Answer Limits of a Function: Proving c ≠ 0 & No Limit at x→0

[amilapsn]

Let $f(x)=\begin{cases}1 &if \ x=\frac{1}{n} \ where \ n\epsilon \mathbb{Z}^{+}\\ 0 & \mbox{otherwise}\end{cases}$
(i) Show that $c\neq 0$ then $\lim_{x \to c}f(x)=0$

(ii) Show that $\lim_{x \to 0}f(x)$ does not exist.
I attempted to answer the question:

I think we have to show that:

$c\neq 0 \Rightarrow \lim_{x \to c}f(x)=0$

We know that,

$c\neq 0 \Leftrightarrow |c|>0$

$\lim_{x \to c}f(x)=0$ means: $\forall \epsilon>0 \ \exists \delta \ s.t. \forall x\epsilon \mathbb{R}$

$0<|x-c|<\delta \Rightarrow |f(x)|<\epsilon$

But I don't understand how to develop the proof any further... I think the question should begin with the statement: $c\neq 0$
Please help me.
I have thought something of proving the first part but it's seems impossible to prove the second part.

 

[BvU]

$0<|x-c|<\delta \Rightarrow |f(x)|<\epsilon$ looks good. Especially that the first $<$ isn't a $\le$ (which it indeed isn't, fortunately). So thos $f$ are 0 and for sure $<\epsilon$

But I seem to remember that "$\lim_{x \to c}f(x)=0$ exists" means a little bit more than you state, namely something with $f(c)$ ... If you have that, you have part b too.

[edit] all wrong (see below). Whole post withdrawn. Sorry folks !

 

[Dick]

Let $f(x)=\begin{cases}1 &if \ x=\frac{1}{n} \ where \ n\epsilon \mathbb{Z}^{+}\\ 0 & \mbox{otherwise}\end{cases}$
(i) Show that $c\neq 0$ then $\lim_{x \to c}f(x)=0$

(ii) Show that $\lim_{x \to 0}f(x)$ does not exist.
I attempted to answer the question:

I think we have to show that:

$c\neq 0 \Rightarrow \lim_{x \to c}f(x)=0$

We know that,

$c\neq 0 \Leftrightarrow |c|>0$

$\lim_{x \to c}f(x)=0$ means: $\forall \epsilon>0 \ \exists \delta \ s.t. \forall x\epsilon \mathbb{R}$

$0<|x-c|<\delta \Rightarrow |f(x)|<\epsilon$

But I don't understand how to develop the proof any further... I think the question should begin with the statement: $c\neq 0$
Please help me.
I have thought something of proving the first part but it's seems impossible to prove the second part.

Just try drawing a picture. $0<|x-c|<\delta$ means every point in an interval around $c$ except for $c$. If $c \ne 0$ then you can make an interval small enough that you don't include any of the points of the form $1/n$, except possibly $c$. If $c=0$ then you can't, since $1/n$ can be made as close as you like to 0.

 

[amilapsn]

Thank you for the advice @Dick. The drawing makes sense, but, I don't know the way to prove the 1st part rigorously. Would the proof look like follows:
$0<|x-c|<\delta \Leftrightarrow x\in (c-\delta,c+\delta)\backslash \{c\}$
$c\neq 0 \Rightarrow x\in (c-\delta,c+\delta)\backslash \{c\}\mbox{ where the interval does not include any value in the form of \frac{1}{n}}$...

 

[Dick]

Thank you for the advice @Dick. The drawing makes sense, but, I don't know the way to prove the 1st part rigorously. Would the proof look like follows:
$0<|x-c|<\delta \Leftrightarrow x\in (c-\delta,c+\delta)\backslash \{c\}$
$c\neq 0 \Rightarrow x\in (c-\delta,c+\delta)\backslash \{c\}\mbox{ where the interval does not include any value in the form of \frac{1}{n}}$...

You need to say what the value of $\delta$ is. If $c \ne 0$ doesn't have the form $\frac{1}{n_0}$ where $n_0 \in Z^+$ then you can just pick $\delta=inf(\{|c-\frac{1}{n}|: n \in Z^+\})$. Right? That will be nonzero, yes? What if it does?

 

[Ray Vickson]

$0<|x-c|<\delta \Rightarrow |f(x)|<\epsilon$ looks good. Especially that the first $<$ isn't a $\le$ (which it indeed isn't, fortunately). So thos $f$ are 0 and for sure $<\epsilon$

But I seem to remember that "$\lim_{x \to c}f(x)=0$ exists" means a little bit more than you state, namely something with $f(c)$ ... If you have that, you have part b too.

I'm not sure what you are attempting to say here, but it sounds wrong. You can have a perfectly good $\lim_{x \to c} f(x)$ even if $f(c)$ does not exist---that is, if you are speaking of functions in general, rather than the specific $f(x)$ in the question.

 

[BvU]

Oops, I am all wrong here. Didn't look at the function close enough. Thanks for noticing.

 

[amilapsn]

You need to say what the value of $\delta$ is. If $c \ne 0$ doesn't have the form $\frac{1}{n_0}$ where $n_0 \in Z^+$ then you can just pick $\delta=inf(\{|c-\frac{1}{n}|: n \in Z^+\})$. Right? That will be nonzero, yes? What if it does?

$\forall c\in [1,\infty)\ \delta=inf(\{|c-\frac{1}{n}|: n \in Z^+\})=c-1$ .
Thus, when $c=1, \delta$ will be $0$. But $\delta$ cannot be $0$, right? To avoid this issue, I think, we should account for the case of $c=1$ separately, by choosing $\delta=\frac{1}{3}$ or some other value so that the interval $(1-\delta,1+\delta)\backslash\{1\}$ will not include values in the form of $\frac{1}{n}$. Right? Please, correct me if I am wrong.


Thank you again @Dick..



P.S. I have a great need of a good textbook on real analysis including good problems. Would you kindly suggest me one?
(Please, pardon me if I bother you.)

 

[Dick]

$\forall c\in [1,\infty)\ \delta=inf(\{|c-\frac{1}{n}|: n \in Z^+\})=c-1$ .
Thus, when $c=1, \delta$ will be $0$. But $\delta$ cannot be $0$, right? To avoid this issue, I think, we should account for the case of $c=1$ separately, by choosing $\delta=\frac{1}{3}$ or some other value so that the interval $(1-\delta,1+\delta)\backslash\{1\}$ will not include values in the form of $\frac{1}{n}$. Right? Please, correct me if I am wrong.


Thank you again @Dick..
P.S. I have a great need of a good textbook on real analysis including good problems. Would you kindly suggest me one?
(Please, pardon me if I bother you.)

You've got the right idea. You want to choose an interval around c that doesn't include any values of the form 1/n except for possibly c itself. But you need to say something about other cases besides c in $[1,\infty)$, yes? And that leads you to saying why there is no limit at c=0. I don't know very many real analysis books. I used Rudin and I liked it. Depends on your taste. You might ask over the "Science and Math Textbooks" forum. I'm sure you get a ton of opinions.

 

[amilapsn]

For any other c value except 0, we are able to find a $\delta$ value. But for c=0, $\delta=inf(\{|c-\frac{1}{n}|:n\in \mathbb{Z}^+\})=0$ But this can't happen. But for any other $\delta>0$ value there will be a $x_0$ value in the form of $\frac{1}{n}$ . Would the above statements be sufficient to conclude that $c\ne 0\Rightarrow \lim_{x\to c}f(x)=0$ ?

p.s.
I think by slightly changing the definition of $\delta$ we can reduce the number of cases we have to consider.
$\delta=inf(\{|c-\frac{1}{n}|:n\in \mathbb{Z}^+\}\backslash \{0\})$ . (Is it okay to change the definition like that?)
For the case of $c=0$ , there won't be a $\delta$ , because, the set $\{|c-\frac{1}{n}|:n\in \mathbb{Z}^+\}\backslash \{0\}$ will be $\emptyset$ . So for any other value except 0 for c, will give us a $\delta$ value which is the result in part (i)

 

[Dick]

For any other c value except 0, we are able to find a $\delta$ value. But for c=0, $\delta=inf(\{|c-\frac{1}{n}|:n\in \mathbb{Z}^+\})=0$ But this can't happen. But for any other $\delta>0$ value there will be a $x_0$ value in the form of $\frac{1}{n}$ . Would the above statements be sufficient to conclude that $c\ne 0\Rightarrow \lim_{x\to c}f(x)=0$ ?

p.s.
I think by slightly changing the definition of $\delta$ we can reduce the number of cases we have to consider.
$\delta=inf(\{|c-\frac{1}{n}|:n\in \mathbb{Z}^+\}\backslash \{0\})$ . (Is it okay to change the definition like that?)
For the case of $c=0$ , there won't be a $\delta$ , because, the set $\{|c-\frac{1}{n}|:n\in \mathbb{Z}^+\}\backslash \{0\}$ will be $\emptyset$ . So for any other value except 0 for c, will give us a $\delta$ value which is the result in part (i)

I know you have the right idea. The mathematical expressions you are using don't do it justice. Just use words to express what you mean. THEN try and translate it into set notation.

 

[amilapsn]

OK. Thanks... I'll try... :-|