Finding Limits Using the Epsilon-Delta Method

[DumpmeAdrenaline]

Homework Statement

Suppose f is defined on [a,b] and that Lim x->c f(x)=L for some c in (a,b). Choose any two numbers m and M subject to the condition m<L<M. Show that there exists a number δ such that m<f(x)<M provided that 0< |x-c| <δ

Relevant Equations

Lim x->c f(x)=L

Lim x->c f(x)=L means that for a given ϵ we can find a δ such that when |x-c|<δ-> |f(x)-L|<ϵ. To satisfy the criterion m<f(x)<M we choose ϵ=min (L-m, M-L) and for that ϵ we determine a δ.
m<f(x)<M
m-L<f(x)-L<M-L
|m-L|<|f(x)-L|<|M-L|
|L-m|<|f(x)-L|<|M-L|
|L-m|<|L|+|m|
|f(x)-L|<|f(x)|+|L|
|M-L|<|L|+|M|
Putting this altogether we obtain
|m|<|f(x)|<|M|
Is this proof complete?
How is the epsilon delta method helping us in finding the Limiting value which is L in this case? In the epsilon delta we say that we are content if we could be sure that the difference between the f(x) and L is ϵ which I don't know how we got in the first place. We would in turn be interested in knowing how much x would deviate from c to be and still have f(x) in the required range. In short, the methods goal is solving an inequality to meet an implication.

 

[andrewkirk]

Is this proof complete?

I can't see any proof in the post.
The problem asks you to prove that under certain conditions we have m<f(x)<M. So you need a series of logically valid steps, starting with the premises you were given and ending with that inequality. But you have started by asserting that inequality, in your 3rd line. So you certainly can't claim to have proved it.

 

[FactChecker]

You went wrong somewhere. You stopped with a comparison of absolute values whereas the real proof should not have absolute values as the final result. Also, your result with absolute values can be wrong if m is a negative number with |m| > |M|.

 

[DumpmeAdrenaline]

Lim x->c f(x)=L means that for a given ϵ we can find a δ such that when |x-c|<δ-> |f(x)-L|<ϵ=> -ϵ<f(x)-L<ϵ
We are also given that m<L<M
ϵ≤M-L and ϵ≤L-m
We are not sure whether M-L≤L-m or L-m≤M-L. Suppose ϵ=min(L-m, M-L)=L-m. This makes the range of outputs bounded above by M and below by m. Since L-m is the minimum of the two intervals.Therefore, L-m<M-L
m-L<f(x)-L<L-m<M-L
Adding L to both sides does not change the signs of the inequality.
m<f(x)<L

 

[anuttarasammyak]

Lim x->c f(x)=L exists means f(x) is continuous around c, whether x=c is included or excluded which does not matter. For any tiny $\delta$>0 there exists $\epsilon$>0 such that
$$0< |x-c| < \epsilon$$
$$|f(x)-L| < \delta$$
You may choose $\delta$ so that
$$\delta < M-L$$ and
$$\delta < L-m$$

 

[SammyS]

@anuttarasammyak :
You have the roles of $\epsilon$ and $\delta$ reversed relative to what's given in the problem statement as well as what is generally used in the textbooks I'm familiar with.

Moreover, the problem statement does not imply that the function, f, is continuous at all.

 

[SammyS]

How is the epsilon delta method helping us in finding the Limiting value which is L in this case? In the epsilon delta we say that we are content if we could be sure that the difference between the f(x) and L is ϵ which I don't know how we got in the first place. We would in turn be interested in knowing how much x would deviate from c to be and still have f(x) in the required range. In short, the methods goal is solving an inequality to meet an implication.

Are you still so puzzled by how the $\varepsilon$ - $\delta$ definition of limit can be used to solve a problem such as this?

Your work in post #4 comes pretty close to touching on the correct points, but if you are merely reworking a solution to a similar problem without understanding what's going on, you won't get much out of the exercise.

 

[SammyS]

Lim x->c f(x)=L means that for a given ϵ we can find a δ such that when |x-c|<δ-> |f(x)-L|<ϵ=> -ϵ<f(x)-L<ϵ

You were given that $\displaystyle \lim_{x\to c} f(x) = L$. You then gave the $\varepsilon$ - $\delta$ definition for limit. That's good.
Some details: You should state that both $\varepsilon \ge 0$ and $\delta \ge 0$. Also we don't care about the value of $f(c)$ at all. - Might not be close to $L$. In general it might not be defined at all.
So that should state . . . when $0 < |x-c| < \delta$ . . .

We are also given that m<L<M
ϵ≤M-L and ϵ≤L-m
We are not sure whether M-L≤L-m or L-m≤M-L. Suppose ϵ=min(L-m, M-L)=L-m. This makes the range of outputs bounded above by M and below by m. Since L-m is the minimum of the two intervals. Therefore, L-m<M-L
m-L<f(x)-L<L-m<M-L

All of the above can be taken care of quite simply with
Let $\varepsilon = \text{min}(L-m, M-L)$. This gives you both $\varepsilon \le M-L$ and $\varepsilon \le L-m$ , the latter of which gives: $m-L\le -\varepsilon$.

Put those together with what you have at the end of your limit definition

-ϵ<f(x)-L<ϵ

 

[anuttarasammyak]

@anuttarasammyak :
You have the roles of $\epsilon$ and $\delta$ reversed relative to what's given in the problem statement as well as what is generally used in the textbooks I'm familiar with.

Moreover, the problem statement does not imply that the function, f, is continuous at all.

Thanks. I find a counter example to my misunderstanding.

f(c)=1
For other x
f(x)=0 for rational x
f(x)=x-c for irrational x
Then L=0 and f(x) is not continuous at all.

[EDIT]Say f(c)=0, f(x) is continuous at x=c by definition though it contradicts my primitive intuition.