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8

Assuming you have the Mean Value Theorem at your disposeal, that's the easiest way to show that the limit equals $0$: with $g(x) = \sin\sqrt x$, we have $$\sin\sqrt{x+1}-\sin\sqrt x = g(x+1)-g(x) = ((x+1)-x)g'(\xi) = \frac{\cos\sqrt\xi}{2\sqrt\xi}$$ for some $x\le\xi\le x+1$. In particular, $$|\sin\sqrt{x+1}-\sin\sqrt x| = ... 7 A simpler argument is to write f(x)=\sqrt{x+1}-\sqrt{x} and note that since \lim\limits_{x\to \infty}f(x)=0 and \sin is uniformly continuous, we have$$\lim\limits_{x\to \infty}\sin(\sqrt{x}+f(x))-\sin(\sqrt{x})=0.$$Why does this work? Let g be a uniformly continuous function, f a function such that \lim\limits_{x\to\infty}f(x)=0 and h an ... 7 Well it might be easier to start with the version of stokes theorem you probably know best, the fundamental theorem of calculus: \int_a^b df = f(b) - f(a) (when applicable). A sketch of (a) proof is that \int_a^b df = \lim_{N \to \infty} \Sigma_1^N (x_i - x_{i-1})df(y_i), where y_i \in [x_{i-1}, x_i]. By the mean value theorem, we can choose y_i \in ... 6 You cannot swap the n around, that is, inside and outside the limit. The step$$\left (\lim_{n \to \infty} n^{\frac{1}{n}} \right)^n=\lim_{n \to \infty} \left ( \left (n^{\frac{1}{n}} \right)^n \right )$$is wrong. The n outside the LHS is fixed, while the n inside, in the RHS, varies like the other n. The result would be correct, if you said, say, ... 5 There are less complicated examples (at least this won't require a page! :)). Try$$f(x,y) = \begin{cases} 1\,, & x = k/q \text{ and } y = \ell/q \text{ for some integers } k,\ell \text{ with $q$ prime} \\ 0\,, & \text{otherwise}\end{cases}\,.$$Then each horizontal/vertical line segment contains (at most) finitely many discontinuities of f. But ... 4 Let's first present Stokes' theorem symbolically:$$\int_{\partial\Omega}\omega=\int_\Omega d\omega$$Intuitively, \Omega is some 'domain', with boundary \partial\Omega, and \omega is some 'function' with derivative d\omega. So the theorem is roughly stating "the integral of a function on a boundary is the integral of the derivative on the ... 4 When a function has zero derivative within a volume, its values are determined by sources outside the volume, which are measured only by values of the field on the boundary. Got that? Here's the mathematical explanation. There are many functions f that obey the basic differential equation \nabla F = 0. We could be talking about a scalar field, or ... 4 One can define a function f:[a,b]\to\Bbb R to be regulated if there exists a sequence \{s_n\} of step functions such that s_n\to f uniformly. Recall that if s is a step function with constants c_i and intervals of partition [x_{i-1},x_i] for say i=1,\dots, r we define its integral as over [a,b] as$$\int_a^b s=\sum_{i=1}^r c_i\Delta x_i$$... 4 Suppose to the contrary that for n\ge N,$$ n\left(\frac{1+a_{n+1}}{a_n}-1\right)\lt1 $$Then, for n\ge N,$$ \frac{a_{n+1}}{n+1}\lt\frac{a_n}{n}-\frac1{n+1} $$It can be shown by induction that$$ \frac{a_{n+k}}{n+k}\lt\frac{a_n}n-\sum_{j=1}^k\frac1{n+j} $$Since the harmonic series diverges, at some point \dfrac{a_{n+k}}{n+k}<0, contrary to the ... 3 Note that \frac{1}{\sin x} blows up as x approaches 0 from the right. So the limit does not exist. Even if we allow infinite limits, the limit does not exist, for our function approaches 1 as x approaches 0 from the left. Remark: If the question is changed to \lim_{x\to 0^-} (limit as x approaches 0 from the left), then the limit does ... 3 Your test of whether a_n \gt a_{n+1} looks odd. Since a_n, a_{n+1} \gt 0, you can test whether a_n^2\gt a_{n+1}^2 and this comes out as$$\frac{n}{(n+4)^2}\gt \frac {n+1}{(n+5)^2}$$It is then simple to check (clear fractions, since you are multiplying by positive numbers) that this is true for n \ge 4, but not for n=3 - which may be why the ... 3 You can take the natural logarithm of the fraction and use your idea.$$\lim_{x\to +\infty} \dfrac{e^{x^2}}{10^{x}}=e^{ \lim_{x\to +\infty} \ln\dfrac{e^{x^2}}{10^{x}}}$$A much simpler solution is to observe that 10 <e^3 and hence$$ \dfrac{e^{x^2}}{10^{x}}> \dfrac{e^{x^2}}{e^{3x}}$$Added: To clarify why I can interchange the exponential and ... 3 Here's a construction of a counterexample. The idea is based on a simple construction for series that is given on p. 350 of the centenary edition of G. H. Hardy's A Course of Pure Mathematics. The idea in Hardy's example is to construct a sequence \{a_n\} of positive numbers such that \sum a_n < \infty but na_n \not\to 0 as n\to\infty; to do this ... 3 This result may help you: Let \mathbb{F}:(a, b)\rightarrow \mathbb{R} that is continuous on the bounded open interval (a, b) then the two limits given by F(a +) = \lim_{x\to a^{+}} F(x), F(b -) = \lim_{x\to b^{-}} F(x) exists iff F is uniformly continuous on (a, b). This result has been given in the book "The calculus integral by Brian S. ... 2 Let f : \left\{ \begin{array}{ccc} \ell^{\infty} & \to & \mathbb{R} \\ (x_n) & \mapsto & \limsup\limits_{n \to + \infty} x_n \end{array} \right. and g : \left\{ \begin{array}{ccc} \ell^{\infty} & \to & \mathbb{R} \\ (x_n) & \mapsto & \liminf\limits_{n \to + \infty} x_n \end{array} \right.. Show that f and g are ... 2 You need to use the Mean Value theorem in several variables, as you already stated it: For x,y\in U there exists s which lies on the line segment between x and y ( this line segment is contained in U, since it is convex), such that:$$f(x)-f(y)=\nabla f(s)\cdot (x-y)\Rightarrow\text{( Cauchy-Schwarz )}|f(x)-f(y)|\leq|\nabla f(s)|\cdot |x-y|$$... 2 Write the expression above as$$\lim_{x \to 0}\left [ 1 + \sum_{k=1}^n \log{(1+k x)}\right ]^{1/x}$$Note that n remains finite, so we can Taylor expand the logs to get$$\lim_{x \to 0}\left [ 1 +\sum_{k=1}^n k x\right]^{1/x} = \lim_{x \to 0}\left(1+\frac12 n (n+1) x\right)^{1/x} = e^{n(n+1)/2}$$Note that the error term in the Taylor expansion is ... 2 The symbol o(|h|) is like a variable for a function g(h) that satisfies$$\lim_{h \rightarrow 0} \frac{g(h)}{|h|} = 0.$$Multiplying any such function by a constant gives you another function in that class, so we write co(|h|) = o(|h|). Similarly, adding any two functions in that class gives another function in that class, so o(|h|) + o(|h|) = ... 2 One way to think about problems of such type is in terms of linear interpolation. Define your function f(n)=a_n>0 let f to be linear between [n,n+1]. Then, f\in L_1 is equivalent to \sum_{n=1}a_n<\infty whereas f'\in L_1 is equivalent to \sum_{n=1}|a_{n+1}-a_n|<\infty. Now it is clear that the standard example of series mentioned by ... 2 In the context of finding derivatives (Question 2), the only problem with$$f(x+h)-f(x)\over h$$is that at h=0 it gives you 0/0. And the only way to fix that is to find an h in the numerator to cancel that h in the denominator; or, to finesse the problem by doing manipulations to bring it to a form already handled, as in the proof of the product ... 2 You have to be a bit carefull with that comparison. In complex analysis, you really have (for holomorphic functions) that the values on the boundary completely determine the values within. My interpretation of why this works there is that being differentiable is a much stronger property for complex functions than for real ones. For complex functions, it ... 2 One-dimensional counterexample. Let U=(-1,1), X(x)=x^{1/3},$$X^n(x)=(x^2+\epsilon_n^2)^{-1/3}(x+(-1)^n \epsilon_n) \tag1$$where you can choose \epsilon_n>0 converging to 0 as fast as you want. Consider the solution curve of X^n beginning at 0, denoted x_n(t). If n is even, then x_n'(0) is positive, which makes x_n increasing for all ... 2 I think you are trying to ask for the distinction between defining the Riemann integral as a limit of supremum and infimum partitions vs just using a fixed family of partitions, in this case something akin to the dyadic intervals as Osgood does. The point is that both methods yield precisely the same result when the function f is continuous on the whole ... 2 What does \sup(s_n) mean ? In this context, I believe that they mean \sup_{k\geq n}(s_k) so that the expression for the upper limit is \lim_{n\rightarrow \infty} \sup_{k\geq n}(s_k). Note that you can have some numbers in your sequence larger than the upper limit. In fact, every number in the sequence can be larger than the upper limit, take the ... 2 Yes. It is enough to have v:=\alpha'(t_0)\ne 0. Then in a neighbourhood of t_0, the map t\mapsto \langle \alpha(t)-\alpha(t_0),v\rangle is injective and smooth and has a smooth inverse around 0 with 0\mapsto t_0. Let g\colon \mathbb R\to\mathbb (a,b) be a smooth function that coincides with this inverse near 0. Define$$V(x)=f(g(\langle ...

2

Set $y = -x$. Thus, we have: $x^x = (-y)^{-y}$ when $y \ge 0$ Or equivalently, $$x^x = \frac{1}{(-y)^{y}}$$ Which can be simplified to: $$x^x = \frac{1}{(-1)^{y} y^y}$$ Therefore, the domain of $x^x$ consists of both reals and complex numbers depending on the value of $(-1)^y$ or to be more precise depending on the value of $y$.

2

We know that $3$ lengths can form a right angled triangle iff those lengths satisfy $a^2+b^2=c^2$. But no matter what we choose for $m, \; n$ and $\lambda$. $$(2\lambda mn)^2+(\lambda(m^2-n^2))^2=4\lambda^2m^2n^2+\lambda^2m^4-2 \lambda^2m^2n^2+\lambda^2n^4$$ $$=\lambda^2m^4+2\lambda^2m^2n^2+\lambda^2n^4=(\lambda(m^2+n^2))^2$$ as we wanted. So we can ...

2

Considering the second derivative, it is sufficient to prove that $$(x+1) \log (1 + 1/x) - 1 \le \frac{1}{\sqrt{x}}$$ We will show that, $$\log (1 + 1/x) \le \frac{1}{\sqrt{x}}$$ This implies, for $0 \lt x \le 1$, that $$(x+1) \log (1 + 1/x) - 1 \le \frac{x+1}{\sqrt{x}} - 1 = (\sqrt{x} - 1) + \frac{1}{\sqrt{x}} \le \frac{1}{\sqrt{x}}$$ We can easily ...

2

HINT: One direction is very easy. For the other, assume that the sets $E[f\le c]$ and $E[f\ge c]$ for $c\in\Bbb R$ are closed. Then the sets $\{x\in F:f(x)>c\}$ and $\{x\in F:f(x)<c\}$ are relatively open in the subspace topology on $F$ for each $c\in\Bbb R$. This implies that if $a,b\in\Bbb R$ with $a<b$, then \{x\in F:a<f(x)<b\}=\{x\in ...

2

This is a consequence of the Axiom of Countable Choice, which the author is taking for granted (actually, the author is taking stronger assumptions for granted, based on section 6 of the book). The idea is as follows: If $\xi\in A$, then there is a constant sequence of points of $A$ converging to $\xi$. Otherwise, since $\xi$ is the supremum of $A$, ...

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