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Question Number 69607 Answers: 0 Comments: 4

without using lhospital please prove that lim_(x→0) ((x−sin x)/x^3 ) = (1/6) I want every method possible because someone challenge me

$$\boldsymbol{{without}}\:\boldsymbol{{using}}\:\boldsymbol{{lhospital}}\:\boldsymbol{{please}} \\ $$$$\boldsymbol{{prove}}\:\boldsymbol{{that}} \\ $$$$\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{lim}}}\:\frac{\boldsymbol{{x}}−\boldsymbol{{sin}}\:\boldsymbol{{x}}}{\boldsymbol{{x}}^{\mathrm{3}} }\:=\:\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\boldsymbol{{I}}\:\boldsymbol{{want}}\:\boldsymbol{{every}}\:\boldsymbol{{method}} \\ $$$$\boldsymbol{{possible}}\:\boldsymbol{{because}}\:\boldsymbol{{someone}} \\ $$$$\boldsymbol{{challenge}}\:\boldsymbol{{me}}\: \\ $$

Question Number 69557 Answers: 1 Comments: 0

Question Number 69620 Answers: 0 Comments: 1

Question Number 69545 Answers: 0 Comments: 2

Question Number 69538 Answers: 0 Comments: 0

Hello Verry Nice Day for You Find Σ_(k≥0) (1/((8k+1)^2 ))

$${Hello}\:{Verry}\:{Nice}\:{Day}\:{for}\:\:{You} \\ $$$${Find}\:\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{8}{k}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$

Question Number 69535 Answers: 3 Comments: 1

Question Number 69586 Answers: 0 Comments: 0

x^5 −x^4 −x^3 −x^2 −x−1=0

$${x}^{\mathrm{5}} −{x}^{\mathrm{4}} −{x}^{\mathrm{3}} −{x}^{\mathrm{2}} −{x}−\mathrm{1}=\mathrm{0} \\ $$

Question Number 69502 Answers: 3 Comments: 2

∫((3sinx+4cosx)/(4sinx+3cosx))dx

$$\int\frac{\mathrm{3sinx}+\mathrm{4cosx}}{\mathrm{4sinx}+\mathrm{3cosx}}\mathrm{dx} \\ $$

Question Number 69500 Answers: 2 Comments: 0

Question Number 69496 Answers: 0 Comments: 5

Question Number 69494 Answers: 1 Comments: 3

Question Number 69493 Answers: 0 Comments: 0

Question Number 69589 Answers: 0 Comments: 0

x^3 +px−ry+qz+a=0 y^3 +rx+qy−pz+b=0 z^3 −qx+py+rz+c=0 solve for x,y,z, in terms of p,q,r, a,b,c.

$${x}^{\mathrm{3}} +{px}−{ry}+{qz}+{a}=\mathrm{0} \\ $$$${y}^{\mathrm{3}} +{rx}+{qy}−{pz}+{b}=\mathrm{0} \\ $$$${z}^{\mathrm{3}} −{qx}+{py}+{rz}+{c}=\mathrm{0} \\ $$$${solve}\:{for}\:{x},{y},{z},\:{in}\:{terms}\:{of} \\ $$$${p},{q},{r},\:{a},{b},{c}. \\ $$

Question Number 69479 Answers: 0 Comments: 2

to Sir Aifour: we can construct polynomes of both 3^(rd) and 4^(th) degree in a way that the constants are ∈Z or ∈Q and the solutions are not trivial i.e. (t−α)(t+(α/2)−(√β))(t+(α/2)+(√β))=0∧t=x+(γ/3) ⇔ x^3 +γx^2 −(((3α^2 )/4)+β−(γ^2 /3))x−((α^3 /4)+((α^2 γ)/4)−αβ+((βγ)/3)−(γ^3 /(27)))=0 or the more complicated with sinus/cosinus (x−α−(√β)−(√γ)−(√δ))(x−α−(√β)+(√γ)+(√δ))(x−α+(√β)−(√γ)+(√δ))(x−α+(√β)+(√γ)−(√δ))=0 where all constants ∈Q if (√(βγδ))∈Q I could not find a similar construction for a polynome of 5^(th) degree, where the 5 roots are of comparable complexity [(x−a)(x−b−ci)(x−b+ci)(x−d−ei)(x−d+ei) doesn′t count] maybe you should at first focus on this

$$\mathrm{to}\:\mathrm{Sir}\:\mathrm{Aifour}: \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{construct}\:\mathrm{polynomes}\:\mathrm{of}\:\mathrm{both}\:\mathrm{3}^{\mathrm{rd}} \:\mathrm{and} \\ $$$$\mathrm{4}^{\mathrm{th}} \:\mathrm{degree}\:\mathrm{in}\:\mathrm{a}\:\mathrm{way}\:\mathrm{that}\:\mathrm{the}\:\mathrm{constants}\:\mathrm{are} \\ $$$$\in\mathbb{Z}\:\mathrm{or}\:\in\mathbb{Q}\:\mathrm{and}\:\mathrm{the}\:\mathrm{solutions}\:\mathrm{are}\:\mathrm{not}\:\mathrm{trivial} \\ $$$$\mathrm{i}.\mathrm{e}. \\ $$$$\left({t}−\alpha\right)\left({t}+\frac{\alpha}{\mathrm{2}}−\sqrt{\beta}\right)\left({t}+\frac{\alpha}{\mathrm{2}}+\sqrt{\beta}\right)=\mathrm{0}\wedge{t}={x}+\frac{\gamma}{\mathrm{3}} \\ $$$$\Leftrightarrow \\ $$$${x}^{\mathrm{3}} +\gamma{x}^{\mathrm{2}} −\left(\frac{\mathrm{3}\alpha^{\mathrm{2}} }{\mathrm{4}}+\beta−\frac{\gamma^{\mathrm{2}} }{\mathrm{3}}\right){x}−\left(\frac{\alpha^{\mathrm{3}} }{\mathrm{4}}+\frac{\alpha^{\mathrm{2}} \gamma}{\mathrm{4}}−\alpha\beta+\frac{\beta\gamma}{\mathrm{3}}−\frac{\gamma^{\mathrm{3}} }{\mathrm{27}}\right)=\mathrm{0} \\ $$$$\mathrm{or}\:\mathrm{the}\:\mathrm{more}\:\mathrm{complicated}\:\mathrm{with}\:\mathrm{sinus}/\mathrm{cosinus} \\ $$$$ \\ $$$$\left({x}−\alpha−\sqrt{\beta}−\sqrt{\gamma}−\sqrt{\delta}\right)\left({x}−\alpha−\sqrt{\beta}+\sqrt{\gamma}+\sqrt{\delta}\right)\left({x}−\alpha+\sqrt{\beta}−\sqrt{\gamma}+\sqrt{\delta}\right)\left({x}−\alpha+\sqrt{\beta}+\sqrt{\gamma}−\sqrt{\delta}\right)=\mathrm{0} \\ $$$$\mathrm{where}\:\mathrm{all}\:\mathrm{constants}\:\in\mathbb{Q}\:\mathrm{if}\:\sqrt{\beta\gamma\delta}\in\mathbb{Q} \\ $$$$ \\ $$$$\mathrm{I}\:\mathrm{could}\:\mathrm{not}\:\mathrm{find}\:\mathrm{a}\:\mathrm{similar}\:\mathrm{construction}\:\mathrm{for} \\ $$$$\mathrm{a}\:\mathrm{polynome}\:\mathrm{of}\:\mathrm{5}^{\mathrm{th}} \:\mathrm{degree},\:\mathrm{where}\:\mathrm{the}\:\mathrm{5}\:\mathrm{roots} \\ $$$$\mathrm{are}\:\mathrm{of}\:\mathrm{comparable}\:\mathrm{complexity} \\ $$$$\left[\left({x}−{a}\right)\left({x}−{b}−{c}\mathrm{i}\right)\left({x}−{b}+{c}\mathrm{i}\right)\left({x}−{d}−{e}\mathrm{i}\right)\left({x}−{d}+{e}\mathrm{i}\right)\right. \\ $$$$\left.\mathrm{doesn}'\mathrm{t}\:\mathrm{count}\right] \\ $$$$\mathrm{maybe}\:\mathrm{you}\:\mathrm{should}\:\mathrm{at}\:\mathrm{first}\:\mathrm{focus}\:\mathrm{on}\:\mathrm{this} \\ $$

Question Number 69478 Answers: 0 Comments: 1

lim_(n→∞) sin(n𝛑) = ?

$$\:\:\underset{\boldsymbol{{n}}\rightarrow\infty} {\boldsymbol{{lim}sin}}\left(\boldsymbol{{n}\pi}\right)\:=\:? \\ $$

Question Number 69462 Answers: 0 Comments: 1

find the equation of the circle which ends one of the diameters of two points p_1 (−2,3) and p_2 (4,5)

$$ \\ $$$${find}\:{the}\:{equation}\:{of}\:{the}\:{circle}\:{which}\:{ends}\:{one}\:{of}\:{the}\:{diameters}\:{of}\:{two}\:{points}\:{p}_{\mathrm{1}} \left(−\mathrm{2},\mathrm{3}\right)\:{and}\:{p}_{\mathrm{2}} \left(\mathrm{4},\mathrm{5}\right) \\ $$

Question Number 69460 Answers: 0 Comments: 1

find the equation of the circle whose center is the origin and touches the line 3x−4y−15=0

$${find}\:{the}\:{equation}\:{of}\:{the}\:{circle}\:{whose}\:{center}\:{is}\:{the}\:{origin}\:{and}\:{touches}\:{the}\:{line}\:\mathrm{3}{x}−\mathrm{4}{y}−\mathrm{15}=\mathrm{0} \\ $$

Question Number 69459 Answers: 0 Comments: 0