Hi,
I am having trouble understanding the concept of delta and epsilon. Please explain the following problem. Thank you!!
Given the function f(x)=2-(1/x), find delta such that if 0< Ix-2I < delta, then the absolute value of f(x)-1 is < 0.1
You want
|f(x)-1| < 0.1
|2-(1/x)-1)| < 0.1
|1 - 1/x| < 0.1
Now, let x = 2+δ
|1 - 1/(2+δ)| < 0.1
|2+δ-1| < 0.1(2+δ)
|δ+1| < 0.2 + 0.1δ
we know that δ+1 > 0, so that means
δ+1 < 0.1 + 0.1δ
0.9δ < -0.9
δ < -1
This makes no sense. Since f(1/2) = 0, I think you want |x - 1/2| < δ
So, go through the steps and you should come up with a small value for δ. You might consider this page:
http://www.wolframalpha.com/input/?i=%7C2-1%2Fx%7C+%3C+0.1
To understand this problem, let's break it down and identify the key concepts involved.
The problem involves the use of delta and epsilon to express the limit of a function. These concepts are commonly used in calculus to define the behavior of a function as it approaches a particular value or point.
In this case, we are given the function f(x) = 2 - (1/x) and we need to find a value for delta such that if the difference between x and 2, denoted as |x - 2|, is less than delta, then the absolute value of f(x) - 1, denoted as |f(x) - 1|, is less than 0.1.
To achieve this, we will systematically follow these steps:
Step 1: Rewrite the given inequality in terms of f(x) and x.
|f(x) - 1| < 0.1 can also be expressed as -0.1 < f(x) - 1 < 0.1.
Step 2: Substitute the expression for f(x) into the inequality.
-0.1 < 2 - (1/x) - 1 < 0.1
Step 3: Simplify the inequality by combining like terms.
-0.1 < 1 - (1/x) < 0.1
Step 4: Isolate the term containing (1/x) by subtracting 1 from all parts of the inequality.
-1.1 < - (1/x) < -0.9
Step 5: Divide all parts of the inequality by -1 to change the signs.
0.9 < 1/x < 1.1
Step 6: Take the reciprocal of all parts of the inequality to get rid of the fraction.
1/1.1 < x < 1/0.9
Step 7: Simplify the fractions.
10/11 < x < 10/9
Step 8: Determine the range of x-values that satisfies the inequality.
For any x such that 10/11 < x < 10/9, the absolute value of f(x) - 1 will be less than 0.1.
Therefore, delta in this problem can be chosen as the minimum difference between 10/9 and 10/11, which is 1/99 or approximately 0.01. This means that if we take any value of x within the interval (10/11, 10/9), the difference between x and 2 will be less than 0.01, ensuring that the absolute value of f(x) - 1 is less than 0.1.