Spivak, Ch. 5 Limits, Problem 3 viii: Prove a limit of a function

[zenterix]

Homework Statement

I'm doing all the problems in Calculus by Spivak. I am on problem 3 of chapter 5 on Limits, and some of the items don't have solutions. I am interested in two particular items of this problem, $vii$ and $viii$

Relevant Equations

We are asked to determine the limit $l$ for the given function and value of $a$, and to prove that it is the limit by showing how to find a $\delta$ such that $|f(x)-l|<\epsilon$ for all $x$ satisfying $0<|x-a|<\delta$.

The solution for this problem is not available in the solution manual.

Consider item $vii$, which specifies the function $f(x)=\sqrt{|x|}$ with $a=0$

Case 1: $\forall \epsilon: 0<\epsilon<1$

$$\implies \epsilon^2<\epsilon<1$$
$$|x|<\epsilon^2\implies \sqrt{|x|}<\epsilon$$

Case 2: $\forall \epsilon: 1\leq \epsilon < \infty$

$$\epsilon\leq\epsilon^2 \implies |x|<\epsilon^2\implies \sqrt{|x|}<\epsilon$$

Therefore

$$\forall \epsilon>0,|x-0|<\epsilon^2\implies|\sqrt{|x|}-0|<\epsilon$$

$$\implies \lim_\limits{x\to a} \sqrt{|x|}=0$$

Is this correct?Now consider item $viii$, which specifies the function $f(x)=\sqrt{x}$ near $x=1$

$$|x-1|=|(\sqrt{x})^2-1^2|=|(\sqrt{x}+1)(\sqrt{x}-1)|\leq|\sqrt{x}+1||\sqrt{x}-1|$$

Assume $|\sqrt{x}-1|<1$

$$\iff -1<\sqrt{x}-1<1\implies 1<\sqrt{x}+1<3$$

$$\iff |\sqrt{x}+1|<3$$

$$\iff |x-1|<3|\sqrt{x}-1|<3\epsilon$$

In words: Assume that $f(x)$ is less than 1 unit away from the value $1$. From this follows that $|x-1|<3|\sqrt{x}-1|<3\epsilon$. However, because all the steps involved an $\iff$, we could actually start from assuming $|x-1|<3|\sqrt{x}-1|<3\epsilon$ and reach $|\sqrt{x}-1|<1$.

In any case the conclusion is that

$$\forall \epsilon>0,|x-1|<3\epsilon\implies |\sqrt{x}-1|<min(1,\epsilon)<\epsilon$$

$$\implies \lim_\limits{x \to 1}\sqrt{x}=1$$

Is this correct?

 

[PeroK]

Homework Statement:: I'm doing all the problems in Calculus by Spivak. I am on problem 3 of chapter 5 on Limits, and some of the items don't have solutions. I am interested in two particular items of this problem, $vii$ and $viii$
Relevant Equations:: We are asked to determine the limit $l$ for the given function and value of $a$, and to prove that it is the limit by showing how to find a $\delta$ such that $|f(x)-l|<\epsilon$ for all $x$ satisfying $0<|x-a|<\delta$.

The solution for this problem is not available in the solution manual.

Consider item $vii$, which specifies the function $f(x)=\sqrt{|x|}$ with $a=0$

Case 1: $\forall \epsilon: 0<\epsilon<1$

$$\implies \epsilon^2<\epsilon<1$$
$$|x|<\epsilon^2\implies \sqrt{|x|}<\epsilon$$

Case 2: $\forall \epsilon: 1\leq \epsilon < \infty$

$$\epsilon\leq\epsilon^2 \implies |x|<\epsilon^2\implies \sqrt{|x|}<\epsilon$$

Therefore

$$\forall \epsilon>0,|x-0|<\epsilon^2\implies|\sqrt{|x|}-0|<\epsilon$$

$$\implies \lim_\limits{x\to a} \sqrt{|x|}=0$$

You don't need case 2. For $\epsilon \ge 1$ just use, for example, the $\delta$ you found for $\epsilon = 1/2$.

Note that in general we only need to prove the case $0 < \epsilon < \epsilon_0$ for some $\epsilon_0$. If it works for all such $\epsilon$ it cannot fail for larger $\epsilon$.

 

[PeroK]

Now consider item $viii$, which specifies the function $f(x)=\sqrt{x}$ near $x=1$

$$|x-1|=|(\sqrt{x})^2-1^2|=|(\sqrt{x}+1)(\sqrt{x}-1)|\leq|\sqrt{x}+1||\sqrt{x}-1|$$

Assume $|\sqrt{x}-1|<1$

$$\iff -1<\sqrt{x}-1<1\implies 1<\sqrt{x}+1<3$$

$$\iff |\sqrt{x}+1|<3$$

$$\iff |x-1|<3|\sqrt{x}-1|<3\epsilon$$

In words: Assume that $f(x)$ is less than 1 unit away from the value $1$. From this follows that $|x-1|<3|\sqrt{x}-1|<3\epsilon$. However, because all the steps involved an $\iff$, we could actually start from assuming $|x-1|<3|\sqrt{x}-1|<3\epsilon$ and reach $|\sqrt{x}-1|<1$.

In any case the conclusion is that

$$\forall \epsilon>0,|x-1|<3\epsilon\implies |\sqrt{x}-1|<min(1,\epsilon)<\epsilon$$

$$\implies \lim_\limits{x \to 1}\sqrt{x}=1$$

This doesn't look quite right. There are two approaches:

a) You can use the trick above and prove things for $0 < \epsilon < 1$.

b) You can put the bound on delta: $\delta = min\{ 1, \epsilon/3 \}$.

What you've done looks like supporting calculations for a proof, but you haven't put everything together correctly.

 

[zenterix]

$\forall \epsilon>0$, we are looking for $\delta>0:|x-1|<\delta \implies|\sqrt{x}-1|<\epsilon$

The domain of $f$ is $[0,+\infty)$ so $x\geq 0$.

Consider $0<\delta \leq 1$

$$|x-1|<\delta \implies 1-\delta<x<1+\delta \implies x \geq 0$$

$$|f(x)-1|=|\sqrt{x}-1|=\frac{|\sqrt{x}-1||\sqrt{x}+1|}{|\sqrt{x}+1|}=\frac{|x-1|}{|\sqrt{x}+1|}\leq |x-1|<\delta<1$$

because $\sqrt{x}+1|>1, \forall x$

$$\implies \forall \epsilon >0, \delta<min(1,\epsilon) \implies |x-1|<\delta \implies |\sqrt{x}-1|<|x-1|<\epsilon$$

$$\implies \lim\limits_{x \to 1} \sqrt{x}=1$$

 

[PeroK]

It's more usual in these proofs to do it like this:

Let $\epsilon > 0$ and choose $\delta = min(1, \epsilon)$. Then $0 < |x - 1| < \delta \ \Rightarrow \ |\sqrt x - 1| < \epsilon$. Hence $\lim_{x \rightarrow 1} \sqrt x = 1$.

Note that it's $0 < |x - 1| < \delta$ and not $|x - 1| < \delta$. Do you know why?

Also, something like this:

$$\implies \forall \epsilon >0, \delta<min(1,\epsilon) \implies |x-1|<\delta \implies |\sqrt{x}-1|<|x-1|<\epsilon$$

$$\implies \lim\limits_{x \to 1} \sqrt{x}=1$$

Is not a valid set of implications. Do you see why?

 

[zenterix]

Note that it's 0<|x−1|<δ and not |x−1|<δ. Do you know why?

In general, if we write $|x-1|<\delta$ this means that $1-\delta<x<1+\delta$.

If $\delta>1$ then $x$ can be negative.

If we write $0<\delta<min(1,\epsilon)$ this implies that $x\geq 0$ (as I showed in the previous post).

Indeed I forgot to put in the $0<$ part. If we write $0<\delta<min(1,\epsilon)$, then we don't have to write $0<|x-1|<\delta$, we can write just $|x-1|<\delta$, correct?

Is not a valid set of implications. Do you see why?

Let me try to rewrite it:

$$\forall \epsilon>0, \exists \delta : 0<\delta<min(1,\epsilon) \implies (|x-1|<\delta \implies |\sqrt{x}-1|<\epsilon)$$

Is this correct? The part that seemed fishy was the part around the specification of $\delta$. I now specify that "for any $\epsilon>0$ I can choose a $\delta$ such that...", ie "there exists a $\delta$ such that...". I am not sure, though, if it was in fact literally incorrect before? Please tell me where it was literally incorrect.

 

[PeroK]

In general, if we write $|x-1|<\delta$ this means that $1-\delta<x<1+\delta$.

If $\delta>1$ then $x$ can be negative.

If we write $0<\delta<min(1,\epsilon)$ this implies that $x\geq 0$ (as I showed in the previous post).

Indeed I forgot to put in the $0<$ part. If we write $0<\delta<min(1,\epsilon)$, then we don't have to write $0<|x-1|<\delta$, we can write just $|x-1|<\delta$, correct?

Okay, it was just a slip.

Let me try to rewrite it:

$$\forall \epsilon>0, \exists \delta : 0<\delta<min(1,\epsilon) \implies (|x-1|<\delta \implies |\sqrt{x}-1|<\epsilon)$$

Is this correct? The part that seemed fishy was the part around the specification of $\delta$. I now specify that "for any $\epsilon>0$ I can choose a $\delta$ such that...", ie "there exists a $\delta$ such that...". I am not sure, though, if it was in fact literally incorrect before? Please tell me where it was literally incorrect.

This is not right. There's a difference between a definition and a proof based on that definition. You shouldn't try to write your proof as though it were a definition.

The proof finds a specific $\delta$. If you really want to write it in those terms it should be:

We have shown that:
$$\forall \epsilon>0: \delta = min(1,\epsilon) \ and \ 0 < |x-1|<\delta \implies |\sqrt{x}-1|<\epsilon$$
Hence $$\lim_{x \rightarrow 1} \sqrt x = 1$$.
I prefer this:

Let $\epsilon > 0$ and choose $\delta = min(1, \epsilon)$. Then $0 < |x - 1| < \delta \ \Rightarrow \ |\sqrt x - 1| < \epsilon$. Hence $\lim_{x \rightarrow 1} \sqrt x = 1$.

Because that's the way mathematics is done - or, at least, the way I do it.

 

[zenterix]

Because that's the way mathematics is done - or, at least, the way I do it.

I've been thinking that maybe I should read a book or take a course on mathematical logic, to really learn to write mathematics correctly (is that the best way to learn the actual "grammar" of math, if I can call it that?).

I actually understand what you mean with the points you brought up. Most of the time when I write the proofs, I have a lingering doubt about the way to write them. It's not the logic part that bugs me for the most part (though sometimes that too, for example when I show a contradiction, I struggle to identify which part of an argument was contradicted; or what the negation of some multi-part mathematical statement is), it's the formalism. I want to go through the formal training of the language used.

Any suggestions? I'm about to take a MOOC on this, in parallel with my current project of finishing the entire Spivak book.

 

[PeroK]

I've been thinking that maybe I should read a book or take a course on mathematical logic, to really learn to write mathematics correctly (is that the best way to learn the actual "grammar" of math, if I can call it that?).

I actually understand what you mean with the points you brought up. Most of the time when I write the proofs, I have a lingering doubt about the way to write them. It's not the logic part that bugs me for the most part (though sometimes that too, for example when I show a contradiction, I struggle to identify which part of an argument was contradicted; or what the negation of some multi-part mathematical statement is), it's the formalism. I want to go through the formal training of the language used.

Any suggestions? I'm about to take a MOOC on this, in parallel with my current project of finishing the entire Spivak book.

It's just practice. Formal mathematical logic might take you further into symbolic manipulation. Three important things, IMO, are:

1) Being able to interpret formal mathematical statements and definitions. In particular, to really understand what it is you have to prove.

2) Being able to manage the flow of logic in a proof and focus on the core argument.

3) Being able to construct counterexamples.

Note that there is a difference between developing a proof and what you eventually write down.

In this thread, for eaxmple. What you want to prove is:
$$\forall \epsilon > 0: \ \exists \delta > 0 \dots$$
What you have to do to prove that is: Let $\epsilon > 0$ and find $\delta$ (which probably but not always depends on $\epsilon$) ...

Your work is far from bad, so you just need to practise as much as possible. And sharpen everything up a bit.

 

[zenterix]

Note that it's 0<|x−1|<δ and not |x−1|<δ. Do you know why?

Just to be extra precise about the absolute value issue. Isn't it always implicit that there is the $0<$ portion? Absolute value is always non-negative, by definition. To make sure $x \geq 0$ it is only necessary to write

$$|x-1|<\delta\leq 1$$

 

[PeroK]

Just to be extra precise about the absolute value issue. Isn't it always implicit that there is the $0<$ portion? Absolute value is always non-negative, by definition. To make sure $x \geq 0$ it is only necessary to write

$$|x-1|<\delta\leq 1$$

In general, you must avoid the limit point $x$ itself. The limit must be independent of the function value at that point. In this case, you must avoid $x = 1$.

 

[zenterix]

It seems to be tricky to show that $\lim\limits_{x \to a} \sqrt{x}=\sqrt{a}$ for a general $a$.

More specifically, I think I can prove it for $a \geq 1$, but having trouble with $0 < a < 1$.

The proof I gave for $a=1$ is valid for $a \geq 1$ with slight modifications:

Let $\epsilon>0$, choose a $\delta$ such that $0<\delta<a$ and $0<\delta<\epsilon$

Then,

$$|x-a|<\delta<a \implies -a<-\delta<x-a<\delta<a \implies 0<a-\delta<x<a+\delta \implies x>0$$

$$|f(x)-l|=|\sqrt{x}-\sqrt{a}|=\frac{|\sqrt{x}-\sqrt{a}||\sqrt{x}+\sqrt{a}|}{|\sqrt{x}+\sqrt{a}|}=\frac{|x-a|}{|\sqrt{x}+\sqrt{a}|}$$

If $a \geq 1$ then $|\sqrt{x}+\sqrt{a}| \geq 1$

$$\implies |\sqrt{x}-\sqrt{a}|\leq |x-a|<\delta<\epsilon$$

Therefore, we've shown that:

$$\forall \epsilon>0:0<\delta<\epsilon\text{ and }\delta<a,|x-a|<\delta\implies |\sqrt{x}-\sqrt{a}|<\epsilon$$

$$\implies \lim\limits_{x \to a} \sqrt{x}=\sqrt{a}, a \geq 1$$

The question now is what happens if $a < 1$?

 

[PeroK]

What about, for $a \ne 0$, $\delta = min(a, \epsilon \sqrt a)$?

 

[zenterix]

What about, for a≠0, δ=min(a,ϵa)?

I didn't succeed with this suggestion.

Here is an alternative way I found:

Let

(i) $0<a<1$
(ii) $\delta<min(a,1-a) \leq 0.5$.

$$|x-a|<\delta \implies a-\delta<x<a+\delta$$
$$0<a\leq 0.5 \implies \delta < a \implies 0<x<2a<1$$
$$0.5 \leq a < 1 \implies \delta < 1-a \implies 0<2a-1<x<1$$

$$|x-a|=|(\sqrt{x}+\sqrt{a})(\sqrt{x}-\sqrt{a})|=|(\sqrt{x}+\sqrt{a})||(\sqrt{x}-\sqrt{a})|<\delta$$
$$\implies |(\sqrt{x}-\sqrt{a})|<\frac{\delta}{(\sqrt{x}+\sqrt{a})}<\frac{\delta}{x+a}<\frac{\delta}{2a-\delta}$$

where I have used the following
$$0<x<1 \implies x<\sqrt{1}$$
$$0<a<1\implies a<\sqrt{a}$$
$$a-\delta<x<a+\delta\implies 2a-\delta<x+a<2a+\delta$$

We've shown that if $0<a<1$ and $\delta<min(a,1-a)$ and $|x-a|<\delta$ then $|\sqrt{x}-\sqrt{a}|<\frac{\delta}{2a-\delta}$.

Therefore, $\forall \epsilon>0$, we can choose a $\delta$ such that $\frac{\delta}{2a-\delta}\leq \epsilon \implies \delta \leq \frac{2a\epsilon}{1+\epsilon}$.

$$\delta< min(\frac{2a\epsilon}{1+\epsilon},min(a,1-a)), |x-a| < \delta \implies |\sqrt{x}-\sqrt{a}|<\frac{\delta}{2a-\delta}<\epsilon$$

Thus, $\lim\limits_{x \to a} \sqrt{x}=\sqrt{a}$, $0<a<1$

 

[PeroK]

It isn't that complicated.
$$|\sqrt x - \sqrt a| = \frac{|x -a|}{\sqrt x + \sqrt a} < \frac{|x -a|}{\sqrt a}$$

 

[zenterix]

If $0<a<1$, then $\forall \epsilon>0$ choose $\delta = \sqrt{a}\epsilon$. Then $|x-a|<min(a,\delta) \implies |\sqrt{x}-\sqrt{a}|<\epsilon$

My god, so simple.

 

[PeroK]

If $0<a$, then $\forall \epsilon>0$ choose $\delta = \sqrt{a}\epsilon$. Then $|x-a|<min(a,\delta) \implies |\sqrt{x}-\sqrt{a}|<\epsilon$

That works, in fact, for all $a > 0$.