Solve Excercise 9 in AP Calculus: Find DELTA for |f(x)-L| < E

In summary, the conversation is about solving a limit question in Exercise 9 where a positive number E and the limit L of a function f at a are given. The goal is to find a number DELTA such that the absolute value of f(x) - L is less than E if the absolute value of x - a is less than DELTA. The given exercise is \lim_{x \rightarrow 4} 2x = 8; E = 0.1. The first step in solving this question is to write down what is known and what needs to be proven.
  • #1
Pseudo Statistic
391
6
Here's a limit question which I can't seem to figure out how to do, if someone can explain how to do this one question, I'd greatly appreciate it, as I barely understand this:

In Excercise 9, a positive number E and the limit L of a function f at a are given. Find a number DELTA such that |f(x) -L| < E if 0 < |x-a| < DELTA

Excercise 9:
[tex]\lim_{x \rightarrow 4} 2x = 8; E = 0.1[/tex]

Thanks.
 
Physics news on Phys.org
  • #2
Write down what you know and what you need to prove:

Find a [itex]\delta>0[/itex], such that:
[tex]|2x-8|<0.1[/tex] whenever [tex]|x-4|<\delta[/tex]

That's almost always the first step. Simply write it down. The answer may already be obvious from the above expression.
 
  • #3


To solve Exercise 9, we first need to understand the definition of a limit. The limit of a function f at a point a is the value that f approaches as x gets closer and closer to a. In other words, as x gets closer to a, the value of f(x) gets closer to the limit L.

In this exercise, we are given that the limit of 2x as x approaches 4 is 8, and we are asked to find a number DELTA such that |f(x)-L| < E if 0 < |x-a| < DELTA, where E is a positive number.

To find DELTA, we can use the definition of a limit. We know that as x approaches 4, the value of f(x) gets closer to 8. So, we can say that for any value of x that is close enough to 4, the difference between f(x) and 8 will be less than E.

Mathematically, we can write this as: |f(x)-8| < E

Now, we need to find a value of DELTA such that if |x-4| < DELTA, then |f(x)-8| < E. This means that we need to find a range of values for x that are close enough to 4 so that the difference between f(x) and 8 is less than E.

To do this, we can use the fact that the limit is 8. This means that if we choose a value of DELTA such that |x-4| < DELTA, then we can guarantee that |f(x)-8| < E.

So, for any value of E, we can choose DELTA to be a small enough value such that |x-4| < DELTA. This will ensure that |f(x)-8| < E.

In this exercise, E = 0.1. So, we can choose DELTA to be any value less than 0.1. For example, we can choose DELTA = 0.05.

Therefore, the number DELTA such that |f(x)-L| < E if 0 < |x-a| < DELTA is DELTA = 0.05.

I hope this explanation helps you understand how to solve Exercise 9 in AP Calculus. Remember, when finding DELTA, we need to choose a value that is small enough so that the difference between
 

What is Exercise 9 in AP Calculus?

Exercise 9 in AP Calculus is a problem involving finding the value of DELTA in order to satisfy a given epsilon value for the limit of a function, where the limit is approaching a specific value L.

What is DELTA in calculus?

In calculus, DELTA is a symbol that represents a small change or difference in a variable. It is typically used in the context of limits, where DELTA x represents a small change in the x variable.

How do you solve Exercise 9 in AP Calculus?

To solve Exercise 9 in AP Calculus, you will need to use the definition of a limit in order to find the appropriate value of DELTA. This involves setting up an equation with the given epsilon value, the limit value L, and the function f(x), and then solving for DELTA.

What is the definition of a limit in calculus?

The definition of a limit in calculus states that the limit of a function f(x) as x approaches a specific value L is equal to L if and only if for any given epsilon value, there exists a corresponding delta value such that the absolute value of f(x)-L is less than epsilon when x is within a certain distance of L.

Why is Exercise 9 important in AP Calculus?

Exercise 9 is important in AP Calculus because it helps students understand the concept of limits and how to manipulate the definition of a limit in order to solve for DELTA. This is a fundamental skill that is necessary for solving more complex problems in calculus and other areas of mathematics.

Similar threads

Replies
1
Views
1K
Replies
14
Views
1K
Replies
9
Views
917
Replies
16
Views
2K
Replies
24
Views
2K
  • Calculus and Beyond Homework Help
Replies
9
Views
805
  • Calculus and Beyond Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Differential Equations
Replies
1
Views
767
Back
Top