I have tried a proof $\;\lim\limits_{x\to0}x^2=1\,.\,$ I know it is wrong but could you tell me where?

Question

I have tried a proof $\;\lim\limits_{x\to0}x^2=1\,.\,$ I know it is wrong but could you tell me where ?

By $\,\delta\,,\,\varepsilon\,$ definition $\;|f(x)-l|<\varepsilon\,$, $\;|x-0|<\delta$

1. $-\varepsilon<x^2-1<\varepsilon$
2. $-\varepsilon+1<x^2<\varepsilon+1$
3. $\sqrt{(-\varepsilon+1)}<x<\sqrt{(\varepsilon+1)}$

So I take $\,\delta\,$ as $\,\sqrt{(1 - \varepsilon)}\,.$

Answer 1

In your step 3,

3. $\sqrt{1-\epsilon} < x < \sqrt{1+\epsilon}$

This is a region roughly centered around $x=1$. In particular, when $\epsilon$ is very small, you cannot put e.g. $x=0.001$ to satisfy the inequality. When $\epsilon<1$, $x=0$ doesn't belong to the set.

The set $|x|<\delta$ is a small region around $0$. If you put $\delta = \sqrt{1-\epsilon}$ then that means you are considering $|x|<\sqrt{1-\epsilon}$ i.e. $$-\sqrt{1-\epsilon} < x < \sqrt{1-\epsilon}.$$ In fact these two regions are completely disjoint.

Hope this graph sends the point home:

Answer 2

We need to assume that $\epsilon<1$, and this is without loss of generality.

Next you have that:

4. $\sqrt{(-\epsilon + 1)}-1 < x-1 < \sqrt{(\epsilon + 1)}-1$

Take $\delta=\min\{\sqrt{(\epsilon + 1)}-1,1-\sqrt{(-\epsilon + 1)}\}$.

In fact setting $$ 1-\sqrt{(-\epsilon + 1)} =\frac{\epsilon}{1+\sqrt{(1-\epsilon)}}>\frac{\epsilon}{3}, $$ we obtain $$ |x-1|<\delta=\frac{\epsilon}{3}\Longrightarrow |x^2-1|=|x-1||x+1|\le |x-1| (|x-1|+2)\le \delta (2+\delta)< 3\delta =\epsilon. $$