Prove that limit does not exist using delta-epsilon

Question

$$\lim_{x\to1} \frac{1}{(x-1)^2}$$ 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?

Answer 1

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.

Answer 2

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!

Answer 3

x approaches 1 as x-1 approaches 0 as (x-1)^2 approaches 0 as 1/(x-1)^2 approaches infinity.