Help with epsilon-delta proof that 1/(x^2) is continuous at a point.

Question

I'm trying to prove that $\lim_{x \to x_0} \frac{1}{ x^2 } = \frac{1}{ {x_0}^2 }$. I know this means that for all $\epsilon > 0$, I must show that there exists a $\delta > 0$ such that $\left | x - x_0 \right | < \delta \Rightarrow \left | \frac{1}{ x^2 } - \frac{1}{ {x_0}^2 } \right | < \epsilon$. Since $f(x) = \frac{1}{x^2}$ is an even function, I'll restrict the domain to $x>0$. After some manipulation of $\left | \frac{1}{ x^2 } - \frac{1}{ {x_0}^2 } \right | < \epsilon$ I have $\left | x - x_0 \right | < \epsilon\frac{x^2{x_0}^2}{x+{x_0}}$. I know that I can't just let $\delta = \epsilon\frac{x^2{x_0}^2}{x+{x_0}}$ since $\delta$ should not depend on $x$, so I need something else. This is where I'm stuck. I think I need to choose an upper bound for $\left|x-x_0\right|$ which depends on $x_0$ and find a sufficient $\delta$ from this restriction, but I'm not sure how to go about this.

Answer 1

Take several steps.

First, choose a $\delta_0 > 0$ such that for $\lvert x-x_0\rvert \leqslant \delta_0$ you can bound the expression $\dfrac{x+x_0}{x^2x_0^2}$ by an expression depending only on $x_0$. A choice of the form $\delta_0 = c\cdot x_0$ is the canonical way.

Then use the bound involving only $x_0$ - let's call it $b(x_0)$ - to get a $\delta_1 > 0$, namely $\delta_1 = \dfrac{\epsilon}{b(x_0)}$, and choose $\delta := \min \{ \delta_0,\delta_1\}$ to obtain the desired

$$\lvert x-x_0\rvert < \delta \implies \left\lvert \frac{1}{x^2} - \frac{1}{x_0^2}\right\rvert < \epsilon.$$