Function being continuous at a point I've been looking at the $\epsilon-\delta$ arguments for determining whether a function is continuous at a point. I'm really stuck on how to choose your $\epsilon$. Specifically lets look at the function, $f(x)=\frac{2}{3x}$ being continuous at $x=1$.
Proof: Given $\epsilon$ > 0 there exists $\delta>0$ s.t $ |f(x)-f(1)|< \epsilon$, whenever 0$<|x-1|<\delta$. Well, $|f(x)-f(1)|=|\frac{2}{3x}-\frac{2}{3}|=|\frac{2-2x}{3x}| = |\frac{2(x-1)}{3x}| = \frac{2}{3} |\frac{x-1}{x}|\leq\frac{2}{3}(\frac{1}{|x|}+|1|)$. We need $\frac{1}{|x|}<\delta$. Suppose $\frac{1}{|x|}<1 \Rightarrow 1<|x|$.....
 A: With the inequality
$$\frac23 \left\lvert \frac{x-1}{x}\right\rvert \leqslant \frac23\left(\frac{1}{\lvert x\rvert} + 1\right),$$
you can't make the right hand side arbitrarily small anymore. From the right hand side, you cannot prove the continuity. The usual way to deal with an expression like the one on the left hand side is to first introduce an $\varepsilon$-independent bound on $\delta$, for example, whatever $\varepsilon$ is, we will always choose $\delta \leqslant \frac12$. Then that constraint gives you $\lvert x\rvert = \lvert 1 - (1-x)\rvert \geqslant 1 - \lvert 1-x\rvert \geqslant 1 - \delta \geqslant \frac12$. Plugging that into the left hand side, you get
$$\frac23 \left\lvert \frac{x-1}{x}\right\rvert \leqslant \frac23 \frac{1}{1/2} \lvert x-1\rvert = \frac43 \lvert x-1\rvert.$$
Now you have an inequality that allows to find a $\delta$ easily for any given $\varepsilon > 0$. Adding our global constraint, we find

For $\varepsilon > 0$ given, choose $\delta = \min \left\{\frac34\varepsilon,\, \frac12\right\}$, then $\lvert x-1\rvert < \delta \Rightarrow \lvert f(x)-f(1)\rvert < \varepsilon$.

