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Consider a function f : R → R which is differentiable andsatisfies limx→∞ f 0 (x) = 0. Prove that we have limx→∞ (f(x + 1) −f(x...

Question

Consider a function f : R → R which is differentiable andsatisfies limx→∞ f 0 (x) = 0. Prove that we have limx→∞ (f(x + 1) −f(x)) = 0.

Consider a function f : R → R which is differentiable and satisfies limx→∞ f 0 (x) = 0. Prove that we have limx→∞ (f(x + 1) − f(x)) = 0.



Answers

Prove that the function $f$ given by $$f(x)=|x-1|, x \in \mathbf{R}$$ is not differentiable at $x=1$

We want to prove that if f of X is differential at X equals zero and f of X is always lesson or equal to zero for all X and F of zeros equals zero, then f prime of zero must be equal to zero. So let's graph it to kind of look at this. So first thing we know is that f of X is less than or equal to zero for all X. So we're never gonna have a graphic comes into the positive part. I'm of the of the, um quadrants. Never. It's never gonna go above y equals zero. And this means that we could have a graph that looks like this. Or we could have a graph that, you know, kind of looks like this. Or it could look like this and just, like, touch the x axis and come back down. And it looks like, in fact, it does touch the X axis f of zero has to equal zero. So our options for going to and from this then it could come up and just stay there at X equals zero. Or it could come up and come down, or it could have come from Ah, Y equals zero and then go down. But in any case, we're always gonna have a moment at X equals zero where the graph is flat because it's either going to be like a peak like that. Where is going to be somewhere along a straight line that eventually kind of curves downward? Or maybe it's one that curved upward and go straight. But in any case, because it is the highest point, it possibly can be. The Tanja line has to be flat. So because F of X cannot be greater than zero and and f of zero is equal to zero, uh, X equals zero is ah, high point and will have a slope of zero.

For the following problem, we want to define F by affects being equal to eat the negative one over X squared. And we're also going to let f X equals zero when X is equal to zero. So we want to show that F is differentiable um on all values of X. So if we look at crime extra enough, that is the case. And even though it shows right here that zero is undefined, we know that it is defined and that's because we just didn't define f of zero equals zero for this particular problem, but we know that to be the case. And um So it's going to be what we end up finding. We can also see that f prime of zero is equal to zero using the definition of the derivative. Because we see that the point at that value equals zero and it has no change. We see that this is a horizontal tangent line at the given point.

They want us to find the derivative of Y. Equals F. To the G. Where F. And G. Are both functions of acts. And we're gonna use uh log, rhythmic log arrhythmic differentiation here. So take the natural log on both sides. So that's you know whenever we see powers like this, that's kind of the first thing you should think. So we get natural look of why? Because we can bring this power down times. She tends the natural look of F differentiate both sides. We get one over Y. Dy dx. So now we need to use the product rule. So we've heard this term G Derivative of this time is one over F. The F. D. X. And then the derivative this term is D. GDX and times this term natural large breath. And so now we can multiply two by Y. So we get to whitey X. And why? It's just F. Times G. Race to the G. So we get raised the G. Times G over F. D. F. B. X. Plus the efforts the G. Natural log of F. G. D. X. And we can see one power here. We can subtract one off here. So we get G after the G -1 -1 kanji as a function of acts um hopes that the F. D. G. That would be an interesting um problem because she is a functional right? Anyway the F. D. X. And then this thing we just leave alone after the G. Um Natural like a half times the G. D. X. So again this is kind of a power rule kind of thing for you know when you have the power is a function and this is a function. So they let's see here. Yeah. Um Yes. And this is you know X. To the finding the natural, finding the derivative of X. Dx. Not an easy thing to do. But if you use this formula then we're all you'll be able to do it

To prove this. We are going to find the limits As a church tends to zero of um If of X plus H minus F. Of X. All divided by age. Okay, so F F X plus H is equals two X plus H. That's it. Times G. X plus H. And F of X have already defined. So let's make the substitution. So then this becomes the limit At age 10- zero. Uh Okay. Of um X plus H times she of X plus age minus X. G. Of X. Okay. And then we divide this by H. So now we want to prove that this is differentiate will at zero. So we have to let X equals to zero. So we substitute X to be zero. So this will be the limit at age 10-0. Um off. So it's 0-plus eight which is going to be a church. Then we have G exit sara. So this will be a change minus zero. All divided by H. And this is equals two H times G. Of H. All right, divided by H. And we can see that these two are actually going to cancel and then we are left with so I left with the limit as H tends to zoo uh of Uh sorry I forgot the limit here. So it's the limited h tends to zero. This is equals to the limit in exchange 20 of G. F. H. So at this point now we can actually substitute um H H equals to zero. Okay, so this is going to be um G off the sarah. So we know that this exists because um G is continuous. She is continuous. It zero. So um GF zero must exist. So this is true. Um Okay, F is differentiable at zero. Yeah. So to find the derivative of if in terms of all with respect to X who are going to use the product rule, we were going to treat this as a constant first. So we're going to have X. And then we find the derivative of G. So it's going to be cheap prime uh X of X plus now we treat G of X as a constant, so the derivative of X is one. And then we're going to have um G of X unchanged. Right? So now we are looking for the derivative of if we're X s equals to zero and this is equals two. So we substitute X equals to zero. So to be zero times G prime zero plus G of zero, and this will be equal to G of zero. The exact same G that we got earlier. Yeah. Mhm.


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