epsilon delta limit proof

To show that $\displaystyle \lim_{\varepsilon\to0+}\frac1{V(B(\vec{x_0},\varep silon))}\iiint_{B(\vec{x_0},\varepsilon)}f(\vec{x} )\,dV = f(\vec{x_0})$ , you need to show that the difference $\displaystyle \frac1{V(B(\vec{x_0},\varepsilon))}\iiint_{B(\vec{ x_0},\varepsilon)}f(\vec{x})\,dV - f(\vec{x_0})$ can be made arbitrarily small, provided that $\varepsilon$ is small enough. To get something "arbitrarily small", you usually try to make it smaller than some given $\varepsilon$ . But we already have an $\varepsilon$ here, so we had better choose another letter, say $\eta$ . Let's try to make

$\displaystyle \left|\frac1{V(B(\vec{x_0},\varepsilon))}\iiint_{B (\vec{x_0},\varepsilon)}f(\vec{x})\,dV - f(\vec{x_0})\right| < \eta,$ where $\eta>0$ is given.

The first thing to notice is that $\displaystyle \frac1{V(B(\vec{x_0},\varepsilon))}\iiint_{B(\vec{ x_0},\varepsilon)}f(\vec{x_0})\,dV = f(\vec{x_0})$ (because when you integrate the constant $f(\vec{x_0})$ over a ball, you get that constant times the volume of the ball. Therefore

$\displaystyle \left|\frac1{V(B(\vec{x_0},\varepsilon))}\iiint_{B (\vec{x_0},\varepsilon)}f(\vec{x})\,dV - f(\vec{x_0})\right| = \left|\frac1{V(B(\vec{x_0},\varepsilon))}\iiint_{B (\vec{x_0},\varepsilon)}\bigl(f(\vec{x}) - f(\vec{x_0})\bigr)\,dV\right|.$

Next, the absolute value of an integral is always less than the integral of the absolute value (of the function in the integral). Therefore

$\displaystyle \left|\frac1{V(B(\vec{x_0},\varepsilon))}\iiint_{B (\vec{x_0},\varepsilon)}\bigl(f(\vec{x}) - f(\vec{x_0})\bigr)\,dV\right| \leqslant \frac1{V(B(\vec{x_0},\varepsilon))}\iiint_{B(\vec{ x_0},\varepsilon)}\bigl|f(\vec{x}) - f(\vec{x_0})\bigr|\,dV.$

Now, notice that if we could arrange that $\bigl|f(\vec{x}) - f(\vec{x_0})\bigr| < \eta$ for all $\vec{x}\in B(\vec{x_0},\varepsilon)$ , then that last integral would be less than $\eta$ , as required. But that condition, $\bigl|f(\vec{x}) - f(\vec{x_0})\bigr| < \eta$ for all $\vec{x}\in B(\vec{x_0},\varepsilon)$ , follows from the given fact that $f$ is continuous at $\vec{x_0}$ (can you see why?).