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Prove (with an &- & proof) that lim (x3 _ 1) = 7 172...

Question

Prove (with an &- & proof) that lim (x3 _ 1) = 7 172

Prove (with an &- & proof) that lim (x3 _ 1) = 7 172



Answers

Prove the limit statements $$\lim _{x \rightarrow 3}(3 x-7)=2$$

Alright for this problem, we want to prove that the limit as X approaches infinity of x cubed is equal to infinity. Using the definition uh specifically definition seven. So that means that we need to show that for every possible for every positive M. So we have em greater than zero that there exists a N greater than zero. Such that if X is greater than capital, N then F of X is greater than capital. So we can start out with our assumption that and is less than capital act and it's less than Cadillacs. Excuse me. Uh and is less than X. So if n is less, if N is less than X, then we can expect exponentially eight this expression without changing the ordering of it. So we can have that N cubed is going to be less than execute, which well, N cubed we can write as just being itself for the moment but X cubed is our F of X. Now we can see that fx will always be greater than N. Cute. So we can say that N cubed is equal to M. Or alternatively that N equals M to the power of 1/3. The showing that we can always find an for any excuse me showing that we can find an end for every value of M greater than zero.

From number 22. Let's absolutely fingers and see you and F of X is equal to X squared minus seven when a seven expressed will over X minus three. We're l is April to negative one, so every backs plus one will be smaller than absolutely if, um negative one minus absolute smaller than F of X and every X is equal toe extra square minus seven plus well over X. My name three and X minus seven can be controlled as H minus three and X minus four over. Excellent. A three negative one plus absolutely so is equal to X minus four. So X minus three will be bigger than negative. Absolute and the smoother than absent. So X minus three Absolute smaller than absence of the winter is equal to toe. Absolutely so for our pit ready Absolutely, it is. Delta is ableto absence on that your ex Ah x minus three um absolutes. A smaller Banda Delta, then f of X plus one is wonder than an absolute

Already. We gotta prove the limit is X approaches. Zero come, experts is three of three. X ninety seven is equal to two. All right, so what we need to do, we need to find a delta. So some stuff for any absolute Gruden zero sense of fixing up so long absolutely cannot just be arbitrary. Greater than zero. We need to find a delta such that ex mines three Whenever x Maestri's Listen, Delta, we need three X minus seven. Mine is, too to be less than Absalon. But this actually isn't so bad here because if we just rewrite this, this's three x finest nine. Listen, Absalon, But if we factor out of three, we see the X minus three for any given choice of excellent. Absolute is going to be less than Absalon over three. So Interstate Delta Bi Absalon over three, so indulge his absolute over three f of X minus two is less than absolute

This is problem number thirty of this tour. Calculus, eighth edition, Section two point four Prove this statement using the absolute out. The definition of a limit element is ex Purchase two of the function X squared plus two recommended seven. Is he quota one. The Epsilon Delta definition of limite states that for every absolute greater than zone, there must be a delta greater than zero such that if the absolutely of the difference between X and A is less indulge him in the value of the difference between the function and its limit. In this case, one is less than Epsilon. We begin with this second statement, this inequality use our function X squared plus two x minus seven. Subject its limit one and make the absolutely of that quantity lesson. Absolutely. If we simplify the inside yet X squared plus two x minus eight absolutely of that quantity Lesson Absalon. We can factor inside here. Thanks plus four X minus two all within one absolute values sign. And we can split each of these terms into two separate absolutely terms. At this point, we can't directly solve for this exact inequality explains to That's the inequality or the absolute value of interest since we are approaching, too. Um, So we will find the relationship in Delta Absalon through, Ah, an estimation or ah, gas for the delta Valium until we are consistent enough to make a statement on the Delta and absolute definitions. So our first step will be to because we are centered about two. We're approaching two very closely, and one is a very small number relative to the entire function. And so this is going to be our first condition. Ah, this is equivalent to saying they get a one is less than X minus two is lesson one. And if we as sixty each cider get five is less than X plus for is less than seven. At this point, we see that the reason we added six eyes to have a an understanding of what this term may be equal to, or maybe related to, so that we can focus on again this absolutely of X minus two. Here we have that X Plus four is between five and seven for convenience. We are going to modify this lower bound. Since experts for is greater than five, we can choose any number less than five. And the inequality remains Experts for is greater than negative seven in this case. And the reason we chosen seven is to do this change from a non absolute value, double inequality to an absolute value inequality. So this expos, for absolutely you can be written as the absolutely of this thiss explicit order is less than seven. And if this is true, we're going to make a direct institution into our current on equality down here we're on assumed that we can replace that this with seven and we're going to be choosing in A and Absalon that's consistent with this on his greater than this love side. And so we end up getting the absolute value of the quantity expense too, is listen, Absalon over seven. Now, from this first inequality, the difference between X and what X is approaching. This is sad, is the delta and what we have here is we made an assumption of, ah, what the absolutely of explains to would be less than we. We're able to determine an absolute based on this assumption. And now we can make a statement about the Delta. We have a delta can either be equal to one or absolute over seven. However, in order for this work that we did, you know, in order for our argument to be withheld, we need to make sure that each of these are satisfied. We are not sure what absolute ins, but we know that for any Absalom, um, the choice of Delta needs to satisfy both of these on inequalities. So what we do is we have to make sure we choose the smallest of these doctor values. And the way that we write this the notation is adults is equal to the minimum of either one or absolution or seven. So it's absolute. Over seven is greater than one disfunction will choose. One is the value for Delta, and they will confirm our definition of limit improves our statement. If we choose an absolute that smaller, this absolute over seven, that's smaller than one, then this minimum function will choose that term and set for Delta. And again, it will be consistent with our argument. Therefore, through the absolute off Delta definition of limit, we have proved this statement and showed that this, um, it's equal to one


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