Bolz said:Homework Statement
If f(x) = 3x+1 en assume δ > 0. Assume ε>0.
Give a δ > 0 with the following property :
|x-1|< δ => |f(x) - f(1)| < ε
Homework Equations
The Attempt at a Solution
|f(x) - f(1)| < ε
<=> |3x+1-(3*1+1)| < ε
<=> |3x-3| < ε
<=> |x-1| < ε/3
|x-1| < δ
|x-1| < ε/3
=> δ=ε/3>0
What am I doing wrong?
pasmith said:Nothing. Your solution is correct.
HallsofIvy said:Then it would have been better to ask "Is this correct" rather than "What am I doing wrong"! You are not going to do very well assuming that you cannot do the work.
The purpose of finding delta in terms of epsilon-delta definition is to precisely define the concept of a limit in calculus. This definition allows us to rigorously prove that a function approaches a certain value as the input approaches a specific value.
In the epsilon-delta definition, delta represents the distance between the input value and the limit value, while epsilon represents the maximum difference between the output value and the limit value. In other words, delta specifies the range in which the input can vary, and epsilon specifies the acceptable range for the output.
The main steps in finding delta in the epsilon-delta definition are: 1) Start with the epsilon, which represents the desired precision or accuracy. 2) Use algebraic manipulation to isolate delta in terms of epsilon. 3) Simplify the resulting expression to find the specific value of delta that satisfies the definition.
No, delta cannot be any value in the epsilon-delta definition. It must be a positive number that is chosen carefully to satisfy the definition. This value is often dependent on both the function being evaluated and the chosen value for epsilon.
Finding delta in the epsilon-delta definition provides a precise and rigorous way to prove the existence of a limit. By finding a specific value for delta that satisfies the definition, we can show that the function approaches a specific limit value as the input approaches a certain value, providing evidence for the existence of a limit.