Prove that limit does not exist using delta-epsilon $$\lim_{x\to1} \frac{1}{(x-1)^2}$$
I'm trying to prove that this limit does not exist. Here is my attempt:
Given $L > 0$, we want to prove$$\exists\epsilon>0, \forall\delta, 0<|x-1| < \delta \land |\frac{1}{(x-1)^2} - L| > \epsilon $$
Let $\epsilon = 1$ and fix $L$.
Then
$$|\frac{1}{(x-1)^2}| > L + 1 \implies \frac{1}{x-1} > \sqrt{L+1} $$
$$\implies x-1 \leq \frac{1}{\sqrt{L+1}}$$
Let $\delta = \frac{1}{\sqrt{L+1}}$
Is this sufficient as a proof?
 A: 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.
A: The limit of interest is $\infty$.  To prove this, we let $\epsilon>0$, however large, be given.  Then, we have for $x\ne 1$
$$\frac{1}{(x-1)^2}>\epsilon$$
whenever, $0<|x-1|<\delta =\frac{1}{\sqrt{\epsilon}}$.  And we are done!
A: x approaches 1 as x-1 approaches 0 as (x-1)^2 approaches 0 as 1/(x-1)^2 approaches infinity.
