Understanding the Implication in the Proof of a Theorem

[MIB]

This theorem I understand but I was only stuck in a simple implication in the proof

Theorem : If f is continuous at b and ${lim}_{x \rightarrow a} g(x) = b$ ,
then , ${lim}_{x \rightarrow a} f(g(x)) = f(b)$ .

proof

since f is continuous at b then , Given $\epsilon > 0$ , there exists $\delta_1 > 0$ such that ,

if $0 < |y-b|<\delta$ then $|f(x)-f(b)|<\epsilon$ ... 1

and since ${lim}_{x \rightarrow a} g(x) = b$ , then ther exist δ such that ,

if $0 < |x-a|<\delta$ then $|g(x)-b|<\delta_1$

it is easy to show that f is defined on some open interval containing a , the proplem is in the following implication ,
that is $|g(x)-b|<\delta_1$ implies $|f(x)-f(b)|<\epsilon$

Here I see that this true only if g(x) is in the domain of f to replace it with y that is there is some y such that y=g(x) where g(x) must takes all values in some interval containing b ( Right) , I see tat is true because because we can make $|g(x)-b|<\epsilon$ for arbitrary ε , that is even for very small ε so we must have then g(x) takes all values on some interval containing b . (Right)

Thanks

sorry I posted before I finished

Mod note: For absolute values, just use two | characters.

 

[Deveno]

help i would give but

 

[gb7nash]

Theorem : If f is continuous at b and $\stackrel{{\lim}}{{x\rightarrow a}}$

This makes no sense. If you want to state a limit in tex, use:

\lim_{x \to a}

inside the tex tags.

 

[MIB]

This makes no sense. If you want to state a limit in tex, use:

\lim_{x \to a}

inside the tex tags.

I corrected the post .

Thanks

 

[MIB]

Here I see that this true only if g(x) is in the domain of f to replace it with y that is there is some y such that y=g(x) where g(x) must takes all values in some interval containing b ( Right) , I see tat is true because because we can make $|g(x)-b|<\epsilon$ for arbitrary ε , that is even for very small ε so we must have then g(x) takes all values on some interval containing b . (Right)

Thanks

sorry I posted before I finished

Mod note: For absolute values, just use two | characters.

Now I think that all what I say in the end was useless but I am not sure , the definition says that if a function f is defined on some open containing a except possibly at a , then we write

$\lim_{x \to a} f(x) = L$

if for every ε > 0 , there exists positive number δ such that

if $0< |x-a|<\delta$ then $|f(x)-L|<\epsilon$

here in definition it says that for any x which satisfy the first inequality x must be element of the domain of f . so I didn't need to justify that g(x) belong to the domain of f because if $|g(x)-b|<\delta_1$ , g(x) which satisfy this inequality will automatically element of dom(f) because any y such that $0 < |y-b|<\delta_1$ , that is the reason that we can replace y with g(x) where there is g(x) do the same job as y , so we have
whenever $0 < |x-a|<\delta$ , $|g(x)-b|<\delta_1$ which implies $|f(g(x))-f(b)|<\epsilon$ ( I wrote it wrong in my first post sorry I am not used to write on computer )

please tell me if I am understanding the definition well .

Thanks

 

[MIB]

Now I think that all what I say in the end was useless but I am not sure , the definition says that if a function f is defined on some open containing a except possibly at a , then we write

$\lim_{x \to a} f(x) = L$

if for every ε > 0 , there exists positive number δ such that

if $0< |x-a|<\delta$ then $|f(x)-L|<\epsilon$

here in definition it says that for any x which satisfy the first inequality x must be element of the domain of f . so I didn't need to justify that g(x) belong to the domain of f because if $|g(x)-b|<\delta_1$ , g(x) which satisfy this inequality will automatically element of dom(f) because any y such that $0 < |y-b|<\delta_1$ , that is the reason that we can replace y with g(x) where there is g(x) do the same job as y , so we have
whenever $0 < |x-a|<\delta$ , $|g(x)-b|<\delta_1$ which implies $|f(g(x))-f(b)|<\epsilon$ ( I wrote it wrong in my first post sorry I am not used to write on computer )

please tell me if I am understanding the definition well .

Thanks

IS This argument is right

 

[RandomMystery]

"Given ϵ>0 , there exists δ1>0 "
Where did ϵ and δ come from?
I don't recall learning this at all when learning limits.

 

[mahmoud2011]

"Given ϵ>0 , there exists δ1>0 "
Where did ϵ and δ come from?
I don't recall learning this at all when learning limits.

I don't understand you . How did you learn limits ?

 

[HallsofIvy]

"Given ϵ>0 , there exists δ1>0 "
Where did ϵ and δ come from?
I don't recall learning this at all when learning limits.

The definition of "$\lim_{x\to a} f(x)= L$" is

"Given $\epsilon> 0$, there exist $\delta> 0$ such that if $0< |x- a|< \delta$, then $|f(x)- L|< \epsilon$."


If you don't know the definition of "limit" you can't prove anything about them! So, if you did not learn that definition, what definition did you learn?

 

[MIB]

another question please , when I want to prove statements like that

$\lim_{x \to a} f(x) = \lim_{h \to 0} f(a+h)$

MUST I prove the two following statements or one of them is enough

1 - Assume that $\lim_{x \to a} f(x) = L$ , and peove that given ε > 0 , there exists δ > 0 such that ,

if $0 < |h| < \delta$ then $|f(a+h) - L | < \epsilon$

2- Assume that $\lim_{h \to 0} f(a+h) = L$ , and prove that given ε > 0 , there exists δ > 0 such that ,

if $0 < |x-a| < \delta$ then $|f(x) - L | < \epsilon$

the roof of these statements is easy .

I see it is important to prove the two statements , But I want to be sure , so please help me , if one of them is enough tell me .

 

[HallsofIvy]

Assuming the theorem actually says "If either of these limits exists, then the other also exists and they are equal" you need to prove both of the statements. If, on the other hand, you were given that a specific one exists, you would only need to prove one, depending upon which was given to exist. Of course, if you are not given that either exists, the statement may not be true.

 

[RandomMystery]

Well, we just learned that the

lim x -> a

notation simply means that the x value is approaching a, but it isn't a. So it is the value right next to a.

I guess it's just a practical and less formal way of saying the same thing, although simply saying that x minus a is greater than zero but less than some random symbol doesn't seem to prove of anything (at least to me)
For example:

"Given ϵ>0, there exist
δ>0 such that if
0<|x−a|<δ, then
|f(x)−L|<ϵ."

Okay, so δ can equal 9999 and ϵ can equal 42?

That would mean that there are over 9000! values for the lim x -> a.


The problem is that something is being implied about δ and ϵ that isn't stated in the proof, making it null for anyone who doesn't know that the

limit of δ and ϵ is zero (which is what I'm implying)

which is silly since this proof of the limit requires using a limit, which you would then have to proof, which then...!

 

[mahmoud2011]

Well, we just learned that the

lim x -> a

notation simply means that the x value is approaching a, but it isn't a. So it is the value right next to a.

I guess it's just a practical and less formal way of saying the same thing, although simply saying that x minus a is greater than zero but less than some random symbol doesn't seem to prove of anything (at least to me)
For example:

"Given ϵ>0, there exist
δ>0 such that if
0<|x−a|<δ, then
|f(x)−L|<ϵ."

Okay, so δ can equal 9999 and ϵ can equal 42?

That would mean that there are over 9000! values for the lim x -> a.The problem is that something is being implied about δ and ϵ that isn't stated in the proof, making it null for anyone who doesn't know that the

limit of δ and ϵ is zero (which is what I'm implying)

which is silly since this proof of the limit requires using a limit, which you would then have to proof, which then...!

I can't understand what you want to say . This definition you use that is that we can make f(x) close to L by making x sufficiently close to a , is not a formal definition , it is used only in textbooks when it introduce you to the idea of limits , although you can think limits in this way , but you can't depend it in proving theorems , try to use it to prove that

$\lim_{x \to a} f(x)g(x) = \lim_{x \to a} f(x) \lim_{x \to a} g(x)$
assuming $\lim_{x \to a} f(x)$ and $\lim_{x \to a} g(x)$ exists .

note that we have a problem with your definition that is we don't know how close we must be to a such that f(x) is close to L . here the formal definition

The definition of "$\lim_{x\to a} f(x)= L$" is

"Given $\epsilon> 0$, there exist $\delta> 0$ such that if $0< |x- a|< \delta$, then $|f(x)- L|< \epsilon$."

the definition can be interpreted i many ways I will leave the geometric interpretions now

it says that if we choose any ε no matter how ε is small , , we can find δ such that if the distance between x and and a is less than δ the distance the distance between f(x) and L is less than ε .

Try to interpret it using interval notation , using this definition you can prove the known limit laws . I will make an example , but not know because I am busy .

 

[mahmoud2011]

In the attachments I wrote to you interpretions for the definition , in addition to the proof of uniqueness the value f(x) approaches as x approaches .