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$$\lim_{x\to 3} x^2 = 9$$

Hi, I'm new to this community(and to calculus) and english isn't my first language, so I'm sorry if I'm not able to properly convey this problem I have. If I'm doing anything wrong, could someone please correct me?

So, I know there has to be a $\delta>0$ for all $\epsilon>0$ given so that $0<|x-3|<\delta \implies |x^2-9|<\epsilon$

And I also know that $|x^2-9|<\epsilon$ can be written as $|x-3||x+3|<\epsilon$

The author(Stewart) used this method before, but I'm kind of confused...

First, he used a positive constant $C$ so that $|x+3|<C$ I'd imagine that he's trying to "restrict" $|x+3|$ by restricting the interval... $|x-3|<1 \implies 2<x<4 \implies 5<x+3<7$

Then he chose $C = 7$, which I guess comes from the fact that it's the upper bound of x+3, so that $|x-3||x+3| < 7|x-3| < \epsilon \implies |x-3| < \frac{\epsilon}{7}$

And that $\delta = \min(1,\frac{\epsilon}{7})$ so that if $\epsilon > 7$, it still works.

I think I understand each individual part... But, for some reason, I feel something is missing. Why is that? Even after doing most of the exercises, this still feels alien to me. Maybe I should try reading another book as well? I've bought Spivak book(and it might take a while for it to arrive), does it teaches the $\epsilon - \delta$ differently?

What should I do so that I get comfortable with it?

Thanks in advance!

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Any positive constant $C$ works for the proof you want to show. First, he set $\delta = 1$. Then, by algebra, we have $2 < x < 4$ and $5 < x + 3 < 7$. Finally, play around with $|x^2 - 9| < \varepsilon$ and set $\delta$, the minimum. –  NasuSama Feb 23 at 15:52
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And Yes! By all means do not stick to Stewart. Please read Spivak. –  Ishfaaq Feb 23 at 16:38
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I don't know what you feel like you're missing. When doing these kinds of exercises, I do think that it's important to pay special attention to what the delta depends on, i.e. convergence versus uniform convergence. Spivak is good book. Professor Su at Harvey Mudd College also has some amazing calculus lectures (on Youtube and on an interactive site) that really help with understanding. –  rocinante Feb 23 at 16:45

2 Answers 2

up vote 4 down vote accepted

Here's another method that might get you understanding the problem better (one almost directly copied from Spivak).

$$|x^2 - 9| = |x - 3||x + 3| = |x - 3||(x - 3) + 6| \le |x - 3|(|x - 3| + |6|) = |x - 3|^2 + 6 |x - 3|$$

Now if we chose $\delta \le 1$ then $\delta^2 \le \delta$ and $\delta ^2 + 6\delta \le 7\delta. $.

Now since we know that $|x - 3| \lt \delta$, $ |x - 3|^2 + 6 |x - 3| \lt \delta^2 + 6\delta \le 7\delta$. So if we chose $\delta = \frac {\epsilon}{7} $ we are through. But not so fast.

Since we are the ones supplying $\delta$ we can make it as small as we please. But we also have to account for given $\epsilon$ values which can be extremely small too (well below $1$). Hence we cannot get away with saying choose $\delta \le 1$. We must also account for values like $\epsilon = \frac 1 {10^9}$. The way to get around that is to say

$$\delta = \text{Min} \{ 1 , \frac {\epsilon}{7}\}$$

If the given $\epsilon$ is such that $ \frac {\epsilon}{7} \gt 1$ then we choose any quantity less than or equal to $1$. If not we choose $\delta = \frac {\epsilon}{7}$ which also satisfies the condition $\delta \le 1$ so all our arguments hold. Hope thiis helped.

Now, I am no expert but I did not enjoy reading Stewart during my first year. Yes it is a standard Text but that's the problem. Spivak is a more independent book and teaches you mathematics as opposed to sticking to a curriculum. It is a great book to build a foundation in analysis and logic. So make sure you read through the chapters on inequalities and limits in Spivak and keep working exercises. All of this isn't too hard to conquer. Just needs some time.

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There are three roughly independent big ideas going on here.


The first big idea is that of trying to understand the behavior of $x^2$ near the point $x=3$. There are a number of different paths there, and a number of different ways to actually write a precise statement. The one I'm going to use is to note the exact representation

$$ x^2 = 9 + (x+3) (x-3) $$

and the approximate representation

$$ x^2 \approx 9 + 6 (x-3) $$

The relevant aspect we're going to use for the proof is that the value at $3$ is $9$. The predominant description of how the function varies near $3$ is that the relative size of $x-3$ varies dramatically, but $x+3$ just hovers near $6$ without there being much relative error in that approximation.

So, as $x \to 3$, the convergence happens because the $x-3$ factor vanishes, while the $x+3$ factor just mucks around near $6$, rather than doing something (like diverging to $+\infty$) to counteract the vanishing of $x-3$.

Incidentally, once you get this proof down, it would be a useful exercise to try to devise a different proof based on the representation

$$ x^2 = 9 + 6(x-3) + (x-3)^2 $$


The second big idea is simply the mechanics of how $\epsilon-\delta$ proofs work and what they express. IMO, the predominant obstacle here is simply adapting to the necessary thought structure.

However, one key idea that we will use is that if some value of $\delta$ forces "good enough" approximations, then making $\delta$ even smaller will ensure that your approximations are still good enough.


The third big idea is the methods of translating intuition to arithmetic. Recall the key ideas were that $x-3$ vanishes as $x \to 3$, but $x+3$ doesn't grow large. The size of the $x-3$ part is controlled directly by $\delta$, so what we're missing is a technique to express that $x+3$ isn't growing large.

The usual technique is to just pick some big upper bound on $x+3$ -- say, $100$, and find a way to pick $\delta$ to ensure $|x+3| < 100$. For example, insisting that $\delta < 10$ is good enough

A slightly easier approach would be to do it the other way around: pick, say, $\delta < 5$ first, and then find some upper bound on what value $x+3$ can possibly attain. In this case, $|x-3| < 5$ implies $|x+3| < 8$.

So if we always take care to ensure that $\delta \leq 5$, then we can invoke the bound $|x+3| < 10$ whenever we want.


Surprise fourth idea! This idea isn't a prerequisite to understanding, but it's one that often doesn't get well-emphasized, and people often make problems a lot harder than they need to because they don't quite get this idea. Thus, I felt I should mention it.

Notice how loose I was with my bounds in the previous section. $\delta < 10$ tells us something much stronger than merely $|x+3| < 100$. And $\delta < 5$ was good enough to get $|x+3| < 8$, so why did I say $|x+3| < 10$?

The point is that you don't actually need very good bounds at all -- the only important fact about our bound on $|x+3|$ is that it doesn't grow to $+\infty$. The differences between $100$, $10$, $8$, $7$, or even $6.0001$ are completely insignificant when compared to the thing we actually need to show.

When literally anything will do, the best approach is to do something that is quick, simple, and easy to work with. Don't bother with trying to do a more careful analysis to look for the "best" bounds unless you're working a problem when such things are actually useful.


If you are uncomfortable with any one of the three ideas, then the overall proof is likely to make you uncomfortable, no matter how comfortable you are with the other two parts -- my advice is to try and pin down precisely which aspect of the problem makes you uncomfortable

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+ 1 million internets! This is why my favorite analysis book is the surprisingly unpopular one by Gaskill and Narayanswami. They go through explanations exactly like you did above and have a much more "human" approach to explaining the gritty details Spivak, Rudin, etc. expect students to figure out on their own. –  rocinante Feb 23 at 18:06
    
Thank you, that was very helpful –  ZeroCreativityForNames Feb 23 at 18:15

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