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

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  • $\begingroup$ 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 $$-10=\frac1{0.9-1}>\sqrt{48+1}=7.$$ $\endgroup$ – Martin Argerami Nov 13 '16 at 4:16
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x approaches 1 as x-1 approaches 0 as (x-1)^2 approaches 0 as 1/(x-1)^2 approaches infinity.

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

$$\frac{1}{(x-1)^2}>\epsilon$$

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

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  • $\begingroup$ What happens to the $L$? $\endgroup$ – bob Nov 13 '16 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 '16 at 5:06
  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark $\endgroup$ – Mark Viola Nov 17 '16 at 5:00

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