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QUEStionLetf: R-> R defined by ((x) g R > R delined by g(x) h(x) 2x Then Find: g 0 {(32) "(x) 0 (4 00 Prove that h: R-JR is a One-Ote funclionand h: RR d...

Question

QUEStionLetf: R-> R defined by ((x) g R > R delined by g(x) h(x) 2x Then Find: g 0 {(32) "(x) 0 (4 00 Prove that h: R-JR is a One-Ote funclionand h: RR delined

QUEStion Letf: R-> R defined by ((x) g R > R delined by g(x) h(x) 2x Then Find: g 0 {(32) "(x) 0 (4 00 Prove that h: R-JR is a One-Ote funclion and h: RR delined



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Let $f, g$ and $h$ be functions from $\mathbf{R}$ to $\mathbf{R}$. Show that $$ \begin{array}{l} (f+g) \text { o } h=f \circ h+g o h \\ (f \cdot g) \circ h=(f \circ h) .(g \circ h) \end{array} $$

Okay, This question wants us toe. Look at the quantity of the divergence of the magnitude of our times are so I'm just going toe figure out what this new vector is, and then take the divergence of that. So the magnitude of our is just the square root of X squared plus y squared plus C squared. So that's just a scaler. So then when we distribute that in tow, are we get X Times square root X squared plus y squared plus Z squared comma? Why times the magnitude and then z times the magnitude. So now that we have this, we can calculate the divergence of this expression. So the divergence of the magnitude of our times are is just that I ever just that vector we just wrote. So I'll start with an ex derivative of this expression. And due to the symmetry of this factor field, we could just calculate this and then did we can exploit the symmetry and just replace the axe with wires he's needed. So this first derivative is just using the product rule X squared plus y squared plus C squared and then plus X divided by to square root X squared plus y squared plus z squared times two x So this gives us the divergence contribution from the first term, which is just our magnitude plus X squared over the square root of X squared plus y squared plus C squared. And then due to symmetry, we can do the same thing for Cue y and R Z. So Q. Why we're just going to swap that X for why, on top and then for the ourselves, e we're just gonna put a Z on top. So adding all these together gives us our divergence, which is three copies of this square root magnitude. And then why I wrote them separately first is because now we can write them over this common denominator of their magnitude, and we see that the bottom is just the same expression just raised to the 1/2. So this is raised to the two halves and the bottom is raised to the 1/2 so that actually simplifies to another square root. So this ends up is four copies of the square root, and we had an expression for this, which is nothing less than the magnitude of our, and it's exactly what we're supposed to prove

In the problem, we have been given H of X and that is equal to F of G of X. Now age does x equals two. If that's gx into judas, eggs, this is equal to if gas that is f d s o G and to do this. So this is the right hand side of the problem. Now, this is the answer, so it is proved.

So these things are just using the definition of checking if a function is defined as little O. Of age. Um And that means that if we divide the function by age and let h go to zero. Do we get a finite value? So finite or zero? I think it's finite. Just needs to be a finite value. Mm Not sure should look that up. Um And the first problem G is H cube. So we just have the limit as H zero H cube of H. Which is the limit of H zero of HQ which is zero. Which I know at least that is um That says that H cube is little ohh next problem we have G is eight squared over H minus one. So divide that by H you know one of these cancer we get the limit as H course zero over of H over h minus one. Um And that is zero. So that says that a squared over h minus one is a little old age. I think that would also be the case. I think this just needs to be finite because if this is the low of H. I think you know any any M times X. Any line would be a little of age. So then we look at a cube root of age. Well divide that by age we get 1/8 to the two thirds. The limit is that goes to zero as H zero goes infinity. So h cube root of age is not a little O of age. Now they want us to show that the limit of G Of hsh goes zero must be zero for that function to be little O of age. So by definition if okay so if it's little of H then this has this limit has to go to 02 so high we say. So with a limit of G of H as H zero we can write that is the limit of G. Of H over H times age. And by using the you know definitions for limits you can break the limit up into the limit uh H zero G. Of H over H times the limit of H as H zero. Um This is zero but you can't say that this is always zero because um if this thing is is say this thing went to infinity um then we have infinity times zero. We don't know what that is. So but this is also zero. So we know that zero times zero is going to give us zero. So that we know that that limit is zero. Yeah. Now they ask us to kind of do the little oh um you know sums and products of functions. So we have two functions one G two. You want to show that there are some is if the if they're both little over age than their sum is also. So that's just basically taking the definition, splitting it up and noting that both of these are zero so that has to be zero. So this summit zero. So their product is a little more, is a little little chat more. A little more work. So we need is we need to break this up but we need another agent here. So we multiply and divide by age. So we get the limit as H zero of G one of H over H times G two of age or age times H. Now we can break up the limits. So we get this limit. This limit on this line with this limit is obviously zero. But that doesn't make this some of this product always zero because these things need to be finite. And in fact if G and G one and G to our little of age, then these are both both zero. So then this whole thing is zero. So that means this product of those two functions is also a little of each.


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