$\lim \limits_{ x \rightarrow -3} x^2=9$ epsilon-delta proof

Question

So far I have $|x+3|<δ \Rightarrow |x-3||x+3|< \epsilon$ Do I need to bound $|x-3|$? I know that $|x-3|<|x+3|$ for $x>0$, so I can use $\delta=\sqrt{\epsilon}$ if $x>0$ but how can I make sure $x>0$ by bounding $\delta$? I tried the usual way by letting $\delta<1 \Rightarrow x<-2 \Rightarrow |x-3|<|-2-3|=5$ and got $\delta=min\{1,\frac{\epsilon}{5}\}$ but I'm not sure if this is correct...

Answer 1

Hint: Show that for every $\delta$ in $(0,1)$, $$|x+3|\leqslant\delta\implies|x^2-9|\leqslant7\delta.$$

Answer 2

Assume a priori $\delta < 1 $, and bound now $|x+3|$ using triangle inequality:

$$ |x + 3| = |x -3 + 6| \leq |x-3| + 6 < 7$$ and so

$$ |x^2 - 9 | = |x+3||x-3| < 7 \delta $$

Choose $\delta = \min( \frac{\epsilon}{7}, 1 ) $

Answer 3

Solve the inequation $$ |x^2-9|<\varepsilon $$ which is equivalent to $$ \begin{cases} x^2<9+\varepsilon\\ x^2>9-\varepsilon \end{cases} $$ Since you can assume $\varepsilon<9$, the inequalities become $$ \sqrt{9-\varepsilon}<|x|<\sqrt{9+\varepsilon} $$ which is a neighborhood of $-3$, because part of it is the set of $x$ such that $$ -\sqrt{9+\varepsilon}<x<-\sqrt{9-\varepsilon} $$ and $$ -\sqrt{9+\varepsilon}<-3<-\sqrt{9+\varepsilon}. $$ If you really need to find a suitable $\delta$, you can find the smaller between $-3+\sqrt{9+\varepsilon}$ and $3-\sqrt{9-\varepsilon}$. You can do it like this: \begin{align} -3+\sqrt{9+\varepsilon} &\mathrel{?} 3-\sqrt{9-\varepsilon}\\ \sqrt{9+\varepsilon}+\sqrt{9-\varepsilon} &\mathrel{?} 6\\ 9+\varepsilon+2\sqrt{81-\varepsilon^2}+9-\varepsilon &\mathrel{?} 36\\ 2\sqrt{81-\varepsilon^2} &\mathrel{?} 18\\ \sqrt{81-\varepsilon^2} &\mathrel{?} 9\\ 81-\varepsilon^2 &\mathrel{?} 81 \end{align} Since in the last inequality $<$ holds, it holds also in the other inequalities. Hence you can take $$ \delta=-3+\sqrt{9+\varepsilon} $$