# 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?

• The square root of $(x-1)^2$ is not necessarily $x-1$. According to your argument, $$\left|\frac1{(0.9-1)^2}\right|>48+1,$$ so $$-10=\frac1{0.9-1}>\sqrt{48+1}=7.$$ Nov 13, 2016 at 4:16
• @MartinArgerami I don’t think the post is saying that $\left| \frac{1}{(x – 1)^2} \right|$ is greater than $L + 1$; instead, it is saying that if this inequality holds then the two inequalities after it hold as well. In any case, the definition of limit was wrong to begin with, so I am not sure what any of the subsequent statements are supposed to mean. Mar 29, 2022 at 13:46

## 3 Answers

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.

$$$$\lim_{x \to 1} \frac{1}{(x – 1)^2} = Y$$$$

is equivalent to

$$$$\forall(\epsilon > 0) \exists(\delta > 0) 0 < |x – 1| < \delta \implies \left| \frac{1}{(x – 1)^2} – Y \right| < \epsilon.$$$$

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.

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!

• What happens to the $L$?
– bob
Nov 13, 2016 at 5:03
• Bob, there is no "$L$" that is relevant. If $\lim_{x\to x_0}f(x)=\infty$, then for any $\epsilon>0$, however large, there exists a $\delta>0$ such that $|f(x)|>\epsilon$ whenever $|x-x_0|<\delta$. Nov 13, 2016 at 5:06
• OP is clearly using a definition that only allows for finite limits, and apparently such a definition is generally accepted. Mar 29, 2022 at 13:50
• When $0 = |x – 1|$, $\frac{1}{(x – 1)^2}$ is undefined. The inequality should be $0 < |x – 1| < \delta$. Mar 29, 2022 at 14:13
• Would the downvoter care to comment? Mar 31, 2022 at 13:11

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

• -1 Extremely confusing and does not answer the question. Mar 29, 2022 at 13:47