Using the Formal Definition

Each of Exercises...

Question

Using the Formal Definition

Each of Exercises 31–36 gives a function f(x), a point c, and a positive number ε. Find L = lim x→c f(x). Then find a number δ > 0 such that |f(x)−L| < ε whenever 0 < |x−c| < δ.

f(x) = −3x − 2, c = −1, ε = 0.03

Accepted Answer

Welcome back, everyone. For the function F X equals 4 X minus 7, a point C equals 3, and epsilon E equals 0.04, find L which equals the limit of FF X as X approaches C. Subsequently, identify a number delta that's positive, such that the absolute value of F X minus L is less than epsilon whenever the absolute value of X minus C is positive but less than Delta. Now, let's focus on the first part of our problem. Let's first start by finding the value of L. Now we know that F of X equals 4 X minus 7, C equals 3, and epsilon equals 0.04. So now, That means to find the limit as X approaches C of FFX. We only need FFX and C, which means we're finding the limit as X approaches 3 of 4 X minus 7. Now, as X approaches 3, that means by direct substitution, this is going to be 4 multiplied by 3 minus 7, which equals 12 minus 7 or 5. Therefore, L equals 5. Next, we want to find a positive value of delta for the given epsilon, OK? In other words, for the second part of our problem, we need to find, OK, we need to find a value of delta. Such that The absolute value of FF X minus L is less than Epsilon, OK? Whenever the absolute value of X minus C is positive but less than Delta. Now, if we take a good look at the absolute value of X minus L, let's start to build on our problem. Notice that since we know the value of L. We can say this is going to be equal to the absolute value of 4 X minus 7 minus 5. This equals the absolute value of 4 X minus 12, which equals if we factor 0 12 from both terms 4 multiplied by the absolute value of X minus 3. Now why is this helpful? Well, if we think about it, remember we said the absolute value of FX minus F of X minus L must be less than Epsilon. So for this to be less than Epsilon, then we need 4 multiplied by the absolute value of X minus 3 to be less than 0.04. Therefore, the absolute value of X minus 3 then must be less than 0.04 divided by 4, which equals 0.01. But now, remember we said that the absolute value of X minus C must be less than Delta and from our problem, we know that our limit, or sorry, we know that our point C equals 3. So since this is the absolute value of X minus 3. This means then That for our condition where the absolute value of X minus C must be positive but less than Delta, we could say that the absolute value here of X minus 3. OK. Must be positive but less than 0.01. Since we know the absolute value of X minus 3 is less than 0.01 and C equals 3. So what does this mean? Thus we can set delta then to be equal to 0.01 if we compare both inequalities. That means a smaller positive value of delta would also satisfy the conditions, so delta must be equal to 0.01 or a smaller. Positive. Value. So to summarize. Based on our problem, we found out that the limit L is equal to 5 and. And delta must be equal to 0.01 or a smaller value, such that the absolute value of FX minus L is less than epsilon whenever the absolute value of X minus C is positive but less than Delta. Thanks a lot for watching everyone. I hope this video helped.

Using the Formal Definition

Each of Exercises 31–36 gives a function f(x), a point c, and a positive number ε. Find L = lim x→c f(x). Then find a number δ > 0 such that |f(x)−L| < ε whenever 0 < |x−c| < δ.

f(x) = −3x − 2, c = −1, ε = 0.03

Key Concepts

Limit of a Function

Epsilon-Delta Definition of a Limit

Finding δ for Given ε

Watch next