💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!

# Prove the statement using the $\varepsilon$, $\delta$ definition of a limit.$\displaystyle \lim_{x \to 2} x^3 = 8$

## $\lim _{x \rightarrow 2} x^{3}=8$

Limits

Derivatives

### Discussion

You must be signed in to discuss.

Lectures

Join Bootcamp

### Video Transcript

This is problem number thirty two of the Stuart come his eighth edition. Section two point four proved the statement using the absolute Delta definition of a limit the limit as exit purchase, too, of the function X cubed equals eighteen. The absolute delta definition of a limit is that for any number of absolute greater than zero, there is a delta greater than zero such that if the absolute value of the difference between X and A is less than daughter, then the difference between the absolute the function, the absolute value of the function and the limit that difference his list. And it's on. Ah, we will begin by using our second inequality here. And we're going to determine and see if we can find a relationship before Absolute and Delta that is consistent with his definition and allows us to prove ah, the statement. So beginning with the second inequality, the function f X and this problem is X cubed dynasty limit. Delimit value is equal to eight, and this is ah within the answer Valley sign we. Let's let that be less than Absalon. We can factor this function on the inside. It's a difference of cubes. It's executed at minus two. Cubed on this leaves us with X minus two times the quantity X squared plus two x plus for all within the absolutely sign. And then if we separate these two terms ah, they become their own Absolute value terms the absurdity of the quantity X minus two I want to play, but the absolute value with quantity X squared plus two x bliss for him. Now, at this point, can't say much about how absolutely relates to Delta, and we can show that. Bye. Ah, bye. Explicitly assuring what this first inequality becomes for this problem. We have explained us a A, and this problem is too. That's what X is approaching. And this is less than Delta. So our approach involves, um, trying to compare this inequality with this inequality, and we see that we're very close, but we still have to deal with this absolute value of this Xcor. Just two x plus four term. So we're gonna do is we're going to make an assumption we're going to make an assumption regarding X minus two. I'm going to set a restriction Ah, and then see how that helps us out for this problem. So we're going to that strain. Try Black. So we're going to restrict that the difference between X and two is less than a small number. We're going to choose one, and this is a fair restriction because we're approaching too. So we know that the distance between these two numbers is going to be small, and one is a fairly useful, convenient number for now to use. If we write this, we can rewrite it. This upset value as thinking one is less than X minus two without the ups Valya sign, it's less than one. This's Ah, based on the definition of the absolute value, we added two to both sides. We get that the range of X that we have restricted is between one and three. And this helps us in that we can determine maybe what the maximum value this will be. And if we understand what the maximum value. What we can understand is that this restriction that we said at oh highlighted in red at this restriction that we set on X ah is the same thing as X being restricted between the values one and three. If X reaches the maximum value of three. This whole function has a corresponding value and we can use that constant value. Teo, determine Absalon in terms of this constant value, totally so we're gonna do is we're going to used his maximally ofthree We'LL get into the X squared plus two x plus for equation two three squared plus two times three plus four And we see that this function has a maximum value of nineteen. So will you go back to our problem here? We can substitute this dude because of this restriction. Because of this assumption, we can simplify this in equality now by using the value nineteen and it room. And this inequality remains true because we use the maximum value that function. Now we're able to use that constant. Teo, help us determine what an appropriate absolute value would be absolute Over nineteen. Now, we can compare these two and we get a restriction on X minus two. Well, we get the restriction that this difference must be less than absolute or nineteen s. So this is a second restriction that we found. And then the first restriction that we set in order to get this one is that the absolute value with the difference beteen Nixon to must be won to see how the these two are fairly related. Um, and in fact, the with the exact values of dull town Delta, as we can see, is related to both of these. It is going to be either one, which is the first assumption we made or absolute over an eighteen. It's either of these two, but recall the delta must be the minimum of both the minimum of either, so whichever happens to be the least. And what we do is we using minimum function which chooses to smaller these two values, depending on the value of Epsilon. So with this in place, we can show that any of us alone that we choose we choose an up song that makes this value greater disfunction will choose one, since one is less than that and we will see that we can ah satisfied this absolute a delta definition of a limit. At the same time. If we choose and Absalon here that makes this term at less than one, the minimum function will choose the value for Delta and the definition is remains consistent. And so this way we have prove prove in this statement and this limit as expert just to have execute is definitely equal to eight.

Limits

Derivatives

Lectures

Join Bootcamp