# epsilon delta limit proof

• Dec 7th 2010, 10:26 PM
nikcs123
epsilon delta limit proof
http://img130.imageshack.us/img130/8493/29692333.png

We have barely gone over E-D limit proofs in class this year, I kind of understand them and can do them for simple functions, but I don't know where to start on this one.
• Dec 9th 2010, 09:24 AM
Opalg
To show that $\displaystyle \displaystyle \lim_{\varepsilon\to0+}\frac1{V(B(\vec{x_0},\varep silon))}\iiint_{B(\vec{x_0},\varepsilon)}f(\vec{x} )\,dV = f(\vec{x_0})$, you need to show that the difference $\displaystyle \displaystyle \frac1{V(B(\vec{x_0},\varepsilon))}\iiint_{B(\vec{ x_0},\varepsilon)}f(\vec{x})\,dV - f(\vec{x_0})$ can be made arbitrarily small, provided that $\displaystyle \varepsilon$ is small enough. To get something "arbitrarily small", you usually try to make it smaller than some given $\displaystyle \varepsilon$. But we already have an $\displaystyle \varepsilon$ here, so we had better choose another letter, say $\displaystyle \eta$. Let's try to make

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

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

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

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

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

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