$f(x) = (2x-1)/x^2$: proof of continuity using the epsilon-delta definition [closed]

$f(x)=(2x−1)/x^2$ , where $x$ is in real number, except $0$.

I have been struggling with this for a while what seems to be quite trivial. I have started with $$|f(x)−f(c)|=\left|\frac{2x−1}{x^2}−\frac{2c−1}{c^2}\right|$$ and tried to cancel and approximate the terms as much as possible using triangle inequality and the properties of polynomial fractions and somehow squeeze in delta so that i can define it in terms of epsilon. If someone could leave some hints, i would appreciate it thanks.

If $x,c\in\Bbb R\setminus\{0\}$, then\begin{align}f(x)-f(c)&=\frac{2 x c^2-c^2-2 x^2 c+x^2}{c^2 x^2}\\&=\frac{c+x-2cx}{c^2x^2}(x-c).\end{align}Now, suppose that $|x-c|<\frac{|c|}2$. It follows from this that $|x|>\frac{|c|}2$ and that $|x|<\frac{3|c|}2$. Therefore\begin{align}\bigl|f(x)-f(c)\bigr|&<\frac{|c|+\frac32|c|+3c^2}{c^4/4}|x-c|\\&=\frac{10|c|+12c^2}{c^4}|x-c|.\end{align}So, if$$\delta=\min\left\{\frac{|c|}2,\frac{c^4\varepsilon}{10|c|+12c^2}\right\},$$you will have $|x-c|<\delta\implies\bigl|f(x)-f(c)\bigr|<\varepsilon$.