Proove that a function is cointinuous at one point

[Hernaner28]

Hi, I've been trying to proof this with a partner and we couldn't, we tried to apply the definition of limit and continuity but we didn't get anything. This is the problem:

"Let f a function which satisfies $$|f(x)|\leq|x| \forall{x\in{\mathbb{R}}}$$

Proof that is continuous at 0.


Thanks!


PD: I forgot to say that I figured out what f(0) is. And it's obviously 0 because the absolute value cannot be less than 0:

$$|f(0)|<=|0|$$ then $$f(0)=0$$ . But I just don't know how to apply the definition.. do I have to find the epsilon or delta? Thanks for answering!

 

[LCKurtz]

Hi, I've been trying to proof this with a partner and we couldn't, we tried to apply the definition of limit and continuity but we didn't get anything. This is the problem:

"Let f a function which satisfies $$|f(x)|\leq|x| \forall{x\in{\mathbb{R}}}$$

Proof that is continuous at 0.


Thanks!

Have you figured out what value f(0) is? Do that and then show us what you get when you apply the $\epsilon-\delta$ definition for continuity at 0.

 

[Hernaner28]

Oh yeah! I forgot to say that I figured out what f(0) is. And it's obviously 0 because the absolute value cannot be less than 0:

$$|f(0)|<=|0|$$ then $$f(0)=0$$ . But I just don't know how to apply the definition.. do I have to find the epsilon or delta? Thanks for answering!

 

[LCKurtz]

Oh yeah! I forgot to say that I figured out what f(0) is. And it's obviously 0 because the absolute value cannot be less than 0:

$$|f(0)|<=|0|$$ then $$f(0)=0$$ . But I just don't know how to apply the definition.. do I have to find the epsilon or delta? Thanks for answering!

Write out the definition, word for word, for a function f(x) to be continuous at $x=a$. Just copy it from your text. Then put $a=0$ in that statement and see if it is true with what you are given.

 

[Hernaner28]

It should be like this:

$$\forall{\epsilon>0}\textrm{ } \exists{ \delta>0}: \textrm{if } 0<|x-0|<\delta \Longrightarrow{|f(x)-0|< \epsilon}$$

So:
$$|x|< \delta$$
and
$$|f(x)|< \epsilon$$


??

 

[LCKurtz]

It should be like this:

$$\forall{\epsilon>0}\textrm{ } \exists{ \delta>0}: \textrm{if } 0<|x-0|<\delta \Longrightarrow{|f(x)-0|< \epsilon}$$

So:
$$|x|< \delta$$
and
$$|f(x)|< \epsilon$$


??

So given $\epsilon>0$, you have to find a $\delta$ depending on $\epsilon$ such that if $|x|<\delta$ then $|f(x)|<\epsilon$. Can you come up with a $\delta$ which works for your function given what you know about it?

 

[Hernaner28]

No, I cannot come up with a solution! And I'm sure this is really simple! Can delta and epsilon be zero?

 

[emailanmol]

|f(x)|<=|x|

So draw two lines.
y = x and y=-x.


The only place where f(x) can lie is the area enclosed between these two lines.(How?)

What do you see?


The basic idea is that if x<=|a| (let a be positive )

then, x lies between -a and a. (with -a and a included)

If x>=|a| then,

x lies between -infinity to -a and a to infinity
(with -a and a included)

So when |f(x)| <=|x|


what do you get the range of f(x).

What does this show for
x---> 0. ?

 

[Robert1986]

OK, try taking a concrete example and see if you can find a $\delta$ that way. For example, let $f(x) = |x|$ clearly $|f(x)| \leq |x| \forall x \in R$. Now, can you find a $\delta$ such that $|f(x) - 0| < \epsilon$ whenever $|x| < \delta$?If so, can you see how this can be genralised (it is a very short generalsation) to your problem?If you can't come up with a $\delta$ I suggest taking a break from this problem and come back to it in a few hours or tomorrow.

 

[Mark44]

Oh yeah! I forgot to say that I figured out what f(0) is. And it's obviously 0 because the absolute value cannot be less than 0:

$$|f(0)|<=|0|$$ then $$f(0)=0$$ .

You mean f(0) >= 0, not <= 0.
This doesn't necessarily imply that |0| = 0. All that you have shown here is that it can't be smaller than 0. You haven't eliminated or dealt with the possibility that it could be larger than 0.

A simpler explanation uses the idea that the absolute value represents the distance between a given number and zero. Obviously the distance from 0 to 0 is 0, so |0| = 0.

But I just don't know how to apply the definition.. do I have to find the epsilon or delta? Thanks for answering!

 

[Hernaner28]

OK, try taking a concrete example and see if you can find a $\delta$ that way. For example, let $f(x) = |x|$ clearly $|f(x)| \leq |x| \forall x \in R$. Now, can you find a $\delta$ such that $|f(x) - 0| < \epsilon$ whenever $|x| < \delta$?If so, can you see how this can be genralised (it is a very short generalsation) to your problem?If you can't come up with a $\delta$ I suggest taking a break from this problem and come back to it in a few hours or tomorrow.

I made a graphic and analised the function |x| but in that case the delta has to be negative from the left and positive fro the right so the limit is zero. I don't understand what the delta means! You're talking about it as if it was an unique number... I thought these numbers (epsilon and delta) could be chosen freely and one depends on the other but they're not unique, just a range. It's the first time I am proving this kind of things and I know it's basic and will be essential so please don't make me think things I won't realize, I need a demonstration first. I really appreciate your help but I've been trying all day... this is getting me irritated. I cannot think anymore ... lol

 

[LCKurtz]

No, I cannot come up with a solution! And I'm sure this is really simple! Can delta and epsilon be zero?

No. You know that $|f(x)|\le|x|$ for your function. How small does $\delta$ need to be so that if $|x|<\delta$, then $|f(x)|<\epsilon$? It is really easy if you just look at it.

 

[LCKurtz]

Oh yeah! I forgot to say that I figured out what f(0) is. And it's obviously 0 because the absolute value cannot be less than 0:

$$|f(0)|<=|0|$$ then $$f(0)=0$$ . But I just don't know how to apply the definition.. do I have to find the epsilon or delta? Thanks for answering!

You mean f(0) >= 0, not <= 0.
This doesn't necessarily imply that |0| = 0. All that you have shown here is that it can't be smaller than 0. You haven't eliminated or dealt with the possibility that it could be larger than 0.

No, he means $|f(0)|\le 0$ like he wrote and his argument is correct that $f(0)=0$.

 

[Hernaner28]

Alright.. I made a horrible paint drawing which I attached. Is that representation right? I think I'm not imagining well how it works graphically...

 

[Mark44]

No, he means $|f(0)|\le 0$ like he wrote and his argument is correct that $f(0)=0$.

I stand corrected. I was laboring under the false impression that f(x) = |x|.

 

[emailanmol]

Read each of the posts again.
Every member has said some really useful point and arranging these like jigsaws will solve your puzzle.

As i said in my post before,

You get f(x) as lying between x and -x(how?)

Meaning f(2) can lie anywhere -2 and 2

As x tends to 0(from the RHS),

lets say x is at 0.001
f(x) can lie between -0.001 and 0.001.

What do you see when x gets really small as in 0?

Once you have figured this out graphically and logically you can start working back on your method using delta's which will look much easier.

 

[Hernaner28]

Oh yeah! Thanks! I think I understood the condition graphically... take a look at my attached drawing. Now I will analise it and see if I can come up with something. thanks!

 

[Hernaner28]

So the delta has to be zero! Why not?

 

[A. Bahat]

Isn't part of the definition of continuity "there exists a delta greater than zero"? So delta cannot equal zero.

 

[LCKurtz]

No. You know that $|f(x)|\le|x|$ for your function. How small does $\delta$ need to be so that if $|x|<\delta$, then $|f(x)|<\epsilon$? It is really easy if you just look at it.

So the delta has to be zero! Why not?

You were within one step of being done with this problem. Have you given any thought to the line I re-quoted above?

 

[Hernaner28]

Yeah I've thought about it and made the graphic which I attached above. I know neither delta nor epsilon can be zero, I mean that delta can be as small as zero (but not zero) so that way x will tend to zero and so will f(x), why is that wrong'?

 

[LCKurtz]

You are given $|f(x)|\le |x|$. How small does $|x|$ need to be to guarantee $|f(x)|<\epsilon$? It is really trivial.

 

[Hernaner28]

Could be that having $$|f(x)|\leq |x|$$ if $$|x|\neq |f(x)|$$ and $$|x|=\epsilon$$
then $$|f(x)|<\epsilon$$?

 

[LCKurtz]

You want $|f(x)| < \epsilon$. What if $|x|<\epsilon$? Would that do it?

 

[Hernaner28]

Yes I thought about that too.. but what's the point of this? What I wrote before is also right, isn't it?

 

[LCKurtz]

If it's right, it is sufficiently poorly written that it isn't clear to me that you actually understand it yet. Remember what you were trying to do: Given $\epsilon>0$ find a $\delta$ such that if $|x - 0 |<\delta$ then $|f(x) - f(0)|<\epsilon$. Since $f(0)=0$ this reduces to finding saying find a $\delta$ such that if $|x|<\delta$ then $|f(x)|<\epsilon$. Since you are given $|f(x)|\le |x|$, if you choose $\delta =\epsilon$ and take $|x|<\delta$, you have $|f(x)|\le |x|<\epsilon$. That is all there is to it and that is also how it should be written up. You might want to ponder that until you are sure you understand it.

 

[Hernaner28]

Oh yes! It was really trivial! I'm so ashamed ... well, I really thank you for the help. Now that you've showed me how these demonstrations work I will continue with another which states the following:

"Suppose g is continuous at 0, $$g(0)=0$$ and $$|f(x)|\leq |g(x)|$$. Proove f is continuous at 0."

I'll do my best to prove this! I'll write back when I get something!
THANKS VERY VERY MUCH everyone for your help!

 

[Hernaner28]

I've done this so far and I think it's the same situation as above (same demonstration):

We have:

$$\textrm{g continuous at 0 and g(0)=0}\Longleftrightarrow{\displaystyle\lim_{x \to{0}}{g(x)}=g(0)=0} $$
$$ \left . \begin{matrix}{g(0)=0}\\{|f(x)|\leq |g(x)|}\end{matrix}\right \} \Longrightarrow{}f(0)=0$$

And I have to prove that f is cont. at 0.
So I have to fid a delta such that:
$$\forall{\varepsilon>0}\textrm{ } \exists{ \delta>0}: \textrm{if } 0<|x-0|<\delta \Longrightarrow{|f(x)-0|< \varepsilon}$$

And I'm in the same problem as before, right? I just choose δ equal ε . If this is OK then I don't understand why it tells me that g is cont. at 0 because I never used that.

 

[LCKurtz]

If you want to do another problem, it would be best to start a new thread. (Actually that is in the forum rules). I'm about to turn my computer off and watch the NCAA basketball tournament. People are more likely to read a new thread than one that has gone on as long as this.

 

[Hernaner28]

alright! I will make a new thread! Thanks!

 

[emailanmol]

I've done this so far and I think it's the same situation as above (same demonstration):

We have:

$$\textrm{g continuous at 0 and g(0)=0}\Longleftrightarrow{\displaystyle\lim_{x \to{0}}{g(x)}=g(0)=0} $$
$$ \left . \begin{matrix}{g(0)=0}\\{|f(x)|\leq |g(x)|}\end{matrix}\right \} \Longrightarrow{}f(0)=0$$

And I have to prove that f is cont. at 0.
So I have to fid a delta such that:
$$\forall{\varepsilon>0}\textrm{ } \exists{ \delta>0}: \textrm{if } 0<|x-0|<\delta \Longrightarrow{|f(x)-0|< \varepsilon}$$

And I'm in the same problem as before, right? I just choose δ equal ε . If this is OK then I don't understand why it tells me that g is cont. at 0 because I never used that.

Hey.
g(x) is continuos is reauired.
Your proof uptil now hasn't painted the whole picture.

Think it this way.
g(0) is 0.

Now if g(x) wasn't continuos g(0.000001) [taking v small value)could hve had value 50000.

In that case f(0.000001) could have been anywhere between -50,000 and 50,000 which would make f(x) non-continuos like g(x).

The basic method is to realize the f(x) gets sandwiched as it reaches 0.

You can prove this by the graphical method and also the algebraic method (which is more important as you camnot always draw graphs) of your previous question.

The method should be same as what you wrote for your last term.
Just replace x with g(x) everywhere.

Also make sure the algebra method makes sense to you.It's important you realize how you got the answer and ask yourself if the answer has loopholes :-)