proof in lim

[orir]

f and g are functions defined in a Punctured neighbourhood of x_0, and L is a real number.
assuming lim (x->x₀) (f*g)(x)=L

prove that lim (x->x₀) f(x) exists, or lim (x->x₀) g(x) exists.

i don't know how to start. thanks for your help...

 

[HallsofIvy]

Let g(x)= f(x)= 1 if x is irrational, -1 if x is rational.

I am assuming that "f*g" means "f times g". Do you mean, instead, "fog(x)= f(g(x))"

 

[orir]

Let g(x)= f(x)= 1 if x is irrational, -1 if x is rational.

I am assuming that "f*g" means "f times g". Do you mean, instead, "fog(x)= f(g(x))"

yes, i ment "f times g".
but i didn't get your answer.

 

[pka]

yes, i ment "f times g".
but i didn't get your answer.

Here is a variation on Halls' example:
\(\displaystyle f(x) = \left\{ {\begin{array}{rl}{1,}&x\text{ is rational}\\{ - 1,}&\text{else}\end{array}} \right.\) and \(\displaystyle g(x) = \left\{ {\begin{array}{rl}{-1,}&x\text{ is rational}\\{ 1,}&\text{else}\end{array}} \right.\)

Note that \(\displaystyle \forall x,~~(f \cdot g)(x)=-1\).

Ask yourself if \(\displaystyle \displaystyle{\lim _{x \to {x_0}}}f \cdot g(x) = - 1\) is true?

Does either \(\displaystyle \displaystyle{\lim _{x \to {x_0}}}f (x)\) or \(\displaystyle \displaystyle{\lim _{x \to {x_0}}} g(x)\) exists?

 

[orir]

Here is a variation on Halls' example:
\(\displaystyle f(x) = \left\{ {\begin{array}{rl}{1,}&x\text{ is rational}\\{ - 1,}&\text{else}\end{array}} \right.\) and \(\displaystyle g(x) = \left\{ {\begin{array}{rl}{-1,}&x\text{ is rational}\\{ 1,}&\text{else}\end{array}} \right.\)

Note that \(\displaystyle \forall x,~~(f \cdot g)(x)=-1\).

Ask yourself if \(\displaystyle \displaystyle{\lim _{x \to {x_0}}}f \cdot g(x) = - 1\) is true?

Does either \(\displaystyle \displaystyle{\lim _{x \to {x_0}}}f (x)\) or \(\displaystyle \displaystyle{\lim _{x \to {x_0}}} g(x)\) exists?

no, i guess none of them exists. (Am I right? )
but is it just ok saying that, or i need to prove that also someway?

btw, is this the only (and easy) way to solve this? i mean, i wanna know if thinking of using those specific functions is something i should've thought and needed to practice.

 

[pka]

no, i guess none of them exists. (Am I right? )
but is it just ok saying that, or i need to prove that also someway?
btw, is this the only (and easy) way to solve this? i mean, i wanna know if thinking of using those specific functions is something i should've thought and needed to practice.

This is called a counter-example.
It is just one of many possible.
But one is enough to show that the original statement is false.

Thus this not a proof. It is a falsification.

 

[orir]

Here is a variation on Halls' example:
\(\displaystyle f(x) = \left\{ {\begin{array}{rl}{1,}&x\text{ is rational}\\{ - 1,}&\text{else}\end{array}} \right.\) and \(\displaystyle g(x) = \left\{ {\begin{array}{rl}{-1,}&x\text{ is rational}\\{ 1,}&\text{else}\end{array}} \right.\)

Note that \(\displaystyle \forall x,~~(f \cdot g)(x)=-1\).

Ask yourself if \(\displaystyle \displaystyle{\lim _{x \to {x_0}}}f \cdot g(x) = - 1\) is true?

Does either \(\displaystyle \displaystyle{\lim _{x \to {x_0}}}f (x)\) or \(\displaystyle \displaystyle{\lim _{x \to {x_0}}} g(x)\) exists?

and - is this fits the assumption the functions are Punctured neighbourhood? Aren't your f and g defined in every neighbourhood of x₀?

 

[HallsofIvy]

Do you know what "punctured neighborhood" means? If you do you should be able to answer that question yourself.

 

[orir]

Do you know what "punctured neighborhood" means? If you do you should be able to answer that question yourself.

i'd be glad to know if i'm wrong - what i know is that when a function is defined in punctured neighbourhood of X it means that it isn't defined in {X} itself.
am i wrong?

 

[pka]

i'd be glad to know if i'm wrong - what i know is that when a function is defined in punctured neighborhood of X it means that it isn't defined in {X} itself. am i wrong?

A punctured neighborhood of \(\displaystyle a\) is just the set \(\displaystyle \{x:0<|a-x|< \epsilon\}=(a-\epsilon,a)\cup(a,a+\epsilon) \), where \(\displaystyle \epsilon>0\). It is punctured at \(\displaystyle x=a\) by not containing \(\displaystyle a\).

Note that the function \(\displaystyle f(x)=x^2\) is defined in every punctured neighborhood, period.

Whereas, the function \(\displaystyle g(x)=\dfrac{x^2-1}{x-1}\) is defined any punctured neighborhood of \(\displaystyle x=1\) even though \(\displaystyle g\) is not defined at \(\displaystyle x=1\).

Thus saying "a function is defined in punctured neighborhood of \(\displaystyle x=a\) does not imply that the function is not defined at \(\displaystyle x=a\).

 

[orir]

A punctured neighborhood of \(\displaystyle a\) is just the set \(\displaystyle \{x:0<|a-x|< \epsilon\}=(a-\epsilon,a)\cup(a,a+\epsilon) \), where \(\displaystyle \epsilon>0\). It is punctured at \(\displaystyle x=a\) by not containing \(\displaystyle a\).

Note that the function \(\displaystyle f(x)=x^2\) is defined in every punctured neighborhood, period.

Whereas, the function \(\displaystyle g(x)=\dfrac{x^2-1}{x-1}\) is defined any punctured neighborhood of \(\displaystyle x=1\) even though \(\displaystyle g\) is not defined at \(\displaystyle x=1\).

Thus saying "a function is defined in punctured neighborhood of \(\displaystyle x=a\) does not imply that the function is not defined at \(\displaystyle x=a\).

so let me see if i got this clear -it's correct to say that sin(x) is also defined in punctured neighborhood of \(\displaystyle x=0\), even though it has a value in \(\displaystyle x=0\) \(\displaystyle f(x)=0\)?

 

[pka]

so let me see if i got this clear -it's correct to say that sin(x) is also defined in punctured neighborhood of \(\displaystyle x=0\), even though it has a value in \(\displaystyle x=0\) \(\displaystyle f(x)=0\)?

Maybe you should tell us why you think \(\displaystyle f(x)=0\) has anything to do with this.

The function \(\displaystyle \sin(x)\) is defined everywhere. Correct?
Then function \(\displaystyle \sin(x)\) is defined in any neighborhood of \(\displaystyle x=0\) . Correct?
So function \(\displaystyle \sin(x)\) is defined in any punctured neighborhood of \(\displaystyle x=0\) . Correct?

 

[orir]

Maybe you should tell us why you think \(\displaystyle f(x)=0\) has anything to do with this.

The function \(\displaystyle \sin(x)\) is defined everywhere. Correct?
Then function \(\displaystyle \sin(x)\) is defined in any neighborhood of \(\displaystyle x=0\) . Correct?
So function \(\displaystyle \sin(x)\) is defined in any punctured neighborhood of \(\displaystyle x=0\) . Correct?

correct. i'm sorry i'm a nagger, i just thought that it necessarily means that the function is not defined at this point (as if a hole or something)...