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Problem 36 Hard Difficulty

Prove that $ \displaystyle \lim_{x \to 2} \frac{1}{x} = \frac{1}{2} $.

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Video Transcript

This is problem number thirty six of the Stuart Calculus eighth edition. Section two point four proved that the limit is X approaches to of the function. One over X is equal to one half and we are going to prove this using our absolute total definitions for a limit. Absalon is greater than zero. Doubt is good and zero for every Absalon there exists. A delta is such that if the absolute value of the difference between X and A is less than dull town Ah, then the absolutely of the difference between the function and the limit is less than absolute. And let's see, we will ah, first right out the second function here and see what what sort of information? Additional that we need. In addition, that we need to have a relationship, a relationship reaching Absalon and Delta. So we're gonna write absolutely of the function. One of Rex minus the limit is just equal to one half less than Absalon. If we combine the two fractions, this will be to our two x minus one over to X so to minus X over to X is less than Epsilon. And we can separate these two Ah, terms at the absolutely of two minus six is the same as the absolutely explains to. So that does not change the ah equation inequality. And we're left at this point. Absolutely. The difference between X and two divided by that's a value of two acts is less than absolute. And in order for us to be able to compare this with this initial Ah abso value X minus two, which is what a is equal to X X is approaching to lessen Delta. In order for us to compare these two, we need to restrict Ah, this ah range of X in order to figure out what this term may be equal to, and then be able to compare Delta an absolute finally. So our restriction, our assumption is that X minus two. That absolutely amount is gonna be lesson one. So we're assuming they were really close to one. Sorry, I'm really close to two. The difference between the two in less than one this community written is negative. One less than X, minus two. Listen, one on then we're gonna have to teo each term giving us Quan less than ex listen three. So provided that This is our assumption were given a range of X values and we will use these this range to determine the range of one over to X. If we're rewrite this middle inequality as one over the absolutely of two X time's thie, absolutely of X minus two Western up son. Then we really need to know what this about this function ranges between Yeah, And if we do that, we see that the minimum value vexes one. And if you plug that in, we get a one half. So we're just trying to see what the range of this function is. We see that we get one half only put in one on one. When we played in three, we see that we get one sixth and clearly won. Six is less than one half, and we see that this function does range between one over six and one of her too. Eso essentially one over to x one over the absolutely two exes, a maximum value of one half. And we're going to use that fact in our inequality. Replace this term here with its maximum value, which allows us to relate. Finally Ah, absolute and Delta one half times. This quantity less than absolute gives us the absolutely of the different Phoenix and two is less than two Absalom. And now we can compare these two and we see that choosing a value of Delta equal to two Absalon is a a possible is a possibility in terms of being consistent with this absolute out the definition of a limit. But we also recall that along with this, we also made the assumption that the difference between X and two is also lesson one that the distance between X into his lesson one so that must also be considered. And therefore, our final answer is that Delta is the minimum of either these values, whichever is the smallest. That is what Delta should be. And as long as we have this in mind, we have essentially proven that this limit as expert just two of this function is equal to one half provided that yeah, this argument for UPS Absalon Indulgences is given