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We have that $$\left(1+\frac{1}{n}\right)^n=\mathrm{e}^{n\log(1+\frac{1}{n})}=\mathrm{e}^{1-\frac{1}{2n}+{\mathcal O}(n^{-2})}=\mathrm{e}\left(1-\frac{1}{2n}+{\mathcal O}\Big(\frac{1}{n^2}\Big)\right),$$ since $$\log \Big(1+\frac{1}{n}\Big)=\frac{1}{n}-\frac{1}{2n^2}+{\mathcal O}\Big(\frac{1}{n^3}\Big) \quad\text{and}\quad \mathrm{e}^h=1+h+{\mathcal ... 5 Let  L = \lim_{x\to\infty} \left( f(x) + \int_{0}^{x} f(t) \, dt \right)  denote the limit. We easily check that$$ f(x) + \int_{0}^{x} f(t) \, dt = \frac{\frac{d}{dx} \left( e^{x} \int_{0}^{x} f(t) \, dt \right) }{\frac{d}{dx} e^{x}}. $$Since e^{x} \to \infty as x \to \infty, it satisfies the condition of L'hospital's rule and hence$$ ...

3

In a double integral, you are actually integrating a differential two-form: $$\int_R \mathrm{f}(x,y) \ \mathrm{d}x \wedge \mathrm{d}y$$ Here, $\mathrm{d}x$ and $\mathrm{d}y$ are the basis differential one-forms and $\mathrm{d}x \wedge \mathrm{d}y$ is their exterior product.

2

Although bof's answer is correct, perhaps this is what the OP was looking for: Let $\phi:[0,1]\rightarrow A$ be a path.For every $x \in A$, choose an open ball $U_x$ such that $x\in U_x\subset A$. Since $\{\phi^{-1}(U_x)|x\in A\}$ forms a cover of $[0,1]$, by the Lebesgue covering lemma, there exists a partition $0=t_0<t_1<...<t_n=1$ of $[0,1]$ ...

2

Well a "solid" background in single variable and multi-variable calculus should be more than enough for you to make an attempt at learning Real Analysis. I have a very feeble foundation on multi-variable calculus but have taken a decent first semester calculus course and am not finding it difficult to make progress studying Real Analysis on my own. I did ...

2

By the Mean Value Theorem, for some $c\in(a-h,a)$ $$f'(c)=\frac{f(a)-f(a-h)}{h}\tag{1}$$ Then $$f'(a)-f'(c)=\int_c^a f''(t)\,\mathrm{d}t\tag{2}$$ The triangle inequality and $(1)$ and $(2)$ yield \begin{align} |f'(a)| &\le|f'(c)|+|f'(a)-f'(c)|\\[9pt] &=\left|\,\frac{f(a)-f(a-h)}{h}\,\right| +\left|\,\int_c^a f''(t)\,\mathrm{d}t\,\right|\\ ... 1 A point (x, y) is called a local minimum (maximum) if there is a open ball B containing (x, y) and f(z, w) \geq (\leq) f(x, y) for all (z, w)\in B. So in your case (c, y) is not a minimum (maximum) for all y. Note that one candidate of f is f(x, y) = x-c. It is harmonic and is not constant. 1 Since for h>0 and x \in \mathbb{R}, the MVT's, \begin{align} f(x+h)&=f(x)+hf'(x)+\frac{h^2}{2}f''(x+\theta h) \\ f(x-h)&=f(x)-hf'(x)+\frac{h^2}{2}f''(x-\theta' h) \end{align} for \theta,\theta'\in (x-h,x+h). Consequently, f'(x)=\frac{1}{2h}(f(x+h)-f(x-h))+\frac{h}{4}(f''(x+\theta h)-f''(x-\theta' h)) which implies |f'(x)|\le ... 1 In what follows we shall show that whenever a,b\in A, then there exists a polygonal line inside A connecting a and b. In particular, we shall show that the set: S=\{w\in A: \text{$a$ and $w$ are connected with a polygonal line inside $A$}\}, $$is both open and closed with respect to A, and hence S coincides with A, since a\in A and thus ... 1 The simplest way to show this fact is using the Intermediate Value Theorem: If f:[a,b]\to \mathbb R is continuous, and f(a)<\xi<f(b), then there exists a c\in(a,b), such that f(c)=\xi. Further, if f is strictly increasing, then the c4 is unique. So, here f(x)=x^n is continuous and strictly increasing in [0,1+x_0], and$$ ...

1

you can let $y=\lim_{x \to \frac{\pi}{2}} (1+\cos x)^{\tan x}$ then $$\ln y=\ln \lim_{x \to \frac{\pi}{2}} (1+\cos x)^{\tan x} \\ \ln y= \lim_{x \to \frac{\pi}{2}} \ln(1+\cos x)^{\tan x} \\ \ln y= \lim_{x \to \frac{\pi}{2}} \tan x\cdot\ln(1+\cos x) \\ \ln y= \lim_{x \to \frac{\pi}{2}} \frac{\ln(1+\cos x)}{\cot x} \\ \ln y= \lim_{x \to \frac{\pi}{2}} ... 1 h_2(1/2) = (1/2^2)h(2^2/2) = h(2)/4 = h(0)/4 = 0 but h(1/2) = 1/2 so the first claim is not true. The functions h_1, h_2, ... are also sawtooth functions, just scaled down: in the terminology of waves, h_0 (=h) has "wavelength" 2 and "amplitude" 1, while h_n has wavelength 2^{1-n} and amplitude 2^{-n}. So \lim_{n \rightarrow ... 1 A discrete set defined hereby is essentially a formalisation of the general notion of discreteness we have used so soften. The set \{1, 2, 3, ..\} is discrete in the sense that there are "gaps" between any element in the set and all other elements in the set. Defining this using balls: If D is a discrete subset of \Bbb C. Then there is an open ball ... 1 No, because the differentials that appear in an integral are just notation; they only signify what variables are to be integrated over. Consider a sum:$$\sum_{i=1}^n i^2 The $i$ in the bottom of the summation symbol tells you that $i$ is the dummy variable here. You could, in principle (though no one does this), notate integrals the same way: ...

1

Choose 6 distinct points $x_k \in [0,1]$ and form the matrix $V = \begin{bmatrix} 1 & x_1 & \cdots & x_1^5 \\ 1 & x_2 & \cdots & x_2^5 \\ \vdots & & & \vdots \\ 1 & x_6 & \cdots & x_6^5 \end{bmatrix}$, and note that the Vandermonde matrix $V$ is invertible. If $\pi(x) = p_1+p_2x + \cdots + p_6 x^5$, then \$V p ...

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