3

Is this proof valid? Can d be relative to x ? enter image description here

Prog Kid
  • 87
  • You are close, but you need your definition of $\delta$ to not depend on $x.$ First insist that $\delta \le 1$ then $|x+3| \ge 5$ – Doug M Feb 16 '18 at 23:38
  • 1
    You can restrict $x$ to a certain range (say, from $2$ to $4$), and that will allow you to place a constraint on $\delta$ that doesn't depend on $x$. (It will depend on $\varepsilon$, but not $x$.) – Brian Tung Feb 16 '18 at 23:40
  • thanks guys that's what i thought :) – Prog Kid Feb 16 '18 at 23:42

2 Answers2

4

No, $\delta$ cannot depend on $x$. What you need to do is use the "when $x$ is close" part. If $|x-3|<\delta$, then $3-\delta<x<3+\delta$. So if you don't allow $\delta$ to be big (you don't want that), then $x$ is bounded. So, for instance you can take $$ \delta=\min\left\{1,\frac{\varepsilon}7\right\}. $$ where the $7$ comes from the need to bound $|x+3|$ (see below). Now, if $|x-3|<\delta$, then \begin{align} |x^2-9|&=|x-3|\,|x+3|\leq\delta\,|x+3|\leq\delta\,(|x|+3)\leq\delta(3+\delta+3)\\ \ \\ &=\delta(6+\delta)\leq7\delta\leq\varepsilon. \end{align}

Martin Argerami
  • 205,756
2

You need to find a $\delta$ which is independent of $x$ and depends only on $\epsilon.$

Note that $0<\delta<1 \implies 2<x<4$, thus $|x+3|<7$

Try $\delta = \epsilon/7.$

Mohammad Riazi-Kermani
  • 68,728