Proving continuity with epsilon delta I have the function $f:\mathbb{R}\rightarrow \mathbb{R}\:\:f\left(x\right)=x^2-3x$ and it asks me to prove continuity in point $\:x_o=0$ using the epsilon-delta definition.
I know that in order to do so I must find that for any $\epsilon >0$, there exists a $\delta >0$ so that for any $x\in \mathbb{R}$ with $\left|x-x_o\right|<\delta $ we have $\left|f\left(x\right)-f\left(x_o\right)\right|<\epsilon \:$
This definition I know by heart, but I don't really understand it. Anyhow, following from past examples, I did this:
I said that $$\left|x-3\right|<1\:for\:x\in \left(2,4\right)$$ and $$\left|x\left(x-3\right)\right|=\left|x\right|\cdot \left|x-3\right|$$
so we have $$\left|x-3\right|\le \left|x\right|+3<7$$ and this together would mean that $$\left|f\left(x\right)-f\left(0\right)\right|=\left|x-3\right|\cdot \left|x\right|\le 7\cdot \left|x\right|$$
So now I'd have that for any $\epsilon >0$, choosing $\delta =min\left\{2,\frac{\epsilon }{7}\right\}$, we have that $\delta\gt 0$ and that for any $x\in \mathbb{R}$ with $\left|x\right|<\delta $ we have: $$\left|f\left(x\right)-f\left(x_o\right)\right|<7\left|x\right|<7\delta \le \epsilon $$
Did I do this right? I still don't understand much from it though.
Edit:
Alright how about this. We see that $\left|f\left(x\right)-f\left(0\right)\right|=\left|f\left(x\right)\right|=\left|x\right|\cdot \left|x-3\right|$ and we know $\left|x\right|<\delta $
Then we can say that $$\left|x-3\right|\le 3+\left|x\right|<\delta +3$$
So we'd have $$\left|f\left(x\right)\right|<\delta \left(\delta +3\right)$$
And the equation $t\left(t+3\right)<\epsilon $ has a solution when $t\in \left(\frac{-3-\sqrt{9+4\epsilon }}{2},\frac{-3+\sqrt{9+4\epsilon }}{2}\right)$
So then for any real $\epsilon\gt 0 $ we'd have a real $$\delta =\frac{-3+\sqrt{9+4\epsilon }}{2}$$ so that for any real x with $\left|x\right|<\delta $ we have $\left|f\left(x\right)-f\left(0\right)\right|=\left|x\right|\cdot \left|x-3\right|<\delta \left(\delta +3\right)<\epsilon $
 A: The definition is saying this:
$f(x)$ is continuous at $x_0$ iff getting sufficiently close to $x_0$  ($|x_0 -x| < \delta$) gets us arbitrarily close to $f(x_0)$  ($\Rightarrow |f(x_0) - f(x)| < \epsilon$).
A hint for epsilon delta proofs - don't set the value of delta in the beginning, just assume $|x-x_0|<\delta$, look how $|f(x)-f(x_0)|$ will look like in terms of delta and then reverse engineer delta to be some handy function of epsilon. Then set delta to be just that (on top of you page) and look like a genius :). 
Now a formal proof of continuity will go somewhat like this:
Let $\epsilon > 0$ arbitrary. If $\epsilon < 18$ take $\delta = \frac{\epsilon}{6}$. Else take $\delta = \sqrt{\frac{1}{2}\epsilon}$. Note that if $\epsilon < 18$ we have $0< \delta < 3$ and else $\delta > 3$.
Now for all x with $|x-x_0| = |x| < \delta$ we have:
$|f(x)-f(x_0)| = |x^2 - 3x| \leq |x^2|+|3x| < \delta^2 + 3\delta$
(at this point I'll start working out what delta should be in relation to epsilon)
Now depending on the case we have:
$\delta^2 + 3\delta \leq 6\delta = \epsilon$ or
$\delta^2 + 3\delta \leq 2\delta^2 = \epsilon$.
And thus $|x-x_0| < \delta \Rightarrow |f(x)-f(x_0)| < \epsilon$ meaning f is continuous in $x_0$ q.e.d.
A: You can take $\delta=\sqrt{\epsilon}$, then if $|x-0|<\delta\implies|x|<\delta=\sqrt{\epsilon}\implies|x|^2<\epsilon$
Now $$|x|^2<\epsilon\implies|x^2|<\epsilon\implies|x^2-0|<\epsilon\implies|x^2-0^2|<\epsilon\implies|f(x)-f(0)|<\epsilon$$
