$$\lim_{x \to 1} \frac{1}{(1-x)^2}=\infty$$
All M>0 exists $\delta>0$ so all x with $0<|x-1|<\delta$ implies $\frac{1}{(1-x)^2}>M$
My proof (It probably incorrect) :
$$\frac{1}{(1-x)^2} > M$$ $$\frac{1}{(1-x)^2} > M$$ $$1>M*(1-x)^2$$ $$\frac{1}{M}>(1-x)^2$$ $$\frac{1}{\sqrt{M}}>1-x$$ $$\frac{1}{\sqrt{M}}-1>1-x-1$$ $$\frac{1}{\sqrt{M}}-1>-x$$ $$-\frac{1}{\sqrt{M}}+1<x$$
$$0<\delta \leq -\frac{1}{\sqrt{M}}+1$$
But $-\frac{1}{\sqrt{M}}+1$ could be $<0$
Any help will be appreciated.
Actually you'd be done by noticing that $(1-x)^2 = |1-x|^2$ and thus the condition $$f(x) > M \Leftrightarrow 1 > M(1-x)^2$$ is fulfilled if ($\Leftarrow$) $$M \delta^2 < 1$$ or ($\Leftrightarrow$) $$\delta < \frac1{\sqrt M}$$ so chose $\delta(M) = \begin{cases} \frac1{2\sqrt M} & M > 0 \\ 1 & M\le 0\end{cases}$ and you are done. (Note that WLOG $M>0$ is allowed so we don't have to chose a $\delta$ for $M\le 0$, but for $M\le 0$, $\delta$ can be chosen arbitrarily anyways.)