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Problem 30 Hard Difficulty

Prove the statement using the $ \varepsilon $, $ \delta $ definition of a limit.

$ \displaystyle \lim_{x \to 2} (x^2 + 2x - 7) = 1 $


$\delta=\min (1, \varepsilon / 7)$


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Video Transcript

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