prove $\lim_{x \rightarrow 3} x^2 =9 $

Question

"The first thing we 'll do is to require that $|x-3|<1$" from Spivak. But why he fix the $\delta$ ? The defintion only allow us to fix the $\epsilon$. Suppose the case $\lim_{x\rightarrow1} 1/x $, fix $\delta=1$ won't work. Instead I try

$|x^2-9|=|x^2-6x+9+6x-18|<|x-3|^2 + 6|x-3|=|x-3|(|x-3|+6)< \epsilon < \epsilon^2/36 +\epsilon$
Since I choose $\delta =\epsilon/6$

Answer 1

I'll explain as I usually do by setting $u=x-3\to 0$

It may appear overworking it with such a simple limit, but keep in mind that our mathematical toolbox is richer regarding limits in $0$ or $\infty$.

Rewriting $|x^2-9|=|(x-3)(x+3)|=|u(u+6)|=\underbrace{|u|}_{\to 0}\times\underbrace{|u+6|}_{\text{bounded}}$

• $|x-3|<\delta$ translates immediately to $|u|<\delta$
• $u\to 0$ so it eventually is less than $1$ leading to $|u+6|<|u|+6<7$ which is bounded.

So we have two conditions to fulfil:

• $|u|<\delta$ with $\delta=\varepsilon$ to get the delta-epsilon proof going on
• $|u|<\delta$ with $\delta=1$ to be able to bound $|u+6|$

Note we have chosen $\delta=1$ but you realize that any positive number is suitable, could have been $\frac 12$ or could have been $1000$.

Both conditions are realized simultaneously when $\delta=\min(1,\varepsilon)$

And you get $|x^2-9|=|u||u+6|< 7\varepsilon$

You can eventually choose another $\delta=\min(1,\frac{\varepsilon}7)$ but this is pure aesthetic to get a bare epsilon at the end, it is not really requested. As long as you have $cst\times \epsilon$ this is fine.

Remark:

Why will you say, do we bother taking $\delta=\min(1,\varepsilon)$, since $\varepsilon$ is so small already, the condition is equivalent to $\delta=\varepsilon$ anyway, isn't it ?

In fact the limit definition does not explicitly says that epsilon is small, it says

$\forall \varepsilon>0,\ \exists \delta>0\ \text{ s.t. }|x-x_0|<\delta\implies |f(x)-\ell|<\varepsilon$

It is "for all" epsilons, possibly one million, possibly one million$^{th}$, though implicitly we are only interested in the small ones.

Therefore this $\delta=\min(1,\varepsilon)$ is just to cope with this definition.