Calculating value of $\delta$ for a function $y=\frac1{(1-x)^2}$ unbounded at $x\rightarrow 1$ The function $y=\frac 1{(1-x)^2}$ is unbounded at $x\rightarrow 1$.
How can I calculate the value of $\delta$ such that $y> 10^6$ if $\left | x-1 \right |< \delta$ ?
Taking $y=f(x)=10^6$, putting in ${y=\frac1{(1-x)^2}}$, we get
$10^6 = \frac1{(1-x)^2}\\
\frac1{(1-x)^2} = \frac1{10^6}\\
x=1-\sqrt{(10^{-6})}$ meaning $x= 0.999$.
As, $\left |x-1\right |< \delta,$ hence $ \left|0.999 -1\right| < \delta,$ or
$10^{-3} < \delta$.
It can imply that,
$ \delta$ must be at least $10^{-3}$ or greater $ \delta \geq 10^{-3}$
(After correction by user:505767)

Pick $\delta$ such that
$$y=\frac{1}{\delta^2}=10^6$$
then at $\delta=±0.001$, $y=10^6$.
So$$|x-1|<\delta=0.001$$
in order to have $$y=\frac{1}{(1-x)^2}>\frac{1}{\delta^2}=10^6$$
