Delta-Epsilon Definition of a Limit...1

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Calculus
Section 1.7

Can you work out 25 in step by step fashion as a guide for me to do a few more on my own?

Screenshot_20220422-230705_Samsung Notes.jpg
 
25.

A function f is said to have a limit L as x approaches c, denoted
upload_2022-4-23_17-48-11.jpeg
, if for every ε>0, there exists a δ>0 such that |x−c|<δ implies |f(x)−L|<ε.

Then, to prove that
upload_2022-4-23_17-48-49.jpeg
, we must show that for any ε>0 there exists δ>0 such that |x−0|<δ implies
upload_2022-4-23_17-49-59.jpeg
.


Proof:
Let ε>0 be arbitrary, and let δ=min{ε,1}.

Suppose |x−0|=|x|<δ. Note that as δ≤1, we have |x|<1, meaning
upload_2022-4-23_17-50-37.jpeg
. Furthermore, as δ≤ε, we have |x|<ε. With those facts:
upload_2022-4-23_17-51-20.jpeg

Thus, for an arbitrary ε>0, we have found a δ>0 such that |x−0|<δ implies
, meaning
upload_2022-4-23_17-51-38.jpeg


.
 
25.

A function f is said to have a limit L as x approaches c, denoted View attachment 2706 , if for every ε>0, there exists a δ>0 such that |x−c|<δ implies |f(x)−L|<ε.

Then, to prove that View attachment 2707 , we must show that for any ε>0 there exists δ>0 such that |x−0|<δ implies View attachment 2708.


Proof:
Let ε>0 be arbitrary, and let δ=min{ε,1}.

Suppose |x−0|=|x|<δ. Note that as δ≤1, we have |x|<1, meaning View attachment 2709. Furthermore, as δ≤ε, we have |x|<ε. With those facts:
View attachment 2710
Thus, for an arbitrary ε>0, we have found a δ>0 such that |x−0|<δ implies
, meaning View attachment 2711

.

Thank you but I will need more time to review what you are saying here.
 
25.

A function f is said to have a limit L as x approaches c, denoted View attachment 2706 , if for every ε>0, there exists a δ>0 such that |x−c|<δ implies |f(x)−L|<ε.

Then, to prove that View attachment 2707 , we must show that for any ε>0 there exists δ>0 such that |x−0|<δ implies View attachment 2708.


Proof:
Let ε>0 be arbitrary, and let δ=min{ε,1}.

Suppose |x−0|=|x|<δ. Note that as δ≤1, we have |x|<1, meaning View attachment 2709. Furthermore, as δ≤ε, we have |x|<ε. With those facts:
View attachment 2710
Thus, for an arbitrary ε>0, we have found a δ>0 such that |x−0|<δ implies
, meaning View attachment 2711

.

I will do my best to do all the even numbers here. If I am wrong or get stuck, I would like your help or guidance.
 

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