Prove that $f(x) =1/x^2$ is continuous on its domain $(0, \infty)$

Question

Prove that the function $\smash{\dfrac{1}{x^2}}$ is continuous on its domain $(0,∞)$.

Proof: $$ \lim_{x \to a} \frac{1}{x^2} = \frac{1}{a^2} $$

After manipulating I would get $$ \frac{(x-a)(x+a)}{x^2 a^2} $$ where $x+a < 1 + 2\lvert a \rvert$

What do I do to the $\smash{\dfrac{1}{x^2 a^2}}$?

Answer 1

Prove that $f(x)=1/x^2$ is continuous on its domain $(0,\infty)$

Let proof for $x\in [1,\infty)$,

We have that, $$\begin{align*} |f(x)-f(c)| &= \left|\frac{1}{x^2}-\frac{1}{c^2}\right| \\ &= \left|\frac{c^2-x^2}{x^2c^2}\right|\\ &= \left|\frac{x+c}{x^2c^2}\right| |x-c|\\ &= \left|\frac{x}{x^2c^2}+\frac{c}{x^2c^2}\right| |x-c|. \end{align*}$$

Now, $|x|\leq x^2[\because x\in [1,\infty)]$ implies that $\frac{|x|}{x^2}\leq 1$ and therefore $$ \frac{|x|}{x^2c^2} \leq \frac{1}{c^2} \leq 1$$

It follows that $|f(x)-f(c)|\leq 2|x-c|$. Hence given $\epsilon>0$ we let $\delta=\epsilon/2$ and if $|x-c|<\delta$ then $|f(x)-f(c)| \lt \epsilon$.

For $x\in (0,1)$, See The answer.