Question: I'm looking at a proof that in the last step states the following:

$$
\frac{1}{\varepsilon} \int\_{y\_{0}}^{y\_{0}+\varepsilon}\left[f\left(u\left(x\_{2}, y\right)\right)-f\left(u\left(x\_{1}, y\right)\right)\right] d y \rightarrow f\left(u\left(x\_{2}, y\_{0}\right)\right)-f\left(u\left(x\_{1}, y\_{0}\right)\right), \quad \text { as } \quad \varepsilon \downarrow 0
$$

Under the assumption that both $f$ and $u$ are continuous. Can anyone explain to me how this results from the fundamental theorem of calculus? The term in the integral looks similar to the derivative of $f$, but $f$ is evaluated at two different $x$-values.

Answer: 
Let$$F(y)=\int\_{y\_0}^yf\bigl(u(x\_2,t)\bigr)-f\bigl(u(x\_1,t)\bigr)\,\mathrm dt.$$Then $F(y\_0)=0$ and therefore$$F'(y\_0)=\lim\_{\varepsilon\to0}\frac{F(y\_0+\varepsilon)}\varepsilon.$$But, by the Fundamental Theorem of Calculus,$$F'(y\_0)=f\bigl(u(x\_2,y\_0)\bigr)-f\bigl(u(x\_1,y\_0)\bigr).$$
Perhaps one could present this problem as an exercise in applying the fundamental theorem of calculus. Nonetheless, it seems to me that using the fundamental theorem of calculus on this problem is a bit of overkill. You can do this pretty directly with the definition of the definite integral.

Let me denote $g(y) = f(u(x\_2,y))-f(u(x\_1,y))$, so you're asked to find
$$\lim\_{\epsilon \to 0} \int\_{y\_0}^{y\_0+\epsilon} g(y) \, dy
$$
Let 
$$m\_\epsilon = \inf\{g(y) \mid y\_0 \le y \le y\_0+\epsilon\}
$$
$$M\_\epsilon = \sup\{g(y) \mid y\_0 \le y \le y\_0+\epsilon\}
$$
These exist because a continuous function is bounded on a closed interval. 

From the definition of the definite integral we have
$$g(m\_\epsilon)\cdot\epsilon \le \int\_{y\_0}^{y\_0+\epsilon} g(y) \, dy \le g(M\_\epsilon) \cdot \epsilon
$$
$$g(m\_\epsilon) \le \frac{1}{\epsilon} \int\_{y\_0}^{y\_0+\epsilon} g(y) \, dy \le g(M\_\epsilon)
$$
By continuity of $g$, 
$$\lim\_{\epsilon \to 0} g(m\_\epsilon) = \lim\_{\epsilon \to 0} g(M\_\epsilon) = g(y\_0)
$$
By the [Panini Theorem](https://en.wikipedia.org/wiki/Squeeze_theorem),
$$\lim\_{\epsilon \to 0} \frac{1}{\epsilon} \int\_{y\_0}^{y\_0+\epsilon} g(y) \, dy = g(y\_0)
$$