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Find an exatple of [unetion with the property that f() is deflined for all = € R- lim f(x) exists but lim f () DNE [or F0....

Question

Find an exatple of [unetion with the property that f() is deflined for all = € R- lim f(x) exists but lim f () DNE [or F0.

Find an exatple of [unetion with the property that f() is deflined for all = € R- lim f(x) exists but lim f () DNE [or F0.



Answers

Find a function $f$ for which $\lim _{x \rightarrow c}|f(x)|$ may exist even though $\lim _{x \rightarrow c} f(x)$ does not exist.

This time they tell us that is Ariel number and that Ah, prime of exists. So the second derivative exists of the limit. Ah, the definition of derivative for the first derivative exists at the point A. And they also tell us Ah, this means that this limit a A plus mining of a times two plus of a vanity coverage must also exists. But they also want us to find its value. Well, I get clients, manipulate this. Remember, we're taking Ah limit here. A a plus. Minus half of it. Oh, great. Minus the limit. Ah, no. Ah, my hope This plus minus. So I could take a limit of individual terms. I've just organized my facts a little bit. Martin nicely here. Well, the H squared is an issue at this point. So let's see what we can do. I we're likely going to have to relate the second derivative when I was training of the difference between two first derivatives. So why don't we work the other way? We know that C minus. Oh, great. Is the definition of ah, first derivative? Yeah. This is equal Teoh prime. And so if we're going to find the second derivative A Then we're going to take the limit. And as a church goes to zero Ah, the first derivative at a my, uh, prime. We can go the other way. A minus. Don't get over a Well, what is this equal to limit? As h goes to zero, we can actually put in our definition of the first derivative, so I'll get another limit Appeared. We're going to get some nice nest in going on. Ah, a my men, everything over each this inputs a little more interest. So read it carefully with we'll end up with, uh, let's see that input a is actually going to be a minus h this time. So we have a plus h minus h. That ends up just being an Afghani, and then we'll minus of a minus h all over it. And don't forget, we're dividing by one more. Eight. All these have common denominators up top so we can combine them and we end up with the limit as h A purchase. Sarah. Well, we have a church. See, uh, a minus after bay, minus effort Bay plus f of a minus. So we got all over it. Uh, let me reiterate this at that age belongs appear under the previous mine and then we have one more eight years. But if you know this is exactly a previous limit. So that pretty cool, simplified limit they gave a plus H minus two with a plus f of a minus over each month happens to be equal to the value of the second derivative of the 0.8 they

They want us, Teoh. What? See? Show the No. They want to find the value of They want us to find the value of the limits as h approach zero of this month and just minus Teoh everything plus, uh, a h all over h squared here is you can try integrating that it a bit. They also tell us that a is the wrong number and the second derivative this. Ah, let's see. Do you remember your form? You leftward derivatives? Well, let's talk about the derivative as a limit. But the second relativism on it. So what function I retaking the limit the limit as age approaches zero of our first derivative. Let's see, a crime of it was EJ mine and stuff of eight over h balance perfectly acceptable. However, we could also let's see, have run it as the limit as h approaches zero f prime of a my message Mine. It's absolutely over each. Yeah, right. Uh, from me. Uh, what is the derivative halftime of a plus age? Well, that one's just let as each approaches here. Ah f prime of a message minus prime. Ah, those This is a limited. Already in the first year of issue. So it's the regular function minus the function of A Yeah, So let's see, all over age. So knowing that we could write the derivative as either one of these Well, the second derivative, we can also read the first year. But it is like that as well. So let's see. We can actually replace this derivative here with Ah, this expression on a roll for this, This one is basically the same expression, but shifted by a minus eight straight. Somebody would have just been a here on then, in minus h here. So we actually have that. The second derivative is equal to limit as age approaches. Zero. I'm gonna pull out of this one over age to the front so we can just deal with this derivative, which is the limit as a Jew approaches zero. Uh uh. A message minus separately. Over Rich. Okay. And now what about this term? We're gonna be so drug in it. It's a when im how we could write it as the other farm, which is shifted to the left by each compared to the other one. So we have half of a bus each my message shifted it. And this one is a minus age. Also shifted all over age. Well, what properties combined man's. Yeah, but lament as h grudges. Zero, uh, for each and then limit h approaches there. Uh, when our property, we can bring them in. But we have, uh, a passage. Well, this is a subtraction. So it's another minus F minus two. Uh uh. Then a plus, because it's a minus in front and a negative right there for us of a manage all the reach. There's no limit property. We have to limits in here. We actually just have a moment as age approaches. 00 half of a message minus thio of plus of a message all over each one. Actually. Let's take a scrolls with Okay. Yeah. They wanted us to compete this limit, but this is just If we knew the second river about it, that's what it is.

All right. So we're gonna use the definition of the derivative which says that F. Prime of A is equal to the limit as a church approaches zero of F. Of a plus H minus F. Of a all over H. To solve the function F of X equals X squared plus four Given a. is equal to one. So the first thing I'm gonna do is off to the side is find F. Of A. So F. Of A. Is now instead of finding A. We're going to find one. So F of one is one squared plus four which is five. And then I'm also going to find F. Of one because that's my a plus H. So I'm gonna plug one plus H in two X squared and add four. So now I can go plug into my definition of the derivative formula. It solved my limit. So it's equal to the limit As a judge approaches zero F. Of A plus H. We said was one plus H squared plus four minus. We said F. Of A which is F. Of one which is five all over. H. To simplify, I'm going to foil one plus age. So we have the limit as H approaches zero, foiling that you get one plus two H plus H squared Plus 4 -5 all over H. And then we have the limit as H approaches zero. We're going to combine like terms so one Plus four is 5 but 5 -5 is zero and then I have to H plus H squared all over H. And again I have an age age squared in an age here so I can take out an H from each of those terms. And I get the limit as h approaches zero Of two plus H because again I canceled the denominator and took one out of each numerator. Now I'm going to actually apply the limits. So the limit as H approaches zero of two plus H is equal to two plus zero or simply to. So the derivative of X squared plus four and X equals one is two.

Hello. So you're gonna let ffx be equal to E to the X. And then the derivative is just while the derivative of F F. Prime of X. We know is by definition that's the limit as H approaches zero of F. Of X plus H minus F of X. All divided by H. So now F prime of zero is going to be equal to the limit. As a church approaches zero of F zero plus H minus F of zero all over age that each of the zero plus H minus zero all over H. Which gives us, well, the limit as H goes to zero of each of the H minus zero is just one all over H. And then we just replace here um H by X. And we get the limit as X approaches zero um of E to the X minus one all over. Ex uh so therefore the limit here um of X as X approaches zero of each of the X minus one over X is the derivative of half of x. So which is equal to each of the X. And that is at Um at X is equal to zero


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