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3 25Letg dd h khe fndins defited by 9lw) = - 2x? , {+X-z*f = (4 pe |+Xh+ I fIs a fucim lat slsfies gl) _ f6) < hl) forallx Watis fic, fl) ? IcXA) 3 h (9 c)5 0) I...

Question

3 25Letg dd h khe fndins defited by 9lw) = - 2x? , {+X-z*f = (4 pe |+Xh+ I fIs a fucim lat slsfies gl) _ f6) < hl) forallx Watis fic, fl) ? IcXA) 3 h (9 c)5 0) Ie limit crtbe determin24 fdm the ibforaliggiven

3 25 Letg dd h khe fndins defited by 9lw) = - 2x? , {+X-z*f = (4 pe |+Xh+ I fIs a fucim lat slsfies gl) _ f6) < hl) forallx Watis fic, fl) ? IcX A) 3 h (9 c)5 0) Ie limit crtbe determin24 fdm the ibforaliggiven



Answers

Finding the Derivative by the Limit Process In Exercises 15-28, find the derivative of the function by the limit process.
$$h(s)=3+\frac{2}{3} s$$

In this problem of limited continuity. We have given that limit ordered pair X and Y approaches towards 00 and Fontanez. Three X. Y divided with two weeks square plus y squared. And now we have to find each limit by approaching 00 along first we have to find the X axis. Then we have to find the value of limit along Y axis. Then the line why it equals to X. Then Along the line wide equal to three x. and along the line. Or we can say along the y equals two access code which is a parabola. So for this, when you put the value, why is it equals to zero? This means we are finding the value of the limit along the X axis. So we have to put y is equal to zero. Now putting Y is equal to zero. So this really would be limit order pair X&Y approaches towards 00. No Minister would be zero. And denominator will be to access where that means the value of the limit is equal to zero. No, we have to find the value of limit along Y axis. That once we have to put X equals zero, putting X is equal to zero. This value would will order pair limit order pair X and Y approaches towards 00. When we put X is equal to zero. New military zero and denominator is only Y square. So this is again zero. Now we have to put y equals two x. So when we put why is equal to X. So this really will be here limit order pair X and Y approaches towards 00 three X multiplied with. Again next. So this is three X squared divided with two X square. And we have to put Y. Is equal to work. So that's where it would be Excess were also. So three x squared. And the value of this limit is here cancel out. And this is a constant which is one. So Value of this Limited one. Now we have to put why is equal to three X. So here limit ordered pair X. And Y approaches towards 00 three X. Multiplied by three X. So this really would be nine X squared divided with two X square plus. Why we have to put three eggs. So this value will be nine X square. Now when we solve at this age limit order pair X. Ny approaches towards 00. And from here we have to cancel out numerator and denominator. From here we have to cancel our Texas. Where from numerator and denominator. So we have left only nine divide with 11, which is the right answer. Now we have to put Y is equal to X. The square. So limit ordered pair X and Y approaches towards the settle settle three X. Multiple vortexes square. So this is three X cubed divided with two x square plus. This value will be extra the powerful because we have to put a bicycle to access square. Now eliminating excess square from numerator and denominator. We would have limit Ordered where X&Y approaches towards 00 three X divided with two plus six square. Now when you put taxes school 20 nominator is equal to zero and denominator job too. So value of this limited equals to zero. And also when you write the answers of party, the answer is zero for part B. The answer is zero for part C. The answer is one For part of the answer is nine divided with 11 and for part of the answer is zero. And from here we also conclude that limit does not exist. Mm. So this is the answer of all the parts.

Easy zero. You obtain the indeterminant form of 00 giving the whatever's in the guarantee them and simplifying. We end up with the limit as H approaches zero off H multiplied by H squared plus nine h plus 27 divided by age so we can cancel it to H is plugging in zero. We don't end up between

Okay, so here we have the function F of x. Y. Z is equal to E. To the X plus Y over one plus Z square. So we then take the limit as H approaches zero of f of one comma two plus H comma three minus F of +123 All over. H. That gives us the limit as H approaches zero of E. To the one plus two plus H over one plus three squared minus E to one plus 2/1 plus three squared all over each. So this is going to be equal to the limit as a church approaches zero of while we get E. Um cubed Times E to the H- E. Cubed both over 10. Um and then all over H. So as H approaches zero um while we can bring you can bring an E cube over 10. We can factor outside the limit. So this is going to be equal to e cubed over 10 and then times the limit as H approaches zero of E. To the H minus one. All over H. Okay, well as it approaches zero E. To the H is going to be equal two. Um So we this is actually gonna be equal to just um we get one here so we get one times E cube over 10, which is equal to E cube over 10 limit here is going to be equal to e cubed over 10.

This is number 70 Section 2.2. Okay, well, they give us three this function, right? It's a great it's a rational function and they ask us to evaluate it, and they ask us to graph it and they ask us to sell it. Algebraic Lee. So there's really 33 parts to this. Um, first thing wants to do is to come up with a lot of different tables. Anytime somebody gives me something and says, I need to come up with a bunch of different values, My first reaction is, let's type it into the calculator and save some effort in time. So I went ahead and did exactly that. So here we are, on the calculator. Go ahead and look where I typed it into y equals. That's the equation from the problem. Okay, so you wanna grafted right? And when you see these weird kind of diagonal verticals that's not really part of the function. It's just it's connecting the highest and lowest value by default. The program is things is just a calculator. It's actually were Nassib would be, but it's not really it's e mean. It's kind of slanted in the drawing anyway. don't get kind of misled by that. But the very interesting thing is like down here around this area, right, Because it crosses a couple times and we've got to do a nice zooming on it. And there's some values here that just don't exist. They're undefined. And the reason that is is because the equation, the way that they gave it to us, is fact herbal, right? You can actually factor it on DSO. As a result, we're seeing these, like, little holes that exist, and we don't really notice them until we look at the table of values. Um, and in the table, we can look and we can see, uh, what exactly we've got at one. And at three, we have two undefined function values, so yeah, this is definitely factual. Those are gonna be where zeros are, anyway. But if you need to put in the number once you have this in the UAE editor, keep in mind you can always go and come up with values by, um as you're looking at this right here, you can just, you know, put in different ones into the actual equation itself. Um, math will take you to the value and then I can say Okay, well, I wanna know what 2.9 is. My you can hit. Enter hand. Upsy escaped. Clear. Um, yeah. So let's go and go back here where I've got my numbers that I did earlier. So here I've got some values of X over here and over here it's approaching on the left and right side of our graph, okay? And they both approach to they're getting closer and closer to do. It's pretty clear there. So the first one asked, This is like, you know, what is it approaching from the left and the right Someone who is good and say that H of X approaches two on both sides. So it's kind of like part day, okay? And it's actually part A and part B. Really, That was the interesting thing to me is when we solve it algebraic Lee, Let's skip to that part. Um, one had worked this out earlier, so we've got the problem, right? And we've got it in its factored for. And then there's some things that we can you know, cross off because they divide the one and we're left with this. Are you here. Okay. And with this one, we would still have an undefined value at one at X equals one, but they're not asking us to evaluate that, asking for us to evaluate it. A sex approaches three so we can use our tried and true techniques. Right? And we can come to two, so the limit is actually to Okay, um, the limits to but you know, is the value, you know, at three actually equal to two, right? What happens when we when we put it back in? Well, let's go back to our table like we can just look at to write It says I'm sorry at three as it about three. It says undefined, even though the limit is too. So there is no value for the actual function. But there's a limit involved, and that's the important difference between limits and, you know, evaluating a function. Hi. Okay, So before I'm done here, here's the interesting part. You've got your answers, but if you look at that goes holes that exist, there was one at right here for, like, you know what would be X equals one. There's one here for X equals three and Those are just little lines with the little, like spots, little points where the lines are not continuous. It's not a continuous function of those points, and the reason they exist is because of these things that are just kind of in their inside here, right, Especially this X minus three. It lives in here even though these things are equal to one another. The introduction of these other factors, even though they divide one, still affect the value of our function as we evaluated right, which is really kind of a cool time. Like you can exclude any point you want from any equation. If you, you know, have a factor in there that has, like, you know, x minus. Whatever it is you're trying to cancel out, which is kind of a cool idea. If you think about it, you can always put a hole in using this technique English probably more than you wanted to know. But that's what we've got here. Part a. And then B. And then this is just the very end. See? So part B is the same approaches to Okay, Thanks


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