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$\lim \limits_{x \to 3} x^2 = 9$

Very simple limit, but I am trying a "new" (for me) method:

by definition:

$\left(\lim \limits_{x \to 3} x^2 = 9\right) \quad = \left(\lim \limits_{x \to 3^+} x^2 = 9\right) \wedge \left(\lim \limits_{x \to 3^-} x^2 = 9\right)$

For either conjunct, if I can show that there exists a function $g: \mathbb{R} \to \mathbb{R}$ such that $g(\epsilon) = \delta$, where $|x - 3| \leq \delta \implies |x^2 - 9| \leq \epsilon$, then I will have proven that the limit exists, as for any given $\epsilon$ I can produce a $\delta$ that satisfies the limit $definition.

$\lim \limits_{x \to 3^+} x^2 = 9$ case:

Since we are approaching 3 from the right, we have that $x > 3$.

$\quad (x > 3) \wedge (|x^2 - 9| \leq \epsilon)$ ...can assume left conjunct and use definition of absolute value

$ = (x > 3) \wedge (x^2 - 9 \leq \epsilon)$ ...algebra...

$ = (x > 3) \wedge (x^2 \leq \epsilon + 9)$ ...law of specialization...

$ \implies x^2 \leq \epsilon + 9$ (CONDITION 1)

We also have:

$\quad x - 3 \leq \delta$ ...algebra...

$= x^2 \leq \delta^2 + 6\delta + 9$ (CONDITION 2)

So now we have to conditions on $x^2$ that must be fulfilled when approaching $3$ from the right in order to satisfy the limit definitions. Recall that we want to find a function relating $\epsilon$ and $\delta$ when approaching $3$ from the right. Let us equate the right hand sides of these conditions, because then we will continue to satisfy condition 1 and 2, while coming up with an $\epsilon$-$\delta$ relationship:

$\quad \epsilon + 9 = \delta^2 + 6\delta + 9$...algebra...

$= (0 = \delta^2 + 6\delta - \epsilon)$

I can now use the quadratic equation to find a meaningful relationship:

$g(\epsilon) = \delta = -3 + \sqrt{9 + \epsilon}$

$\lim \limits_{x \to 3^-} x^2 = 9$ case:

Since we are approaching 3 from the left, we have that $x < 3$. Additionally, I want to put in the condition that $x \geq 0$.

$\quad (x \geq 0) \wedge (x < 3) \wedge (-x^2 + 9 \leq \epsilon)$ ...can assume left conjuncts and use definition of absolute value

$ = (x \geq 0) \wedge (x < 3) \wedge (x^2 + 9 \leq \epsilon)$ ...algebra...

$ = (x \geq 0) \wedge (x > 3) \wedge (x^2 \geq \epsilon - 9)$ ...law of specialization...

$ \implies x^2 \geq \epsilon - 9$ (CONDITION 1)

We also have:

$\quad 3 - x \leq \delta$ ...algebra...

$= x^2 \leq \delta^2 + 6x - 9$ (CONDITION 2)

So now we have to conditions on $x^2$ that must be fulfilled when approaching $3$ from the left in order to satisfy the limit definitions. Recall that we want to find a function relating $\epsilon$ and $\delta$ when approaching $3$ from the left. Let us equate the right hand sides of these conditions, because then we will continue to satisfy condition 1 and 2, while coming up with an $\epsilon$-$\delta$ relationship:

$\quad \epsilon - 9 = \delta^2 + 6\delta - 9$...algebra...

$= (0 = \delta^2 + 6\delta - \epsilon)$

$g(\epsilon) = \delta = -3 + \sqrt{9 + \epsilon}$

So, does this method work?

bzm3r
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1 Answers1

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Looking quickly, it does seems to work, but.. my gosh, all that calculations must have been painful. There's an easier way. Given $\epsilon > 0$, I want $\delta > 0$ such that: $$0 < |x - 3| < \delta \implies |x^2 - 9| < \epsilon$$ First, see that once we've found the $\delta$ that does the job, any $\delta' < \delta$ will do too. So, we can suppose $\delta < 1$. I choose $1$ for simplicity, it could be another number. Fine. Since: $$|x| - |3| < |x - 3| < \delta < 1$$ we get $|x| < 4$, nice. So, we get: $$|x^2 - 9| = |x + 3||x-3| \leq (|x| + 3|)\delta < (4 + 3)\delta$$ by using the bound on $|x|$ and the triangle inequality. So, this way, $\delta = \min\{1, \epsilon/7\}$ does the job. In this question I gave the general method to dealing with $\epsilon-\delta$ proofs for polynomials, you might find it helpful.

Community
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Ivo Terek
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  • The calculations weren't painful, rather, I seem to have a method which will almost always give me a straightforward way to prove limits (and not just for polynomials...), without having to guess too much. So, I am not looking for the "simpler", standard answer, as much as comments on my method. – bzm3r Aug 17 '14 at 03:33
  • I am also curious to know if it is okay to put a caveat on how large an epsilon the "other player" (in the delta-epsilon) game, is allowed to choose. – bzm3r Aug 17 '14 at 03:34
  • The other player can throw in any $\epsilon > 0$, and you'll have to manage to counter it with an appropriate $\delta$. Your solution should not put any restrictions on $\epsilon$. And, with $\delta = -3 + \sqrt{9 + \epsilon}$ this does not happen, it works well. But it would be cleaner to avoid square roots, summing constants, etc, IMO – Ivo Terek Aug 17 '14 at 03:40
  • If you can find a $\delta$ works for $\epsilon = \epsilon_0$, then the same value of $\delta$ will work for all $\epsilon > \epsilon_0$. – JimmyK4542 Aug 17 '14 at 03:55
  • @JimmyK4542 facepalm sometimes I think I am not cut out for math school... – bzm3r Aug 17 '14 at 04:19