# Can there be more than one proof for the limit as x approaches 3 of x squared equal 9?

Can there be more than one proof for this question? An answer has been provided here and I can see that proof is valid: https://www.physicsforums.com/threads/prove-that-limit-as-x-approaches-three-of-x-2-is-equal-to-9.704850/ but I want to know if the following alternate proof is valid too:

0 < |x−3|<δ then

|x−3||x+3|<ε |x−3| < ε/|x+3| thus (preliminary assumption) δ=ε/|x+3|

Proof (substituting δ|x+3|=ε) :

|x−3||x+3|<δ|x+3|

|x−3||x+3|<(ε/|x+3|)*|x+3|

thus

|x−3||x+3|<ε

• You could always do proof by contradiction or contrapositive, although they are logically equivalent to what you have done. – graydad Oct 23 '14 at 19:26
• No, $\delta$ should depend on $\epsilon$ only and not $x$ as well. – Adriano Oct 23 '14 at 19:40

$\delta$ can't depend on $x$!

Assume that $|x-3|<\delta$, where $\delta$ is to be fixed, and is small enough so that $$|x-3|<\delta \implies 0<x<6$$ (so, $\delta < 3$).

Then using what you wrote in your message, when $|x−3| < \delta$: $$|x−3||x+3| \le \delta(x+3)\le 6\delta$$

so taking $\delta = \min(\epsilon / 6,3)$ is convenient. Note that $\delta$ does depend only on $\epsilon$. This is the sightliest modification one can do to your proof to make it correct.

• I don't see why $x+3\le3$. – user84413 Oct 23 '14 at 19:59
• Thanks - this is a nice solution. – user84413 Oct 23 '14 at 20:06
• @user84413 thanks again for the feedback! – mookid Oct 23 '14 at 20:19
• Thank you for that. I have one additional question regarding your solution. I understand intuitively that assuming this : |x−3|<δ⟹0<x<6 (so, δ<3) but I want to know algebraically is that because |x−3|<δ becomes -δ+3<x<δ+3? – Eric Oct 24 '14 at 18:44
• indeed.${{{}}}$ – mookid Oct 24 '14 at 22:50