AsdrubalBeltran
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Continuing with the idea, of @sanath: The second step asserts that $3^m=2^{m-n}$. Here, $2$ and $3$ are prime numbers, so this equation then implies that there is a natural number with two different ...

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Hint: make $u=y^2$ then $du=2y\,dy$

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Yes is correct, remember that $$\frac{d}{dx}\int_{g(x)}^{f(x)}h(t)\,dt=h(f(x))\cdot f'(x)-h(g(x))\cdot g'(x)$$ this is by the second theorem of calculus and by chain rule.

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$$\frac{x+9}{\sqrt{x+9}}=\frac{(x+9)^1}{(x+9)^{\frac{1}{2}}}=(x+9)^{1-\frac{1}{2}}=(x+9)^{\frac{1}{2}}$$

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Hint: Note that $2|a|=|a+b+a-b|\leq|a+b|+|a-b|$ $2|b|=|b+a+b-a|\leq|b+a|+|b-a|=|a+b|+|a-b|$ then $$2|a|+2|b|\leq 2(|a+b|+|a-b|)$$ then $$|a|+|b|\leq |a+b|+|a-b|$$

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The identity is $$\sin^3x=\frac{3\sin x-\sin(3x)}{4}$$ Proof: \begin{align*} 4\sin^3x-3\sin x &= \sin x(4\sin^2x-3) \\ &=\sin x(1-4\cos^2x) \quad \text{Pythagorean identity} \\ &...

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A posibble form is: $$\prod_{i=1}^ni!$$ In other way is $$1^n\cdot2^{n-1}\cdot3^{n-2}\cdot\cdots\cdot(n-1)^2\cdot n^1$$

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If $y=x$ the limit is $0$, but if $x=y^2$ the limit is $\frac{1}{2}$, then the limit don't exist.

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The mistake is: $2\sqrt{4+\sqrt{4^2+\sqrt{4^3+\sqrt{\cdots}}}} \neq \sqrt{4^2 + \sqrt{4^3 + \sqrt{4^4 + \sqrt{\cdots}}}}$ $2\sqrt{4+\sqrt{4^2+\sqrt{4^3+\sqrt{\cdots}}}} = \sqrt{4^2 + \sqrt{4^4 + \... View answer 6 answers 4 votes 390 views 6 votes Hint: Note that $$\binom{2n}{n}=\frac{(2n)!}{(n!)^2}$$ Then the sequence diverges. View answer 6 answers 9 votes 582 views 6 votes Note that if$f(x)=x^n$then $$f'(1)=\lim_{x\to 1}\frac{x^n-1}{x-1}$$ How $$f'(x)=nx^{n-1}$$ then $$f'(1)=n\cdot(1^{n-1})=n$$ View answer 3 answers 3 votes 98 views 5 votes You can make cases$ab=1993-2=1991=11\cdot181ab=1993-3=1990=2\cdot5\cdot199ab=1993-5=1988=4\cdot7\cdot71ab=1993-7=1986=2\cdot3\cdot331$View answer 2 answers 3 votes 177 views 5 votes Note that: $$(x-y)^2\ge 0\implies x^2+y^2\ge2xy\implies x^2+2xy+y^2\ge4xy\implies(x+y)^2\ge4xy$$ View answer 5 answers 9 votes 307 views 5 votes Hint: Note that $$7!<8^7$$ then $$7!<8^7\implies(7!)^\frac{1}{7}<8\implies7!\cdot(7!)^\frac{1}{7}<7!\cdot8\implies(7!)^\frac{8}{7}<8!\implies(7!)^\frac{1}{7}<(8!)^\frac{1}{8}$$ View answer 2 answers 5 votes 572 views Accepted answer 5 votes Hint: firts note that; if$a(b+c)+b(c+d)+c(d+e)+d(e+a)+e(a+b)=0$, then $$(a+b+c+d+e)^2=a^2+b^2+c^2+d^2+e^2$$ Now take$P(x)=x^5+kx^4+rx^3+sx^2+tx+u$, with roots$a,b,c,d,e$then from Viète’s ... View answer 3 answers 5 votes 210 views Accepted answer 5 votes If Supposed that there is$t^t=0$, with$t>0$, if$t>0$then $$\ln{t}\in \mathbb{R}$$ $$t\cdot\ln{t}\in \mathbb{R}$$ $$\ln{t^t}\in \mathbb{R}$$ but $$\ln{t^t}=\ln{0}\in \mathbb{R}$$ is a ... View answer 1 answers 6 votes 197 views Accepted answer 4 votes I like Introduction to set theory View answer 2 answers 2 votes 138 views 4 votes Note that: $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = \frac{1}{a+b+c}$$ $$\frac{ab+ac+bc}{abc}=\frac{1}{a+b+c}$$ $$(ab+ac+bc)(a+b+c)-abc=0$$ $$(a+b)(b+c)(c+a)=0$$ then$a=-b$,$a=-c$or$b=-c$. If$a=-...

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the definition of the rudin book: "A sequence $\{p_n\}$ in a metric space $X$ is said to converge if there is a point $**p \in X**$ with the following property...$$\lim_{n\to\infty }\{p_n\}=p$$ how $... View answer 10 answers 5 votes 2k views 4 votes Number Theory: Structures, Examples, and Problems Titu Andreescu, Dorin Andrica Problem-Solving and Selected Topics in Number Theory, by Michael Th. Rassias View answer 5 answers 3 votes 14k views 3 votes This is equivalent to proof that:$|\{0,1\}^{\mathbb{N}}|=|P(\mathbb{N})|$You can use that if$A⊆\mathbb{N}$, then$f(a)=1$if$a∈A$otherwise$f(b)=0\$, then $$f:P(A)\to\{0,1\}^{\mathbb{N}}$$ is a ...

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Note that: $$2=|1-L+1+L|\leq|1-L|+|1+L|<1$$

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Note that: \begin{array}{} \dfrac{\tan2A}{2}&=&\sin^2160^\circ-\sin160^\circ\sin220^\circ+\sin^2220^\circ\\ \dfrac{\tan2A}{2}&=&(\sin160^\circ-\sin220^\circ)^2+\sin160^\circ\sin220^\...

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An a way is:$$1+\frac{\sqrt{2x^2+1}}{|x|}=(x^2+x-1+1)(1+\sqrt{x^2+2x+3})$$ $$1+\frac{\sqrt{2x^2+1}}{|x|}=(x^2+x-1)(1+\sqrt{x^2+2x+3})+1+\sqrt{x^2+2x+3}$$ $$\sqrt{2x^2+1}-|x|\sqrt{x^2+2x+3}=|x|(x^2+x-1)... View answer 2 answers 7 votes 170 views 3 votes Note that \sqrt[n]{x^n}=|x| with n even:$$(((-4)^2)^{1/2})^{1/2}=(|-4|)^{1/2}=2$$View answer 2 answers 2 votes 47 views 3 votes Note that if$$x\to 2^+\implies |x-2|=x-2$$and if$$x\to 2^-\implies |x-2|=-(x-2)$$View answer 2 answers 2 votes 341 views Accepted answer 3 votes Note that$$r (p - q)^2 + p (q - r)^2 + q (r - p)^2\geq0$$And$$r (p - q)^2 + p (q - r)^2 + q (r - p)^2=(p+q)(q+r)(p+r)-8pqr$$View answer 3 answers 1 votes 372 views 3 votes Hint: Note that:$$3^{2(n+1)} -1=9\cdot3^{2n}-1=(3^{2n}-1)+8\cdot3^{2n}$$View answer 1 answers 2 votes 89 views Accepted answer 3 votes Hint: Suppose that A_1=\max\{A_1,A_2\}, then take t_0, such that \omega_1t_0+\phi_1=\frac{\pi}{2} and take t_1, such that \omega_1t_1+\phi_1=\frac{3\pi}{2}, then:$$f(t_0)>0$$and$$f(...
A bijection is: $$f(m,n)=2^m(2n+1)-1$$