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Determine if the following functions grow faster; slower; or at the same rate as the functionas X =15x" 26x +b.Xln Tx -X13+x43 Ax -1f. Xe 4xsinxThe function 15...

Question

Determine if the following functions grow faster; slower; or at the same rate as the functionas X =15x" 26x +b.Xln Tx -X13+x43 Ax -1f. Xe 4xsinxThe function 15x4 26xgrowsasx - 0 because lim 15x4 X-0 26x +b The function x In Zx - X growsasX -because lim xIn Tx-X X-0The function V 3 +growsas X -because lim X-0v3+d. The functiongrows4X as X ~ because lim X-0The functiongrowsasX =because lim X-0f. The function Xe growsas X0 because Iim Xe 4X X-0

Determine if the following functions grow faster; slower; or at the same rate as the function as X = 15x" 26x + b.Xln Tx -X 13+x4 3 Ax -1 f. Xe 4x sinx The function 15x4 26x grows asx - 0 because lim 15x4 X-0 26x + b The function x In Zx - X grows asX - because lim xIn Tx-X X-0 The function V 3 + grows as X - because lim X-0v3+ d. The function grows 4X as X ~ because lim X-0 The function grows asX = because lim X-0 f. The function Xe grows as X 0 because Iim Xe 4X X-0



Answers

Determine whether the function grows faster than $x^{2}$, at the same rate as $x^{2},$ or slower than $x^{2}$ as $x \rightarrow \infty$. $15 x+3$

Now we're comparing a bunch of functions to X square to see how they grow with respect to X squared. So here we have X squared plus four at now, we can expand this out and that bombs becomes one plus four over X. And as X goes infinity um this goes to one. Um So that means that because this is a constant, that means these two functions grow at the same rate which you kind of expect because you know as X gets large this term is going to be negligible compared to this. And so they both look like growing as as quadratic functions as X gets large. Now we have X to the fifth minus X squared. Um expanding that out, we get x cubed minus one and as X goes infinity that clearly goes to infinity. So this X X to the fifth minus X squared although um grows much faster than X squared. And again we could see that because the highest power here is greater than the highest power here. Um Now we've got a square root, so we have the square root of X squared or X to the fourth plus X squared. Now again we can kind of look at this and say, ok as X gets very large, that's the fourth is gonna get very very big compared to X square, so that's going to be negligible. So this is gonna look like X squared. And so again, we would expect these two things to grow at the same rate. So let's check that out. So we can expand, pull this inside the square root and then pull the limit inside the square root, this becomes one plus one over X squared. The limit of that is one. The square root of that is one. And so indeed they do grill at the same rate. Um Now we have X plus three all squared and again we should see that we'd expect um you know this is a quadratic function and so is this, so if we expand this out and then divide through by this, we get one plus six over X plus nine over X squared. And taking the limit as x goes infinity, this goes to one. And so indeed these to do grow at the same rate. Now we're looking at X. Time for natural log of X. Okay, so this is an interesting function because um in fact we would expect this because natural lagerback's grow slower than X, we would expect this to grow slower than this. So let's see if that's the case. So we take the limit um we can see that in X cancels out. So we get the limit of natural log of X over X. Um Using logic college rule we get that this is the limit of one over X, which indeed goes to zero. And so X X times the natural log of X indeed grow slower than X square. Um Now we have either the X. We want to check that compared to X square and we would expect this to grow faster than X squared because you know either anything to the power of X or anything greater than one to the power of X grows grows very large very quickly. Um So we can use capitals rule, take the derivative appear and we just get this natural log of two term here and here we get to X. That because rule again get to to the X. Times again the natural log of two squared and then this just nominated becomes too and then this clearly goes to infinity. So clearly either the X grows faster than X squared. Mhm. Now this is a little interesting function because now we have something that grows faster than X squared times, something that grows slower the next word. And in fact this grows Um this goes to zero as x goes to infinity. So what we can see here is that if we divide that by X X cubed uh X squared we with the limit of X times E to the minus X. And that's just the limit of X over E. To the X. Capitals rule says that that's the limit of one overeating the X. And that goes to zero. So in fact this grows slower than this because basically in fact the limit of just the limit of um X cubed either the X that limited X cubed times either minus X actually goes to zero also. So Um this whole numerator here goes to zero as x ghost infinity, where this goes to infinity. Now we have um eight X squared, so they left us with an easy one here. Um X squared cancel out. So we get the limit as X scores infinity of eight, which is just eight, so eight X squared grows is the same rate as X squared.

For this problem we want to compare the rate of growth of X Ln of x minus X. To erase the X X approaches infinity. And to do this we have to use this definition And this definition tells us that we have to find the limit as X approaches infinity of X. Ln of x minus X over erase two X. Now the limit as extra purchase infinity of X Ln of x minus X over erase X. This is equal to the limit as X approaches infinity of X times Ln of x minus one over erase the X. And if we plug in infinity we get infinity over infinity. So we have to do loop ethos rule And from here we should get limit as X approaches infinity of the derivative of the numerator. That's X times one over X. Plus we have Ln of x minus one over the derivative of the denominator which is erased two X. And if we simplified as we get limit as X approaches infinity of this is one. So we subtract this by one. We get Ln of X. L left in the numerator over erase two X. And if we plug in infinity we get infinity over infinity. So we have to do lou Beatles rule again and they should equal to the limit as X approaches infinity of the derivative of the numerator which is one over X over derivative of erase to express itself. That's erase the X. And if we simplify this we get the limit as X approaches infinity of one over x times erase X which when evaluated gives us one over infinity, which is equal to zero. So by definition we have X. Ln of x minus X, This gross slower. Then the race to X.

Now let's see here, we're looking at um following slowest growth. Okay we want which one of these functions grows um growing from slowest to fastest. So we basically just need to take a bunch of limits comparing the two functions. So let's compare either the X. A and E to the X over two when we actually already did that. Um Well we did the universe of that in the previous problem. So that gives us the limit as X goes infinity of either the X over to that clearly is infinity. So either the X grows faster than either the X over two. So that's 111 thing we know ordering. And so now we have this interesting function of natural log of X to the X. Um Again we would expect this to grow faster than this because the natural log of X is going to eventually be greater than one. So we can we can pull the exponents now. So we get the limit as X goes to infinity of the natural log of X over E. All to the X. And this clearly goes to infinity as X goes to infinity. And so the natural log of X to the X grows faster than either the X. There's another piece of information. So this goes fashion this this goes faster than this, so that's ordered. And now we need to look at this final X to the X compared the natural log of X to the X. Again, we take the take the uh exponents out here. All right. And what we can see here is that again, this clearly goes to infinity as X ghost infinity. Um because this this grows faster than this and we have that race to the X power. So eventually this is going to go get greater than one. So we're gonna this whole thing is gonna go to infinity. And so that says that X to the X does in fact go faster than natural log of X to the X. So ordering from slow us to fastest, we have either the X over two is slower than E. To the X, which is slower than natural log of X. To the X, which is slower than X. To the X.

For this problem we want to compare the rate of growth of X Q plus three two X squared as X approaches infinity. Now to do that, we have to use this definition which tells us that we have to find the limit as X approaches infinity of X cubed plus three over X squared. Now the limit as X approaches infinity of X cubed plus three over X squared. This is equal to infinity over infinity. So we have to do loop ethos rule and from here we should get limit as X approaches infinity of the derivative with respect to X of the numerator. That's three X squared Over the derivative of the denominator with respect to X That's two X. And we can cancel that X from X squared. So we will have X here and this will be one. So then we will get a limit as X approaches infinity of three x over two. Now evaluating this limit, we get three times infinity over two, which is equal to infinity. And by this definition start ups that X cubed plus three grows faster, then X squared


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