- #1

zenterix

- 475

- 69

- Homework Statement
- I have here a pretty tricky problem from Spivak's Calculus. There is a function that appears for the first time in Chapter 4 on Graphs in problem 19 item v, and then reappears in Chapter 5 on Limits in problem 5.

The function is

##f(x)## = the number obtained by replacing all digits in the decimal

expansion of ##x## which come after the first ##7## (if any) by ##0##.

In Chapter 3 we are asked to "describe as best we can the graph" of this function.

In Chapter 5, we are asked for which numbers ##a## the limit ##\lim\limits_{x\to a}f(x)## exists.

Problem 5 doesn't ask us to prove that the limits exist, however. I would like to know how if someone can provide at least a partial proof that the limit above exists at certain values in the domain.

I have quite a few questions about this interesting function.

- Relevant Equations
- Below you can see a graph of the function as presented in the solution manual

I believe the x-axis is vertical here.

**The graph is composed of**

i)an infinite number of intervals that start on the ##y=x## line and finish at some ##x## with a decimal expansion ending in ##7\bar{9}##. E.g., from ##0.67## to ##0.67\bar{9}## which is considered ##0.68##.

i)

Other examples of intervals would be

$$0.7\text{ to }0.7\bar{9}=0.8$$

$$0.6987\text{ to }0.6987\bar{9}=0.6988$$

$$0.6997\text{ to }0.6997\bar{9}=0.6998$$

**ii)**an infinite number of points on the ##y=x## line; these are points at which ##x## has either no 7's in its decimal expansion or one 7 followed by zeros. ##f## doesn't modify these numbers, as there are no decimal numbers to be removed.

**Note:**For some reason, the solution manual graph has some open intervals, e.g. ##(0.67,0.67\bar{9}]##, ##(0.7,0.7\bar{9}]##. I do not know why this is. For example, take ##x=0.6\bar{9}=0.7##. ##f(0.6\bar{9})=0.6\bar{9}##, so I understand that the point ##(0.6\bar{9},0.6\bar{9})## should be on the graph.

**This would be my first question about the graph.**

Next, I would like to prove that ##\lim\limits_{x\to a}f(x)## exists for any number ##a## that is not on the right (closed) end of one of the aforementioned intervals. These are numbers with a decimal expansion that ends in ##7\bar{9}##.

I am developing my ability to do these proofs, and this one seems quite complicated, I will show my initial work. Please let me know if this route is correct, messy as it may be, and if you see a better way.

**Proposition:**##\lim\limits_{x\to a}f(x)## exists for any number ##a## that is not on the right (closed) end of one of the aforementioned intervals. These are numbers with a decimal expansion that ends in ##7\bar{9}##

**Proof:**

My approach is to consider three cases:

1) ##a##'s decimal expansion contains one digit 7, followed by zeros; ie ##a## is a point on the ##y=x## line but is also part of an interval containing numbers that all have the same value ##f(x)=f(a)##

2) ##a##'s decimal expansion contains at least one digit 7, followed by at least one non-zero digit

3) ##a##'s decimal expansion contains no digit 7, ie ##a## is a single point on the ##y=x## line

__Let me try to prove case 1).__##\forall \epsilon>0##, consider the interval ##A=(a,a+\delta_1)##, ##\delta_1>0##.

Let ##p_1## be the position in which the first digit 7 is located in the decimal expansion of ##a## (tenths is 1, hundredths is 2, etc). Let ##p_2=p_1+1##.

If we choose ##\delta_1=\frac{1}{10^{p_2}}## then ##|x-a|<min(\delta_1, \epsilon) \implies f(x)=a \implies |f(x)-a|=0<\epsilon##

Now consider the interval ##B=(a-\delta_1, a)##.

By assumption, ##a## has one digit 7 followed by zeros.

There does not exist a number ##n<a##, such that there is no number between ##n## and ##a## with no digit 7 in its decimal expansion. In other words, for any number ##n<a##, there is always a number ##n<n_1<a## such that ##n_1## has no digit 7 in its decimal expansion. (I guess I would need to prove this).

Call ##a-n_1=\delta_2<\delta_1##

**What can we now say about ##|f(x)-a|## when ##x \in (a-\delta_2,a)##?**

If ##x## has no digit 7 in its decimal expansion, then ##f(x)=x \implies |f(x)-a|=|x-a|<\delta_2##

If ##x## has one or more digits 7 in its decimal expansion, then

$$a-\delta_2<f(x)\leq x<a \implies -\delta_2<f(x)-a<0<\delta_2\tag{1}$$

$$\implies |f(x)-a|<\delta_2<\delta_1<\epsilon$$

Therefore,

$$0<a-x<\delta_2 \implies |f(x)-a|<\epsilon$$

Also

$$0<x-a<\delta_2 \implies |f(x)-a| <\epsilon$$

$$\implies |x-a|<\delta_2 \implies |f(x)-a| < \epsilon$$

Ie, for ##a## values that satisfy case 1), we have that ##\lim\limits{x \to a} f(x) = a##

Note: in ##(1)## I claimed that ##a-\delta_2<f(x)\leq x##. This is only valid in the very particular case being considered, namely that ##a-\delta_2## has no digit 7's in its decimal expansion, and therefore ##f(a-\delta_2)=a-\delta_2##. For any ##x \in (a-\delta_2,a)##, x either has no digit 7's in its decimal expansion, such that ##f(x)=x>a-\delta_2## or it does have one or more digit 7's in which case ##f(x)## is the value x with zeros after the first digit 7 in the decimal expansion. This must be larger than ##a-\delta_2##, which I think I can prove by contradiction, but I won't do so because this post is too long as it is, maybe separately I can show this.