now, for a given $\varepsilon > 0$ choose $\delta > 0$ so that $\lvert P(w) - P(z)\rvert < \varepsilon$ for $\lVert w - z\rVert < \delta$. By the continuity of $P$ in $z$, that is possible. Then, for $0 < h < \delta$, $$0 \leqslant t \leqslant h \Rightarrow \lVert (z_1 + t,\,z_2) - (z_1,\,z_2)\rVert = t \leqslant h < \delta$$, and we can estimate
$$
\begin{align}
\biggl\lvert \frac{1}{h}\int\limits_0^h P(z_1 + t,\, z_2)\,dt - P(z_1,\,z_2)\biggr\rvert &= \frac{1}{h} \biggl\lvert\int\limits_0^h P(z_1 + t,\,z_2) - P(z_1,\,z_2)\,dt\biggr\rvert\\
& \leqslant \frac{1}{h} \int\limits_0^h \lvert P(z_1+t,\,z_2) - P(z_1,\,z_2)\rvert\,dt\\
&< \frac{1}{h} \int\limits_0^h \varepsilon \,dt = \varepsilon
\end{align}$$
for $h < \delta$.
