As Mark Viola’s answer correctly shows, when you allow infinite limits, this limit is $\infty$. Your continued references to $L$ indicate that you do not allow infinite limits (which is also acceptable), in which case the limit does not exist. But your proof misstates the definition of limit. Below is a valid proof, including the correct definition.
\begin{equation}
\lim_{x \to 1} \frac{1}{(x – 1)^2} = Y
\end{equation}
is equivalent to
\begin{equation}
\forall(\epsilon > 0) \exists(\delta > 0) 0 < |x – 1| < \delta \implies \left| \frac{1}{(x – 1)^2} – Y \right| < \epsilon.
\end{equation}
No such value of $\delta$ exists, though. Per Mark Viola’s answer, we can actually find $\delta$ such that
\begin{align}
0 < |x – 1| < \delta &\implies \frac{1}{(x – 1)^2} > Y + \epsilon\\
&\implies \left| \frac{1}{(x – 1)^2} – Y \right| \not< \epsilon
\end{align}
This is true no matter what $Y$ is, showing that the limit does not exist.