Given that
lim
x→3
(5x − 14) = 1,
illustrate Definition 2 by finding values of δ that correspond to ε = 0.5, ε = 0.1, and ε = 0.05.
So, f(x)→1+ε, that means
5(x)-14 → 1+ε
x → 15+ε
x → 3+ε/5
so, δ = ε/5
To illustrate Definition 2 for the given limit problem, let's first understand what Definition 2 represents. Definition 2 states that a limit L exists as x approaches a value c if, for every positive value of ε, there exists a positive value δ such that if 0 < |x - c| < δ, then |f(x) - L| < ε. In simpler terms, it means that we can get as close as we want to the limit value L by getting sufficiently close to c (within a certain range defined by δ).
In this case, the limit is given as:
lim
x→3
(5x − 14) = 1
We need to find values of δ that correspond to different values of ε. Let's start with ε = 0.5:
|f(x) - 1| < 0.5
To find the corresponding value of δ, we need to manipulate the equation and solve for δ. Let's work through it step by step:
|5x - 14 - 1| < 0.5
|5x - 15| < 0.5
-0.5 < 5x - 15 < 0.5
Now we solve the inequality:
-0.5 + 15 < 5x < 0.5 + 15
14.5 < 5x < 15.5
Divide all sides by 5:
2.9 < x < 3.1
So, for ε = 0.5, we see that if x is within the interval (2.9, 3.1), the corresponding values of f(x) will be within the range of 0.5 units from the limit, which is 1.
Now let's find the values of δ for ε = 0.1:
|f(x) - 1| < 0.1
Following the same steps as before, we manipulate the inequality and solve for δ:
|5x - 14 - 1| < 0.1
|5x - 15| < 0.1
-0.1 < 5x - 15 < 0.1
Solving the inequality:
-0.1 + 15 < 5x < 0.1 + 15
14.9 < 5x < 15.1
Dividing all sides by 5:
2.98 < x < 3.02
So, for ε = 0.1, we need x to be within the interval (2.98, 3.02) in order to ensure that the corresponding values of f(x) are within 0.1 units from the limit, which is 1.
Lastly, let's find the values of δ for ε = 0.05:
|f(x) - 1| < 0.05
Applying the same steps:
|5x - 14 - 1| < 0.05
|5x - 15| < 0.05
-0.05 < 5x - 15 < 0.05
Solving the inequality:
-0.05 + 15 < 5x < 0.05 + 15
14.95 < 5x < 15.05
Dividing all sides by 5:
2.99 < x < 3.01
So, for ε = 0.05, we need x to be within the interval (2.99, 3.01) to ensure that the corresponding values of f(x) are within 0.05 units from the limit, which is 1.
In summary:
- For ε = 0.5: x must be within (2.9, 3.1)
- For ε = 0.1: x must be within (2.98, 3.02)
- For ε = 0.05: x must be within (2.99, 3.01)
These intervals of x values (certain range) indicate how close we need to be to the value x = 3 in order to ensure that the corresponding values of f(x) are within the desired range of ε from the limit value.