A problem from my analysis homework:
Prove directly that the function $\frac{1}{x^2}$ is continuous on $(0,\infty)$.
I think I have the details worked out; I would just like to check that I have the ideas correct.
Proof: Let $ x_0 \in (0, \infty) $ be arbitrary and fix $ \epsilon >0 $. Notice that $$ \left|\frac{1}{x^2} - \frac{1}{x_0^2}\right| \le \frac{(|x| + |x_0|)|x-x_0|}{x^2x_0^2} $$
Let $ \delta = \epsilon\frac{x_0^3}{6} $. Thus if $ |x-x_0| < \delta $, we have $$ \left|\frac{1}{x^2} - \frac{1}{x_0^2}\right| \le \frac{(|x| + |x_0|)|x-x_0|}{x^2x_0^2} < \delta\frac{(|x| + |x_0|)}{x^2x_0^2}<\delta\frac{6}{x_0^3} = \epsilon \quad\square$$
The key part of this proof was the construction of $\delta$, for which I required that $|x| < \frac{x_0}{2}$. Doing so allowed me to say $\frac{(|x| + |x_0|)}{x^2x_0^2} < \frac{6}{x_0^3} $. The part I am unsure about is how we can make this requirement (if it is correct; if not, why not and what is a more justifiable condition?). I know that we are trying to "control" the value of $x$ by making it "close-enough" to $x_0$ but the $\frac{x_0}{2}$ part seems arbitrary. Couldn't we just as easily require $|x| < \frac{x_0}{3}$?
the $\frac{x_0}2$ is required for proving continuity from below.
it is worth noting that continuity from above is somewhat easier to show because (for $\delta \gt 0$): $$ \frac1{x_0^2} - \frac1{(x_0+\delta)^2} = \frac{\delta^2+2\delta x_0}{x_0^2(x_0+\delta)^2} \lt \frac{2\delta^2+2\delta x_0}{x_0^2(x_0+\delta)^2}\lt \frac{2\delta}{x_0^2(x_0+\delta)} \lt \frac{2\delta}{x_0^3} $$ for continuity from below we have: $$ \frac1{(x_0-\delta)^2}-\frac1{x_0^2}= \frac{2\delta x_0-\delta^2}{x_0^2(x_0-\delta)^2} $$ the simplification now requires $\delta \lt \frac{x_0}2$, in which case we have $$ \frac{2\delta x_0-\delta^2}{x_0^2(x_0-\delta)^2} \lt \frac{2\delta x_0}{x_0^2(x_0-\delta)^2} \lt \frac{2\delta x_0}{x_0^2(\frac{x_0}2)^2}=\frac{8\delta}{x_0^3} $$