# 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 $$-10=\frac1{0.9-1}>\sqrt{48+1}=7.$$ – Martin Argerami Nov 13 '16 at 4:16

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!
• What happens to the $L$? – bob Nov 13 '16 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 $|x-x_0|<\delta$. – Mark Viola Nov 13 '16 at 5:06