- #1

DumpmeAdrenaline

- 78

- 2

- 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.

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.