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

[Pseudo Statistic]

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:
\lim_{x \rightarrow 4} 2x = 8; E = 0.1

Thanks.

 

[Galileo]

Write down what you know and what you need to prove:

Find a \delta&gt;0, such that:
|2x-8|&lt;0.1 whenever |x-4|&lt;\delta

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

 

[wais]

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