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Prove $\lim_{x \to 13} \sqrt{x-4} = 3$.

We need to show for all $E> 0$ there exists $D > 0$ such that
if $0 < |x - 13| < D$, then $|\sqrt{x-4} - 3| < E$. Let me write D for delta and E for epsilon please.

Scratch-work here: $\color{darkred}{\text{ I understand the formal proof. Ergo just asking about this. } }$

Note that $|\sqrt{x-4} - 3| = |\sqrt{x-4} - 3|\dfrac{|\sqrt{x-4} + 3|}{|\sqrt{x-4} + 3|} = \dfrac{|x - 13|}{|\sqrt{x-4} + 3|}$.
We can bound $|x - 13|$ for any choice of D, but we need a certain D to also bound $|\sqrt{x-4} + 3|$.

1. Ishfaaq's answer says 'the denominator will definitely cause trouble.' How?
Why won't bounding $|x - a|, |x - 13|$ bound the whole statements?
Beneath pp 83 of Spivak claims this too. How does '$|x + a|$ cause trouble'?

Assume that $D < 1.$

2. What sanctioned assuming $D < 1$? How to know if $1$ is too small, too big? $\\$ In pp 83 of Spivak beneath, $D$ looks randomly chosen?

4. Ishfaaq's last paragraph. $\color{darkred}{\left|{x^2 - a^2}\right| \lt \delta^2 + \left|{}2a\right|\delta. \text{ But we can hardly equate this to ϵ... }}$
Why not? $d^2 + \left|{}2a\right|d = e \iff d(d + |2a|) = e$?

Ishfaaq says $a = 1/2, e = d/2$ is a counterexample. But then $d^2 + 2|a|d < e \iff d^2 + d < d/2 \iff d(d - \frac{1}{2}) < 0 \iff 0 < d < 1/2.$ What founders?

If this works, then any smaller D will also work. Then $|x - 13| < 1 \iff -1 < x - 13 < 1 \iff 12 < x < 14 \iff 8 < x - 4 < 10$
$\implies \sqrt(8) < \sqrt{x - 4} < \sqrt(10) \iff \color{blue}{\sqrt(8) + 3 < \sqrt{x - 4} + 3 < \sqrt(10) + 3} \\ \iff 0 < \color{blue}{\dfrac{1}{\sqrt(10) + 3} < \dfrac{1}{\sqrt{x - 4} + 3} < \dfrac{1}{\sqrt(8) + 3}} $ $\implies \color{green}{\frac{1}{|\sqrt{x - 4} + 3|} < \frac{1}{\sqrt(8) + 3}} $

So then $|x - 13|\color{green}{\dfrac{1}{|\sqrt{x - 4} + 3} < \dfrac{1}{\sqrt(8) + 3}}|x - 13|$ and need this $ < E.$
Thus, we need $|x - 13| < E * \color{green}{(\sqrt{8} + 3)}$. So this is our choice of D. But note that this only works when $D < 1$. Thus, we can take care of both conditions by choosing $D = \min\{1, E(\sqrt{8}+3)\}.$ █

$\color{darkred}{ \text{ 3. All this algebra fazed me. How to graph this to view all this algebra and $D$? Thanks. } }$

enter image description here

Also tried http://www.ocf.berkeley.edu/~yosenl/math/epsilon-delta.pdf.

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Note that $\sqrt{x-4}+3\geqslant3$ hence $|\sqrt{x-4}-3|\leqslant\frac13|x-13|$. Thus, if $|x-13|\leqslant3\varepsilon$ then $|\sqrt{x-4}-3|\leqslant\varepsilon$.

Note that the function $x\mapsto\sqrt{x-4}$ is only defined on $x\geqslant4$ hence the condition $|x-13|\leqslant3\varepsilon$ guarantees that $\sqrt{x-4}$ exists, provided $\varepsilon\leqslant3$ (which is more than enough to conclude). A formula valid for every $\varepsilon$ would be $\delta=\min\{9,3\varepsilon\}$.

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Well the aim is to come up with a $\delta \ $ for any given $\epsilon$, however small. Given an $\epsilon$ - whatsoever - we must provide a corresponding positive quantity $\delta$ such that $\left| \sqrt{x - 4} - 3 \right| \lt \epsilon $ whenever $\left| x - 13 \right| \lt \delta. $

Now, almost always this task is overcome by proving there is a $\delta$ of the form $a \cdot \epsilon$. So that regardless of what we are given for $\epsilon$, the corresponding value for $\delta$ will exist since it is just the multiple of $\epsilon$ times another positive value $a$.

Now, our aim is to bound $\left| x - 13 \right|$ and thereby arrive at an implication for $\left| \sqrt{x - 4} - 3 \right| $. And can you see that the denominator will definitely cause trouble? If we don't state with evidence that $1 /\left| \sqrt{x - 4} + 3 \right|$ will not misbehave - that is adversely affect our inequality - we are providing an insufficient answer.

The choice of $1$ is to make things easy. As in arrive at a multiple of $\epsilon$ for our $\delta$. Try to find a bound for $1 /\left| \sqrt{x - 4} + 3 \right|$ by starting from $\left| x - 13 \right| \lt \delta$ in a general way. You will obtain a bound. But you will be left with terms like $\delta ^2$ and $\sqrt\delta$. This is what we intend to avoid. And if you did what I said you will notice in this case that we will not get anywhere. It doesn't always have to be $1$. It could be $3$ like Did has suggested above. It may seem a little from heaven but it is a trick you will learn from answering many questions.

The example on Spivak could be more enlightening. I'm assuming you have learnt your triangle inequalities from the previous pages.

$\begin{align} \left|{x^2 - a^2}\right| = \left|{x - a}\right|\left|{x + a}\right| & = \left|{x - a}\right|\left|{(x - a) + 2a}\right| \\ & \le \left|{x - a}\right| (\left|{x - a}\right| + \left|{2a}\right|) = \left|{x - a}\right|^2 + \left|{2a}\right|\left|{x - a}\right| \end{align}$

So if we try to begin with $\left|{x - a}\right| \lt \delta$ we will end up with $\left|{x^2 - a^2}\right| \lt \delta^2 + \left|{}2a\right|\delta$. But we can hardly equate this to $\epsilon$ which is why we use the trick to say choose $\delta =\text{ Min} \{1 , \epsilon \}$ because then we can say $\delta \le 1 \implies \delta^2 \le \delta \implies \delta^2 + \left|{}2a\right|\delta \le (\left|{}2a\right| + 1)\delta$ where $(\left|{}2a\right| + 1)$ is but a positive constant. So there corresponds a $\delta = \dfrac {\epsilon}{(\left|{}2a\right| + 1)}$ for any given $\epsilon \gt 0$.

UPDATE: Right. Fair enough. Here goes.

Well the simplest reason why $|x + a|$ and the denominator in your question causes trouble is because it has $x$ in it and hence bounding $|x - a|$ or $|x - 13|$ (which is what we are authorised to do) will not bound the whole statements, $|x^2 - a^2|$ and $\dfrac {|x - 13|}{|\sqrt{x - 4} + 3|}$.

The only quantity we are allowed to bound is the distance between $x$ and and the point at which the limit is evaluated i.e. $|x - a|$ and $|x - 13|$. You need to see why though. This is directly due to the definition. "For any given $E \gt 0$ there exists $D \gt 0$ such that $|f(x) - L| \lt E $ whenever $ |x - a| \lt D$". Our job is to provide $D$ but for any $E$ whatsoever and however small. But since $D$ is ours we can make it as small as we wish. This is what justifies the assertion "Suppose $D \lt 1$". This need not be $1$. You can start with saying suppose $D \lt \frac 1 {10^{100}}$. Wouldn't matter. But choosing $D$ to be smaller than $1$ allows us to say that $D^2 \lt D$ which is very useful and now we come to your third question.

Okay maybe my wording was not good enough there. Yeah I guess you are right. You can always ask the ethereal being who supplies us with the $\epsilon$ to pick himself a $\delta$ such that $\delta^2 + 2|a|\delta \lt \epsilon$. But the existence of such a positive value for $\delta$ is not obvious is it? (- although right now I can't think of a solid counter-example. This is a feeble one - Say $a = \frac 1 2$ and $\epsilon = \frac {\delta}{2}$ then what would you pick for $\delta$ so that our inequality is satisfied?).

But the assertion that picking $\delta = \text{Min} \{ 1 , \epsilon \}$ is very foolproof, clear and trivially simple. It conforms to the above quote every time. It proves there is such a $\delta$ for any given godforsaken $\epsilon$. End of story.

Hope I helped. Not sure if answers this long are allowed. Leave a comment if you need more clarification. Like I said here these concepts are very important. You need a solid base to progress.

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  • $\begingroup$ +Upvote. thanks. I revised my questions. Apprise me please? Answer in answer please? Comments are too small. $\endgroup$ – Analysis Feb 26 '14 at 16:53
  • $\begingroup$ @Tucker Rapu: Read the update. Lemme know. $\endgroup$ – Ishfaaq Feb 27 '14 at 11:13
  • $\begingroup$ I upvote your other questions because can only upvote once. thanks. I revised questions 1 and 4. Apprise me please? Answer in answer please? Comments are too small. $\endgroup$ – Analysis Feb 27 '14 at 12:06
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    $\begingroup$ @Tucker Rapu: Upvoting is not necessary. And frankly is unethical if you ask me. I'll help you as much as I can but you are going to have to grasp this for yourself in the end with some self-exploration. Right tell me how does $\left|{x - a}\right| \lt d \implies \left|{x^2 - a^2}\right| \lt e $ for some $e$ and also tell me if you can prove that given any $e \gt 0$ there exists $d \gt 0$ such that $d^2 + 2\left|{ a }\right|d \le e$. Counterexample is actually, $a \neq 0$ and $ e = d^2$ $\endgroup$ – Ishfaaq Feb 27 '14 at 12:19
  • $\begingroup$ @Ishfaaq: I apprise Ishfaaq here of my beliefs here, becuse my post is long and I can't edit comments, which I need to do because I'm horrible at latex. $|x - a| < d \iff -d < x - a < d \iff -d + a < x < d +a$ $\iff -d + 2a < \color{seagreen}{x + a < d + 2a \implies |x + a| < d + 2a}$. Ergo $|x^2 - a^2| = |x - a|\color{seagreen}{|x + a|} < d\color{seagreen}{(d + 2a)} $. Ergo choose $e = d\color{seagreen}{(d + 2a)}$. Is this unsullied? $\endgroup$ – Analysis Mar 6 '14 at 16:51
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If you found the algebra a bit hard to decipher, maybe this will help.

Assume that $|x-13|<1$ just means that we assume that the distance between $x$ and 13 is less than one. $x$ is the input to your function, it's distance from the number 13 is less than 1. If you want to compute your limit, i.e. what happens to $f(x)$ as $x\rightarrow13$, the distance between $x$ and 13 is going to be less than 1 eventually anyway, so this is not an unreasonable thing to do.

$|x-13|<1\Longleftrightarrow-1<x-13<1$ These two statements mean the same thing. It's easier to work with the equation to the right.

You treat $-1<x-13<1$ like an ordinary equation except instead of doing the same thing to both sides of the equality sign you do the same thing to the left, in the middle and to the right.

$\begin{array}{ccccc} -1 & < & x-13 & < & 1\\ -1+13 & < & x-13+13 & < & 1+13\\ 12 & < & x & < & 14\\ 12-4 & < & x-4 & < & 14-4\\ 8 & < & x-4 & < & 10\\ \sqrt{8} & < & \sqrt{x-4} & < & \sqrt{10}\\ \sqrt{8}+3 & < & \sqrt{x-4}+3 & < & \sqrt{10}+3\\ & \text{flip} & & \text{flip}\\ \frac{1}{\sqrt{8}+3} & > & \frac{1}{\sqrt{x-4}+3} & > & \frac{1}{\sqrt{10}+3}\\ \end{array}$

All these moves are legal. What we have so far is that if the length between your input $x$ and 13 is less than one, i.e $|x-13|<1$, then $\frac{1}{|\sqrt{x-4}+3|} < \frac{1}{\sqrt{8}+3}$.

Now comes the epsilon-delta part. Remember that the limit you want to find is basically the same as asking "what happens to $f(x)$ when my input $x$ gets really close to 13", so let's consider the case where $x$'s distance from 13 is less than 1 (this has to happen eventually).

When $|x-13|<1$, we know from before that $|\sqrt{x-4}-3|=|x-13|\frac{1}{|\sqrt{x-4}+3|}$, so now we have $|\sqrt{x-4}-3|=\text{something less than 1}\cdot\text{something less than }\frac{1}{\sqrt{8}+3}<\frac{1}{\sqrt{8}+3}$

Let's consider when $|x-13|<(\sqrt{8}+3)\epsilon$:

We know from before that $|\sqrt{x-4}-3|=|x-13|\frac{1}{|\sqrt{x-4}+3|}$, so now we have $|\sqrt{x-4}-3|=\text{something less than }(\sqrt{8}+3)\epsilon\cdot\text{something less than }\frac{1}{\sqrt{8}+3}<(\sqrt{8}+3)\epsilon \frac{1}{\sqrt{8}+3}=\epsilon$

Now we choose $\delta=\min\{1,(\sqrt{8}+3)\epsilon\}$. Why? Because there is a flaw in the argument above! What if $\epsilon$ is greater than 1? Then we can't say that $\frac{1}{|\sqrt{x-4}+3|} < \frac{1}{\sqrt{8}+3}$ - but wait! if $\epsilon$ is greater than 1, we already have a $\delta$ that will satisfy our definition: 1.

Example:

If $\epsilon=2.2$, can you find a $\delta$ such that $|x-13|<\delta \Rightarrow |f(x)-3|<2.2?$ Yes, $|x-13|<1$, then $|\sqrt{x-4}-3|<\frac{1}{\sqrt{8}+3}\approx 0.17$

When $\epsilon$ is less than 1, then we know that $\frac{1}{\sqrt{x-4}+3} < \frac{1}{\sqrt{8}+3}$ and we use the $\delta$ we derived using the fact that $|x-13|<1$, namely $\delta=(\sqrt{8}+3)\epsilon$

Think about after $\epsilon=2.2$, when someone tries to pass me an $\epsilon$ like 0.001 so that $(\sqrt{8}+3)\epsilon<1$, then having chosen $|x-13|<\delta=\min\{1,(\sqrt{8}+3)\epsilon\}$, I can repeat my whole argument about how when $|x-13|<1$, then blah blah blah. I have a winning strategy against any player who tries to break me by passing me tiny or large epsilons.

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In Spivak (assuming for the moment $a\neq 0$), when $x$ is close to $a$ then $x$ and $a$ will have the same sign, so |x+a| will be larger than $a$. So there is a (relatively) large factor appearing in a product you want to make small - certainly you want to be able to choose $\epsilon$ much smaller than $a$.

So you have to find some way to control the large factor so it doesn't get in the way of your proof. In Spivak's example, this is easy, but in more complex cases it can be the key to a solution. It is worth noting the idea now, because it will come in useful later.

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