Use $\varepsilon$-$\delta$ to prove $\lim_{x\to 1} \frac{1}{x^2} = 1$

Question

Use $\varepsilon$-$\delta$ to prove that $\lim_{x\to1}$ $\frac{1}{x^2} = 1.$

I set up the proof as follows: Let $\varepsilon>0$ s.t. $0<|\frac{1}{x^2} -1|<\delta.$ which simplifies to $0<|\frac{1-x^2}{x^2}|<\delta$. However I'm not sure where to take this next.

Answer 1

Suppose that $\left|x-1\right|<\frac12$. Then $x>\frac12$ and therefore $x^2>\frac14$. On the other hand,$$|1-x^2|=|x-1||x+1|\leqslant\frac32|x-1|.$$Therefore$$\left|\frac{1-x^2}{x^2}\right|<6|x-1|.$$So, take $\delta=\min\left\{\frac12,\frac\varepsilon6\right\}$.