Can Derivatives Exist at Discontinuities of a Piecewise Function?

[mathmari]

Hey!

Let $f:[-1,1]\rightarrow \mathbb{R}$ be defined by \begin{equation*}f(x)=\begin{cases}\frac{1}{k} & \text{ for } x\in \left (\frac{1}{k+1}, \frac{1}{k}\right ], \ k \in \mathbb{N}\\ 0 & \text{ for } x=x_0=0 \\ \frac{1}{k} & \text{ for } x\in \left [\frac{1}{k}, \frac{1}{k-1}\right ), \ k \in \mathbb{Z}\setminus \mathbb{N}_0, \ \text{ i.e. } (-k)\in \mathbb{N}\end{cases}\end{equation*}

1. Draw $f$
2. Examine at the points $x_0=0$ and $x_n=\frac{1}{n}, n\in \mathbb{N}$, the existence of
   
   View attachment 7648
   
   where $D=\{x\in (-1,1): f'(x)\text{ exists }\}$. Calculate their values, if they exist.

1. How could we draw that function?
   
   Doesn't it depend on how we choose $k$ ?
   
   (Wondering)
2. Do we not have to check first the existence of $f_+'(x_n)$ and $f_-'(x_n)$ and then it follows if $f'(x_n)$ exist?
   
   We have that $$f_+'(x_n)=\lim_{x\rightarrow x_n^+}\frac{f(x)-f(x_n)}{x-x_n}=\lim_{x\rightarrow \frac{1}{n}^+}\frac{f(x)-\frac{1}{n}}{x-\frac{1}{n}}$$ Does it holds that $\displaystyle{\lim_{x\rightarrow x_n^+}f(x)=\frac{1}{n}}$ ? (Wondering)

 

[I like Serena]

[*] How could we draw that function?

Doesn't it depend on how we choose $k$ ?

Hey mathmari!

I don't think we choose $k$. Instead I think we determine $k$ based on $x$.
Suppose we split the interval up in $(\frac 12,1], (\frac 13, \frac 12], ...$, can we find the function values on each of those intervals? (Wondering)

Or suppose $x\in (\frac 1{k+1}, \frac 1k]$, then in which interval will $\frac 1x$ be?
Can we deduce the corresponding $k$ from that? (Wondering)

[*] Do we not have to check first the existence of $f_+'(x_n)$ and $f_-'(x_n)$ and then it follows if $f'(x_n)$ exist?

Sounds like a plan! (Happy)

We have that $$f_+'(x_n)=\lim_{x\rightarrow x_n^+}\frac{f(x)-f(x_n)}{x-x_n}=\lim_{x\rightarrow \frac{1}{n}^+}\frac{f(x)-\frac{1}{n}}{x-\frac{1}{n}}$$ Does it holds that $\displaystyle{\lim_{x\rightarrow x_n^+}f(x)=\frac{1}{n}}$ ?

Let's pick an example.
Suppose we pick $n=3$, then we have that if $x\in (\frac 1{n+1},\frac 1n]= (\frac 14,\frac 13]$ that $f(x)=\frac 1n = \frac 13$.
And if $x\in (\frac 1n, \frac 1{n-1}]=(\frac 13, \frac 12]$ that $f(x)=\frac 1{n-1}=\frac 12$, don't we? (Thinking)

 

[mathmari]

Suppose we split the interval up in $(\frac 12,1], (\frac 13, \frac 12], ...$, can we find the function values on each of those intervals? (Wondering)

If $x\in (\frac 12,1]$ then $f(x)=1$, if $x\in (\frac 13, \frac 12]$ then $f(x)=\frac{1}{2}$, right? (Wondering)

Suppose $x\in (\frac 1{k+1}, \frac 1k]$, then in which interval will $\frac 1x$ be?
Can we deduce the corresponding $k$ from that? (Wondering)

We have the following:
$$\frac{1}{k+1}<x\Rightarrow \frac{1}{x}<k+1 \\ x\leq \frac{1}{k}\Rightarrow k\leq \frac{1}{x}$$ So, we get that $$k\leq \frac{1}{x}<k+1$$
What do we get from here? (Wondering)

Let's pick an example.
Suppose we pick $n=3$, then we have that if $x\in (\frac 1{n+1},\frac 1n]= (\frac 14,\frac 13]$ that $f(x)=\frac 1n = \frac 13$.
And if $x\in (\frac 1n, \frac 1{n-1}]=(\frac 13, \frac 12]$ that $f(x)=\frac 1{n-1}=\frac 12$, don't we? (Thinking)

So, we have that $f\left (\frac{1}{n}\right )=\frac{1}{n}$, $\displaystyle{\lim_{x\rightarrow \frac{1}{n}^-}f(x)=\frac{1}{n}}$, $\displaystyle{\lim_{x\rightarrow \frac{1}{n}^+}f(x)=\frac{1}{n-1}}$, or not? (Wondering)

 

[I like Serena]

If $x\in (\frac 12,1]$ then $f(x)=1$, if $x\in (\frac 13, \frac 12]$ then $f(x)=\frac{1}{2}$, right?

Right. So we can start by drawing the graph from the right, can't we? (Wondering)

We have the following:
$$\frac{1}{k+1}<x\Rightarrow \frac{1}{x}<k+1 \\ x\leq \frac{1}{k}\Rightarrow k\leq \frac{1}{x}$$ So, we get that $$k\leq \frac{1}{x}<k+1$$
What do we get from here?

Isn't $\lfloor\frac 1x\rfloor$ the largest integer less than or equal to $\frac 1x$?
Can we perhaps write $f(x)$ as a single formula? Then we could plot it with Wolfram. (Thinking)

So, we have that $f\left (\frac{1}{n}\right )=\frac{1}{n}$, $\displaystyle{\lim_{x\rightarrow \frac{1}{n}^-}f(x)=\frac{1}{n}}$, $\displaystyle{\lim_{x\rightarrow \frac{1}{n}^+}f(x)=\frac{1}{n-1}}$, or not?

Yep. That is, for $n\in\mathbb N$. (Nod)

 

[mathmari]

Right. So we can start by drawing the graph from the right, can't we? (Wondering)

So, we have the following:

View attachment 7649

right? (Wondering)

Isn't $\lfloor\frac 1x\rfloor$ the largest integer less than or equal to $\frac 1x$?
Can we perhaps write $f(x)$ as a single formula? Then we could plot it with Wolfram. (Thinking)

We have that $\lfloor \frac{1}{x}\rfloor=k$, right? (Wondering)

Yep. That is, for $n\in\mathbb N$. (Nod)

Do we get the following?

\begin{align*}f_+'(x_n)&=f_+'\left (\frac{1}{n}\right )=\lim_{x\rightarrow \frac{1}{n}^+}\frac{f(x)-f\left (\frac{1}{n}\right )}{x-\frac{1}{n}}=\lim_{x\rightarrow \frac{1}{n}^+}\frac{f(x)-\frac{1}{n}}{x-\frac{1}{n}}=\frac{\frac{1}{n-1}-\frac{1}{n}}{\frac{1}{n}-\frac{1}{n}}=\infty\\ f_-'(x_n)&=f_-'\left (\frac{1}{n}\right )=\lim_{x\rightarrow \frac{1}{n}^-}\frac{f(x)-f\left (\frac{1}{n}\right )}{x-\frac{1}{n}}=\lim_{x\rightarrow \frac{1}{n}^-}\frac{f(x)-\frac{1}{n}}{x-\frac{1}{n}}=\frac{\frac{1}{n}-\frac{1}{n}}{\frac{1}{n}-\frac{1}{n}}=1\end{align*}

(Wondering)

 

[I like Serena]

So, we have the following:

right?

We have that $\lfloor \frac{1}{x}\rfloor=k$, right?

Yip yip. (Nod)

Do we get the following?

\begin{align*}f_+'(x_n)&=f_+'\left (\frac{1}{n}\right )=\lim_{x\rightarrow \frac{1}{n}^+}\frac{f(x)-f\left (\frac{1}{n}\right )}{x-\frac{1}{n}}=\lim_{x\rightarrow \frac{1}{n}^+}\frac{f(x)-\frac{1}{n}}{x-\frac{1}{n}}=\frac{\frac{1}{n-1}-\frac{1}{n}}{\frac{1}{n}-\frac{1}{n}}=\infty\\ f_-'(x_n)&=f_-'\left (\frac{1}{n}\right )=\lim_{x\rightarrow \frac{1}{n}^-}\frac{f(x)-f\left (\frac{1}{n}\right )}{x-\frac{1}{n}}=\lim_{x\rightarrow \frac{1}{n}^-}\frac{f(x)-\frac{1}{n}}{x-\frac{1}{n}}=\frac{\frac{1}{n}-\frac{1}{n}}{\frac{1}{n}-\frac{1}{n}}=1\end{align*}

We are not allowed to divide by zero are we? (Worried)

Shouldn't it be:
\begin{align*}f_+'(x_n)&=f_+'\left (\frac{1}{n}\right )
=\lim_{x\rightarrow \frac{1}{n}^+}\frac{f(x)-f\left (\frac{1}{n}\right )}{x-\frac{1}{n}}
=\lim_{x\rightarrow \frac{1}{n}^+}\frac{\frac{1}{n-1}-\frac{1}{n}}{x-\frac{1}{n}}=\infty\\
f_-'(x_n)&=f_-'\left (\frac{1}{n}\right )
=\lim_{x\rightarrow \frac{1}{n}^-}\frac{f(x)-f\left (\frac{1}{n}\right )}{x-\frac{1}{n}}
=\lim_{x\rightarrow \frac{1}{n}^-}\frac{\frac{1}{n}-\frac{1}{n}}{x-\frac{1}{n}}=\color{red}0\end{align*}
Note that we can also observe the latter in the graph, where we can see that the graph is horizontal at the edge points, so that it must have slope 0. (Nerd)

 

[mathmari]

Yip yip. (Nod)

At the graph of post #5, at which $k$ should I stop? (Wondering) Since $\lfloor \frac{1}{x}\rfloor=k$, can we write $f(x)$ as a single formula? (Wondering)

We are not allowed to divide by zero are we? (Worried)

Shouldn't it be:
\begin{align*}f_+'(x_n)&=f_+'\left (\frac{1}{n}\right )
=\lim_{x\rightarrow \frac{1}{n}^+}\frac{f(x)-f\left (\frac{1}{n}\right )}{x-\frac{1}{n}}
=\lim_{x\rightarrow \frac{1}{n}^+}\frac{\frac{1}{n-1}-\frac{1}{n}}{x-\frac{1}{n}}=\infty\\
f_-'(x_n)&=f_-'\left (\frac{1}{n}\right )
=\lim_{x\rightarrow \frac{1}{n}^-}\frac{f(x)-f\left (\frac{1}{n}\right )}{x-\frac{1}{n}}
=\lim_{x\rightarrow \frac{1}{n}^-}\frac{\frac{1}{n}-\frac{1}{n}}{x-\frac{1}{n}}=\color{red}0\end{align*}
Note that we can also observe the latter in the graph, where we can see that the graph is horizontal at the edge points, so that it must have slope 0. (Nerd)

Ahh ok!

That means that $f_+'(x_n)$ doesn't exist and that $f_-'(x_n)$ exist. And since $f_+'(x_n)\neq f_-'(x_n)$, it follows that $f'(x_n)$ does not exist.
Is everything correct? (Wondering) I haven't really understood where $x$ approach at the following limits:

View attachment 7650

(Wondering)

Is it maybe meant that $x$ approaches $x_n$ from the right or the left side? (Wondering)

 

[I like Serena]

At the graph of post #5, at which $k$ should I stop?

I think what you did is fine.
We just need at least enough $k$ to make it clear what the graph looks like.
I imagine that for example $k=-5, ..., +5$ should be fine.

Since $\lfloor \frac{1}{x}\rfloor=k$, can we write $f(x)$ as a single formula?

For positive $x$, we can write $f(x)=\frac{1}{\lfloor\frac 1x\rfloor}$.
In general we can write $f(x)=\frac{\operatorname{sign} x}{\lfloor\frac 1{|x|}\rfloor}$, can't we?
See W|A.
View attachment 7651
Erm, that does leave $f(0)$ undefined, so we still have to define it separately as $0$. (Worried)

Ahh ok!

That means that $f_+'(x_n)$ doesn't exist and that $f_-'(x_n)$ exist. And since $f_+'(x_n)\neq f_-'(x_n)$, it follows that $f'(x_n)$ does not exist.
Is everything correct?

Yep.

I haven't really understood where $x$ approach at the following limits:

Is it maybe meant that $x$ approaches $x_n$ from the right or the left side?

Yes, it's the same thing.
It's just taking into account that the upper limit for $x=1$ is not defined, since that's outside of the domain, although in the second case $x=-1$ seems to be neglected, which is not defined for the lower limit. (Nerd)

 

[mathmari]

I think what you did is fine.
We just need at least enough $k$ to make it clear what the graph looks like.
I imagine that for example $k=-5, ..., +5$ should be fine.

Ah ok! (Nerd)

For positive $x$, we can write $f(x)=\frac{1}{\lfloor\frac 1x\rfloor}$.
In general we can write $f(x)=\frac{\operatorname{sign} x}{\lfloor\frac 1{|x|}\rfloor}$, can't we?
See W|A.

Erm, that does leave $f(0)$ undefined, so we still have to define it separately as $0$. (Worried)

Ah so, we have to consider the point $f(0)=0$ seperately, or not? (Wondering)

Yes, it's the same thing.
It's just taking into account that the upper limit for $x=1$ is not defined, since that's outside of the domain, although in the second case $x=-1$ seems to be neglected, which is not defined for the lower limit. (Nerd)

Ah ok! So, do we have to calculate the derivative of $f$ ? Or how could we calculate else these limits? (Wondering)

 

[I like Serena]

Ah so, we have to consider the point $f(0)=0$ seperately, or not?

Yes. (Nod)

Ah ok! So, do we have to calculate the derivative of $f$ ? Or how could we calculate else these limits?

I believe we just have to recognize that we have already calculated these derivativies, with the additional observation that they are not defined for $1^+$ or $-1^-$. (Nerd)

 

[mathmari]

I believe we just have to recognize that we have already calculated these derivativies, with the additional observation that they are not defined for $1^+$ or $-1^-$. (Nerd)

Do you mean the following?

\begin{align*}\lim_{\substack{x\rightarrow x_n+0\\ x\in D, x_n\neq 1}}f'(x)&=f_+'(x_n) \\ \lim_{\substack{x\rightarrow x_n-0\\ x\in D}}f'(x)&=f_-'(x_n)\end{align*}

(Wondering)

 

[I like Serena]

Do you mean the following?

\begin{align*}\lim_{\substack{x\rightarrow x_n+0\\ x\in D, x_n\neq 1}}f'(x)&=f_+'(x_n) \\ \lim_{\substack{x\rightarrow x_n-0\\ x\in D}}f'(x)&=f_-'(x_n)\end{align*}

Yep. (Nod)

 

[mathmari]

Yep. (Nod)

2. Examine at the points $x_0=0$ and $x_n=\frac{1}{n}, n\in \mathbb{N}$, the existence of

View attachment 7652

where $D=\{x\in (-1,1): f'(x)\text{ exists }\}$. Calculate their values, if they exist.

So, at this subquestion we only have to check the existence of $f_-'(x_n)$ and $f_+'(x_n)$ and everything else implies from that? (Wondering)

But at the beginning of that question it says "Examine at the points $x_0=0$ and $x_n=\frac{1}{n}$...", where do we use the point $x_0=0$ ? (Wondering)

 

[I like Serena]

So, at this subquestion we only have to check the existence of $f_-'(x_n)$ and $f_+'(x_n)$ and everything else implies from that? (Wondering)

But at the beginning of that question it says "Examine at the points $x_0=0$ and $x_n=\frac{1}{n}$...", where do we use the point $x_0=0$ ? (Wondering)

I guess we have to investigate the continuity of $f$ and $f'$ at $x=0$ separately. (Sweating)

 

[mathmari]

I guess we have to investigate the continuity of $f$ and $f'$ at $x=0$ separately. (Sweating)

To check the continuity at $x_0=0$ we have to check whether it holds $$\lim_{x\rightarrow 0^-}f(x)=\lim_{x\rightarrow 0^+}f(x)=f(0)$$ right? (Wondering)

To calculate the limit $\displaystyle{\lim_{x\rightarrow 0^-}f(x)}$ do we check the function if $x$ approaches $\frac{1}{k-1}$ for $k$ a very big negative number, so that it approaches $0$ ? (Wondering)

If yes, we have that as $x$ approaches $\frac{1}{k-1}$ for a very big negative number $k$, so as $x$ approaches $0$ from the left side, then $f(x)$ approaches $\frac{1}{k}$, which also approaches $0$, so $\displaystyle{\lim_{x\rightarrow 0^-}f(x)}=0$.

Similarily, we have that as $x$ approaches $\frac{1}{k+1}$ for a very big natural number $k$, so as $x$ approaches $0$ from the right side, then $f(x)$ approaches $\frac{1}{k}$, which also approaches $0$, so $\displaystyle{\lim_{x\rightarrow 0^+}f(x)}=0$.

We have that $f(x_0)=f(0)=0$.

That means that $f$ is continuous at $x_0=0$.

Is everything correct? (Wondering)

 

[I like Serena]

To check the continuity at $x_0=0$ we have to check whether it holds $$\lim_{x\rightarrow 0^-}f(x)=\lim_{x\rightarrow 0^+}f(x)=f(0)$$ right? (Wondering)

To calculate the limit $\displaystyle{\lim_{x\rightarrow 0^-}f(x)}$ do we check the function if $x$ approaches $\frac{1}{k-1}$ for $k$ a very big negative number, so that it approaches $0$ ? (Wondering)

If yes, we have that as $x$ approaches $\frac{1}{k-1}$ for a very big negative number $k$, so as $x$ approaches $0$ from the left side, then $f(x)$ approaches $\frac{1}{k}$, which also approaches $0$, so $\displaystyle{\lim_{x\rightarrow 0^-}f(x)}=0$.

Similarily, we have that as $x$ approaches $\frac{1}{k+1}$ for a very big natural number $k$, so as $x$ approaches $0$ from the right side, then $f(x)$ approaches $\frac{1}{k}$, which also approaches $0$, so $\displaystyle{\lim_{x\rightarrow 0^+}f(x)}=0$.

We have that $f(x_0)=f(0)=0$.

That means that $f$ is continuous at $x_0=0$.

Is everything correct? (Wondering)

Let's see... we have:
$$\lim_{x\to 0^+} f(x) = \lim_{x\to 0^+} \frac{1}{\lfloor \frac 1x \rfloor} = 0$$
don't we?

Equivalently, we have $\lim_{x\to 0^-} f(x)=0$.

So yes, I think $f$ is continuous at 0 with function value 0. (Nod)

 

[mathmari]

Let's see... we have:
$$\lim_{x\to 0^+} f(x) = \lim_{x\to 0^+} \frac{1}{\lfloor \frac 1x \rfloor} = 0$$
don't we?

Equivalently, we have $\lim_{x\to 0^-} f(x)=0$.

So yes, I think $f$ is continuous at 0 with function value 0. (Nod)

So, do we have to check the continuity at $0$ using this formula of the function (with the floor function) and not the given one? (Wondering)

 

[I like Serena]

So, do we have to check the continuity at $0$ using this formula of the function (with the floor function) and not the given one?

We can also check with the given one as you did correctly. (Nod)

 

[mathmari]

2. Examine at the points $x_0=0$ and $x_n=\frac{1}{n}, n\in \mathbb{N}$, the existence of

View attachment 7653

where $D=\{x\in (-1,1): f'(x)\text{ exists }\}$. Calculate their values, if they exist.

I read again the statement of that subquestion. We have to check the existence of $f'(x_n)$, $f_+'(x_n)$, $f_-'(x_n)$, $\displaystyle{\lim_{\substack{x\rightarrow x_n+0, \\ x\in D, x_n\neq 1}}f'(x)}$, $\displaystyle{\lim_{\substack{x\rightarrow x_n-0\\ x\in D}}f'(x)}$ with $n\in \mathbb{N}_0$. That means that we have to check the existence for $n=0$ and for $n\in \mathbb{N}$, i.e. the existence of the following:
\begin{align*}&f'(x_0), \ \ f_+'(x_0), \ \ f_-'(x_0), \ \ \lim_{\substack{x\rightarrow x_0+0, \\ x\in D}}f'(x), \ \ \lim_{\substack{x\rightarrow x_0-0\\ x\in D}}f'(x) \\ & f'(x_n), \ \ f_+'(x_n), \ \ f_-'(x_n), \ \ \lim_{\substack{x\rightarrow x_n+0, \\ x\in D, x_n\neq 1}}f'(x), \ \ \lim_{\substack{x\rightarrow x_n-0\\ x\in D}}f'(x), \ \ \ n\in \mathbb{N}\end{align*}
or not?

So, it is not meant to check the continuity of $f$ at $x_0=0$, is it?

(Wondering)

 

[I like Serena]

Indeed. (Thinking)

 

[mathmari]

Indeed. (Thinking)

So, we have the following:

• For $n=0$ :
  \begin{equation*}f_+'(x_0)=f_+'\left (0\right )
  =\lim_{x\rightarrow 0^+}\frac{f(x)-f\left (0\right )}{x-0}
  =\lim_{x\rightarrow 0^+}\frac{f(x)-0}{x}=\lim_{x\rightarrow 0^+}\frac{f(x)}{x}\end{equation*}
  We have that $x$ approaches $0$ from the right side, as $x$ approaches $\frac{1}{k+1}$ for a very big natural number $k$, then $f(x)$ approaches $\frac{1}{k}$, which also approaches $0$, so $\displaystyle{\lim_{x\rightarrow 0^+}f(x)}=0$.
  
  So, we get \begin{equation*}f_+'(x_0)=\lim_{x\rightarrow 0^+}\frac{0}{x}=0\end{equation*}
  
  \begin{equation*}f_-'(x_0)=f_-'\left (0\right )
  =\lim_{x\rightarrow 0^-}\frac{f(x)-f\left (0\right )}{x-0}
  =\lim_{x\rightarrow 0^-}\frac{f(x)-0}{x}=\lim_{x\rightarrow 0^-}\frac{f(x)}{x}\end{equation*}
  We have that $x$ approaches $0$ from the left side, as $x$ approaches $\frac{1}{k-1}$ for a very big negative number $k$, then $f(x)$ approaches $\frac{1}{k}$, which also approaches $0$, so $\displaystyle{\lim_{x\rightarrow 0^-}f(x)}=0$.
  
  So, we get \begin{equation*}f_-'(x_0)=\lim_{x\rightarrow 0^-}\frac{0}{x}=0\end{equation*}
  
  So, $f_+'(x_0)$ and $f_-'(x_0)$ exist and they are equal, therefore $f'(x_0)$ also exist and ist equal to $f'(x_0)=f_+'(x_0)=f_-'(x_0)=0$. We have that $\displaystyle{\lim_{\substack{x\rightarrow x_0+0, \\ x\in D}}f'(x)=f_+(x_0)}$ and $\displaystyle{\lim_{\substack{x\rightarrow x_0-0\\ x\in D}}f'(x)=f_-(x_0)}$, so both of these limits exist and they are equal to $0$.
• For $n\in \mathbb{N}$:
  \begin{align*}f_+'(x_n)&=f_+'\left (\frac{1}{n}\right )
  =\lim_{x\rightarrow \frac{1}{n}^+}\frac{f(x)-f\left (\frac{1}{n}\right )}{x-\frac{1}{n}}
  =\lim_{x\rightarrow \frac{1}{n}^+}\frac{\frac{1}{n-1}-\frac{1}{n}}{x-\frac{1}{n}}=\infty\\
  f_-'(x_n)&=f_-'\left (\frac{1}{n}\right )
  =\lim_{x\rightarrow \frac{1}{n}^-}\frac{f(x)-f\left (\frac{1}{n}\right )}{x-\frac{1}{n}}
  =\lim_{x\rightarrow \frac{1}{n}^-}\frac{\frac{1}{n}-\frac{1}{n}}{x-\frac{1}{n}}=0\end{align*}
  So, $f_+'(x_n)$ doesn't exist but $f_-'(x_0)$ does exist, therefore $f'(x_0)$ doesn't exist. We have that $\displaystyle{\lim_{\substack{x\rightarrow x_n+0, \\ x\in D, x_n\neq 1}}f'(x)=f_+(x_n)}$ doesn't exist but $\displaystyle{\lim_{\substack{x\rightarrow x_n-0\\ x\in D}}f'(x)=f_-(x_n)}$ exists and it is equal to $0$.

Is everything correct and complete? Could I improve something? (Wondering)

 

[I like Serena]

[*] For $n=0$ :
\begin{equation*}f_+'(x_0)=f_+'\left (0\right )
=\lim_{x\rightarrow 0^+}\frac{f(x)-f\left (0\right )}{x-0}
=\lim_{x\rightarrow 0^+}\frac{f(x)-0}{x}=\lim_{x\rightarrow 0^+}\frac{f(x)}{x}\end{equation*}
We have that $x$ approaches $0$ from the right side, as $x$ approaches $\frac{1}{k+1}$ for a very big natural number $k$, then $f(x)$ approaches $\frac{1}{k}$, which also approaches $0$, so $\displaystyle{\lim_{x\rightarrow 0^+}f(x)}=0$.

So, we get \begin{equation*}f_+'(x_0)=\lim_{x\rightarrow 0^+}\frac{0}{x}=0\end{equation*}

I'm afraid that this is not quite correct.
We would end up with $\frac 00$, which is not defined, so we cannot deduce that it's zero. (Worried)Instead, given an $x \in (0,1]$, we can suppose that $\frac{1}{k+1}<x\le \frac 1k$ for some $k$, so that we have:
$$\frac{\frac 1k}{\frac 1k} \le \frac{f(x)}{x} < \frac{\frac 1k}{\frac 1{k+1}}$$
Can we find the limits of the left hand side and the right hand side for $k\to \infty$? (Wondering)

 

[mathmari]

I'm afraid that this is not quite correct.
We would end up with $\frac 00$, which is not defined, so we cannot deduce that it's zero. (Worried)

Ah ok!

Instead, given an $x \in (0,1]$, we can suppose that $\frac{1}{k+1}<x\le \frac 1k$ for some $k$, so that we have:
$$\frac{\frac 1k}{\frac 1k} \le \frac{f(x)}{x} < \frac{\frac 1k}{\frac 1{k+1}}$$
Can we find the limits of the left hand side and the right hand side for $k\to \infty$? (Wondering)

So, we have the following:

For $n=0$ :
\begin{equation*}f_+'(x_0)=f_+'\left (0\right )
=\lim_{x\rightarrow 0^+}\frac{f(x)-f\left (0\right )}{x-0}
=\lim_{x\rightarrow 0^+}\frac{f(x)-0}{x}=\lim_{x\rightarrow 0^+}\frac{f(x)}{x}\end{equation*}
Since $x$ approaches $0$ from the right side, we have that $x\in \left (\frac{1}{k+1}, \frac{1}{k}\right ]$ with $k\rightarrow \infty$.

Then we have that $f(x)=\frac{1}{k}$.

We also have that \begin{align*}\frac{1}{k+1}<x\leq \frac{1}{k}&\Rightarrow \frac{1}{\frac{1}{k}}\leq \frac{1}{x}<\frac{1}{\frac{1}{k+1}}\\ & \overset{ \cdot f(x)=\frac{1}{k}>0 }{ \Longrightarrow }\frac{f(x)}{\frac{1}{k}}\leq \frac{f(x)}{x}<\frac{f(x)}{\frac{1}{k+1}}\\ & \Rightarrow \frac{\frac{1}{k}}{\frac{1}{k}}\leq \frac{f(x)}{x}<\frac{\frac{1}{k}}{\frac{1}{k+1}}\\ & \Rightarrow 1\leq \frac{f(x)}{x}<\frac{k+1}{k}\\ & \Rightarrow 1\leq \frac{f(x)}{x}<1+\frac{1}{k}\end{align*}
We have that when $x$ approaches $0$ from the right side, $\frac{1}{k}$ approaches also $0$.

Then when $x\rightarrow 0^+$ we have that $1\leq \frac{f(x)}{x}<1+0\Rightarrow \frac{f(x)}{x}\rightarrow 1$.

So, we get $f_+'(x_0)=1$. \begin{equation*}f_-'(x_0)=f_-'\left (0\right )
=\lim_{x\rightarrow 0^-}\frac{f(x)-f\left (0\right )}{x-0}
=\lim_{x\rightarrow 0^-}\frac{f(x)-0}{x}=\lim_{x\rightarrow 0^-}\frac{f(x)}{x}\end{equation*}
Since $x$ approaches $0$ from the left side, we have that $x\in \left [\frac{1}{k}, \frac{1}{k-1}\right )$ with $k\rightarrow -\infty$.

Then we have that $f(x)=\frac{1}{k}$.

We also have that \begin{align*}\frac{1}{k}\leq x\leq \frac{1}{k-1}&\Rightarrow \frac{1}{\frac{1}{k-1}}<\frac{1}{x}\leq \frac{1}{\frac{1}{k}}\\ & \overset{ \cdot f(x)=\frac{1}{k}<0 }{ \Longrightarrow }\frac{f(x)}{\frac{1}{k}}\leq \frac{f(x)}{x}<\frac{f(x)}{\frac{1}{k-1}}\\ & \Rightarrow \frac{\frac{1}{k}}{\frac{1}{k}}\leq \frac{f(x)}{x}<\frac{\frac{1}{k}}{\frac{1}{k-1}}\\ & \Rightarrow 1\leq \frac{f(x)}{x}<\frac{k-1}{k}\\ & \Rightarrow 1\leq \frac{f(x)}{x}<1-\frac{1}{k}\end{align*}
We have that when $x$ approaches $0$ from the left side, $\frac{1}{k}$ approaches also $0$.

Then when $x\rightarrow 0^-$ we have that $1\leq \frac{f(x)}{x}<1-0\Rightarrow \frac{f(x)}{x}\rightarrow 1$.

So, we get $f_-'(x_0)=1$. So, $f_+'(x_0)$ and $f_-'(x_0)$ exist and they are equal, therefore $f'(x_0)$ also exist and ist equal to $f'(x_0)=f_+'(x_0)=f_-'(x_0)=1$. We have that $\displaystyle{\lim_{\substack{x\rightarrow x_0+0, \\ x\in D}}f'(x)=f_+(x_0)}$ and $\displaystyle{\lim_{\substack{x\rightarrow x_0-0\\ x\in D}}f'(x)=f_-(x_0)}$, so both of these limits exist and they are equal to $1$.
Is everything correct? (Wondering)

 

[I like Serena]

[*] For $n\in \mathbb{N}$:
\begin{align*}f_+'(x_n)&=f_+'\left (\frac{1}{n}\right )
=\lim_{x\rightarrow \frac{1}{n}^+}\frac{f(x)-f\left (\frac{1}{n}\right )}{x-\frac{1}{n}}
=\lim_{x\rightarrow \frac{1}{n}^+}\frac{\frac{1}{n-1}-\frac{1}{n}}{x-\frac{1}{n}}=\infty\\
f_-'(x_n)&=f_-'\left (\frac{1}{n}\right )
=\lim_{x\rightarrow \frac{1}{n}^-}\frac{f(x)-f\left (\frac{1}{n}\right )}{x-\frac{1}{n}}
=\lim_{x\rightarrow \frac{1}{n}^-}\frac{\frac{1}{n}-\frac{1}{n}}{x-\frac{1}{n}}=0\end{align*}
So, $f_+'(x_n)$ doesn't exist but $f_-'(x_0)$ does exist, therefore $f'(x_0)$ doesn't exist. We have that $\displaystyle{\lim_{\substack{x\rightarrow x_n+0, \\ x\in D, x_n\neq 1}}f'(x)=f_+(x_n)}$ doesn't exist but $\displaystyle{\lim_{\substack{x\rightarrow x_n-0\\ x\in D}}f'(x)=f_-(x_n)}$ exists and it is equal to $0$.

Is everything correct and complete? Could I improve something?

Don't we have a special edge case for $n=1$, which brings us to the edge of the domain? (Wondering)
Note that $\frac 1{n-1}$ is not defined in that case, and it doesn't apply anyway.

We also have that \begin{align*}\frac{1}{k}\leq x\leq \frac{1}{k-1}&\Rightarrow \frac{1}{\frac{1}{k-1}}<\frac{1}{x}\leq \frac{1}{\frac{1}{k}}\\ & \overset{ \cdot f(x)=\frac{1}{k}<0 }{ \Longrightarrow }\frac{f(x)}{\frac{1}{k}}\leq \frac{f(x)}{x}<\frac{f(x)}{\frac{1}{k-1}}\\ & \Rightarrow \frac{\frac{1}{k}}{\frac{1}{k}}\leq \frac{f(x)}{x}<\frac{\frac{1}{k}}{\frac{1}{k-1}}\\ & \Rightarrow 1\leq \frac{f(x)}{x}<\frac{k-1}{k}\\ & \Rightarrow 1\leq \frac{f(x)}{x}<1-\frac{1}{k}\end{align*}

That $1\leq \frac{f(x)}{x}<1-\frac{1}{k}$ can't be correct can it?
That's because the right hand side is lower than the left hand side. (Worried)

 

[mathmari]

Don't we have a special edge case for $n=1$, which brings us to the edge of the domain? (Wondering)
Note that $\frac 1{n-1}$ is not defined in that case, and it doesn't apply anyway.

So, do we have to consider the case $n=1$ seperately? (Wondering)

That $1\leq \frac{f(x)}{x}<1-\frac{1}{k}$ can't be correct can it?
That's because the right hand side is lower than the left hand side. (Worried)

In this case $k$ is negative, so the right hand side is bigger than the left hand side, or not? (Wondering)

 

[I like Serena]

So, do we have to consider the case $n=1$ seperately?

I believe so, although to be fair, nothing really changes.
It's more like dotting the i's. (Nerd)

In this case $k$ is negative, so the right hand side is bigger than the left hand side, or not?

Oh yes. (Blush)

 

[mathmari]

So, is at the case $n=0$ everything correct? (Wondering) As regard the case $n\in \mathbb{N}$:

• If $n\neq 1$ we have the following:
  
  Since $x\rightarrow \frac{1}{n}^+$ we have that $x\in \left (\frac{1}{n}, \frac{1}{n-1}\right ]$. Therefore $f(x)= \frac{1}{n-1}$. \begin{align*}f_+'(x_n)&=f_+'\left (\frac{1}{n}\right )
  =\lim_{x\rightarrow \frac{1}{n}^+}\frac{f(x)-f\left (\frac{1}{n}\right )}{x-\frac{1}{n}}=\lim_{x\rightarrow \frac{1}{n}^+}\frac{\frac{1}{n-1}-\frac{1}{n}}{x-\frac{1}{n}}=\infty\end{align*}
  
  Since $x\rightarrow \frac{1}{n}^-$ we have that $x\in \left (\frac{1}{n+1}, \frac{1}{n}\right ]$. Therefore $f(x)= \frac{1}{n}$.
  
  \begin{align*}
  f_-'(x_n)&=f_-'\left (\frac{1}{n}\right )
  =\lim_{x\rightarrow \frac{1}{n}^-}\frac{f(x)-f\left (\frac{1}{n}\right )}{x-\frac{1}{n}}=\lim_{x\rightarrow \frac{1}{n}^-}\frac{\frac{1}{n}-\frac{1}{n}}{x-\frac{1}{n}}=0\end{align*} So, $f_+'(x_n)$ doesn't exist but $f_-'(x_n)$ does exist, therefore $f'(x_n)$ doesn't exist. We have that $\displaystyle{\lim_{\substack{x\rightarrow x_n+0, \\ x\in D, x_n\neq 1}}f'(x)=f_+(x_n)}$ doesn't exist but $\displaystyle{\lim_{\substack{x\rightarrow x_n-0\\ x\in D}}f'(x)=f_-(x_n)}$ exists and it is equal to $0$.
  
  Is everything correct at this case? (Wondering)
  
  $$$$
  
• If $n= 1$ we have the following:
  
  Since $x\rightarrow 1^+$ the function is not defined. So, we cannot calculate $f_+'(x_1)$, right? (Wondering) Since $x\rightarrow 1^-$ we have that $x\in \left (\frac{1}{2}, 1\right ]$. Therefore $f(x)= 1$.
  
  \begin{align*}
  f_-'(x_1)&=f_-'\left (1\right )
  =\lim_{x\rightarrow 1^-}\frac{f(x)-f\left (1\right )}{x-1}=\lim_{x\rightarrow \frac{1}{n}^-}\frac{1-1}{x-1}=0\end{align*} So, $f_+'(x_1)$ doesn't exist but $f_-'(x_1)$ does exist, therefore $f'(x_1)$ doesn't exist. Can we calculate the limits $\displaystyle{\lim_{\substack{x\rightarrow x_n+0, \\ x\in D, x_n\neq 1}}f'(x)=f_+(x_n)}$ and $\displaystyle{\lim_{\substack{x\rightarrow x_n-0\\ x\in D}}f'(x)=f_-(x_n)}$ ? Because $x_n$ must be in $D$ but $1$ is not. (Wondering)

 

[I like Serena]

So, is at the case $n=0$ everything correct?

I believe it's correct.
That is, except that I've taken another look at $\displaystyle{\lim_{\substack{x\rightarrow x_n+0, \\ x\in D, x_n\neq 1}}f'(x)=f_+(x_n)}$ and $\displaystyle{\lim_{\substack{x\rightarrow x_n-0\\ x\in D}}f'(x)=f_-(x_n)}$.
I believe those 2 limits are different after all.
That's because for any $x\in D$, we have $f'(x)=0$, so that is also what the limit to $x_n$ must be.
Note that part of the definition of $D$ is that $f'(x)$ must be defined, and we found that in all those cases we have $f'(x)=0$.
It also means that $\forall k\in \mathbb Z$ we have that $x_k \not\in D$. (Thinking)

The same applies to the other cases for $n$.

As regard the case $n\in \mathbb{N}$:

[*] If $n\neq 1$ we have the following:

Since $x\rightarrow \frac{1}{n}^+$ we have that $x\in \left (\frac{1}{n}, \frac{1}{n-1}\right ]$. Therefore $f(x)= \frac{1}{n-1}$.

Is everything correct at this case?

It's correct, except that my same comment for the limits with respect to $D$ applies.
And I'd just write 'If' insted of 'Since' (twice). (Nerd)

[*] If $n= 1$ we have the following:

Since $x\rightarrow 1^+$ the function is not defined. So, we cannot calculate $f_+'(x_1)$, right? (Wondering)

Since $x\rightarrow 1^-$ we have that $x\in \left (\frac{1}{2}, 1\right ]$. Therefore $f(x)= 1$.

\begin{align*}
f_-'(x_1)&=f_-'\left (1\right )
=\lim_{x\rightarrow 1^-}\frac{f(x)-f\left (1\right )}{x-1}=\lim_{x\rightarrow \frac{1}{n}^-}\frac{1-1}{x-1}=0\end{align*}

All correct.

So, $f_+'(x_1)$ doesn't exist but $f_-'(x_1)$ does exist, therefore $f'(x_1)$ doesn't exist.

I'm a bit unsure about $f'(x_1)$. It seems to me that it does exist and is equal to $f_-'(x_1)$.
That's because if we calculate $f'(x)$ according to $\varepsilon$-$\delta$ definition of a limit, that's what we'll find. (Thinking)

Can we calculate the limits $\displaystyle{\lim_{\substack{x\rightarrow x_n+0, \\ x\in D, x_n\neq 1}}f'(x)=f_+(x_n)}$ and $\displaystyle{\lim_{\substack{x\rightarrow x_n-0\\ x\in D}}f'(x)=f_-(x_n)}$ ? Because $x_n$ must be in $D$ but $1$ is not. (Wondering)

The first limit explicitly excludes $x_n=1$, so we cannot calculate that limit.
We can calculate that second limit though.
Note that all $x_n\not\in D$, but that does not prevent us from calculate the limit to it. (Thinking)

 

[mathmari]

Happy New Year! (Party)

I believe it's correct.
That is, except that I've taken another look at $\displaystyle{\lim_{\substack{x\rightarrow x_n+0, \\ x\in D, x_n\neq 1}}f'(x)=f_+(x_n)}$ and $\displaystyle{\lim_{\substack{x\rightarrow x_n-0\\ x\in D}}f'(x)=f_-(x_n)}$.

Does this equality only hold when the derivatives are continuous at the point $x_0$ ? (Wondering)

I believe those 2 limits are different after all.
That's because for any $x\in D$, we have $f'(x)=0$, so that is also what the limit to $x_n$ must be.
Note that part of the definition of $D$ is that $f'(x)$ must be defined, and we found that in all those cases we have $f'(x)=0$.
It also means that $\forall k\in \mathbb Z$ we have that $x_k \not\in D$. (Thinking)

I haven't really understood that. Could you maybe explain it further to me? (Wondering)

 

[I like Serena]

Happy New Year! (Party)

Happy 2018! (drink)

Does this equality only hold when the derivatives are continuous at the point $x_0$ ?

It certainly holds if the derivative is continuous at a point $x_0$, but it also still holds in some of the cases. (Thinking)

I haven't really understood that. Could you maybe explain it further to me?

Let's pick a point close to 0 that is in $D$.
Say $x=0.095 \in (\frac 1{10}, \frac 1{11})$.
Then $f'(x)=0$ isn't it?

No matter how close we pick $x$ to $0$, as long as it's inside an interval of the form $(\frac 1{k+1}, \frac 1{k})$, which is where $f'(x)$ exists, we have $f'(x)=0$ don't we?
Consequently, the limit is $0$ instead of $1$.

Or more specifically for $n=0$:
$$\begin{aligned}\lim_{\substack{x\rightarrow x_n+0, \\ x\in D, x_n\neq 1}} f'(x)
&=\lim_{\substack{x\rightarrow 0^+, \\x\in D}} f'(x)
=\lim_{\substack{k\to\infty, \\x\in (\frac 1{k+1}, \frac 1k)}} f'(x) \\
&=\lim_{\substack{k\to\infty, \\x\in (\frac 1{k+1}, \frac 1k)}} \lim_{h\to 0} \frac{f(x+h) - f(x)}{h}
=\lim_{\substack{k\to\infty, \\x\in (\frac 1{k+1}, \frac 1k)}} \lim_{h\to 0} \frac{\frac 1k - \frac 1k}{h}
=0
\end{aligned}$$
(Thinking)

 

[mathmari]

Let's pick a point close to 0 that is in $D$.
Say $x=0.095 \in (\frac 1{10}, \frac 1{11})$.
Then $f'(x)=0$ isn't it?

No matter how close we pick $x$ to $0$, as long as it's inside an interval of the form $(\frac 1{k+1}, \frac 1{k})$, which is where $f'(x)$ exists, we have $f'(x)=0$ don't we?
Consequently, the limit is $0$ instead of $1$.

Does this mean that for each $x\in D$ we have that $f'(x)=0$ since $f(x)$ is constant?

 

[I like Serena]

Does this mean that for each $x\in D$ we have that $f'(x)=0$ since $f(x)$ is constant?

Yep. (Nod)

That is, $f(x)$ is constant in each of those disjointed open intervals, although the constant is different on each of those intervals. (Nerd)

 

[mathmari]

Yep. (Nod)

That is, $f(x)$ is constant in each of those disjointed open intervals, although the constant is different on each of those intervals. (Nerd)

Does this mean that every limit of $f'(x)$, when $x\in D$, will be zero?

I mean:

$$\lim_{\substack{x\rightarrow x_0+0, \\ x\in D}} f'(x) =\lim_{\substack{x\rightarrow x_n+0, \\ x\in D, x_n\neq 1}} f'(x) =0$$ for $n\in \mathbb{N}$.

(Wondering)

 

[I like Serena]

Does this mean that every limit of $f'(x)$, when $x\in D$, will be zero?

I mean:

$$\lim_{\substack{x\rightarrow x_0+0, \\ x\in D}} f'(x) =\lim_{\substack{x\rightarrow x_n+0, \\ x\in D, x_n\neq 1}} f'(x) =0$$ for $n\in \mathbb{N}$.

Yep. (Nod)

 

[mathmari]

Yep. (Nod)

But then if $f_+'(x_n), f_-'(x_n)$ and so $f'(x_n)$ (for $n\in \mathbb{N}_0$) exist, do they not have to be equal to $0$ ? (Wondering)