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

  • $\begingroup$ 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 $$-10=\frac1{0.9-1}>\sqrt{48+1}=7.$$ $\endgroup$ Nov 13, 2016 at 4:16
  • $\begingroup$ @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. $\endgroup$ Mar 29, 2022 at 13:46

3 Answers 3


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.


The limit of interest is $\infty$. To prove this, we let $\epsilon>0$, however large, be given. Then, we have for $x\ne 1$


whenever, $0<|x-1|<\delta =\frac{1}{\sqrt{\epsilon}}$. And we are done!

  • $\begingroup$ What happens to the $L$? $\endgroup$
    – bob
    Nov 13, 2016 at 5:03
  • $\begingroup$ 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 $|x-x_0|<\delta$. $\endgroup$
    – Mark Viola
    Nov 13, 2016 at 5:06
  • $\begingroup$ OP is clearly using a definition that only allows for finite limits, and apparently such a definition is generally accepted. $\endgroup$ Mar 29, 2022 at 13:50
  • $\begingroup$ When $0 = |x – 1|$, $\frac{1}{(x – 1)^2}$ is undefined. The inequality should be $0 < |x – 1| < \delta$. $\endgroup$ Mar 29, 2022 at 14:13
  • $\begingroup$ Would the downvoter care to comment? $\endgroup$
    – Mark Viola
    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.

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

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