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)

We have that for $A,B>0$ (i.e. dividing both sides by $AB$)
$$A>B \iff \frac 1 A < \frac 1 B$$
therefore
$$\frac{1}{(1-x)^2}>10^6 \iff (1-x)^2<\frac1{10^6}=10^{-6}$$
form which, using that $\sqrt{A^2}=|A|$ and (taking the square roots both sides)
$$A^2<B \iff |A|<\sqrt B$$
we find
$$(1-x)^2<10^{-6} \iff |1-x|<10^{-3}$$
and $|1-x|=|x-1|$.
In your evaluation it should be
$$x=1-\sqrt{10^{-6}}=1-10^{-3}=0.999$$
which is correct but not much effective and clear to obtain the result.
