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Question Number 73466 Answers: 1 Comments: 0
$${please}\:{explain}\:{this}\: \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}} {{Lim}}\frac{{sinx}}{{x}}\:=\:\mathrm{1}\:\:{by}\:{l}'{hopitals}\:{theorem} \\ $$$$ \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {{Lim}}\:\frac{{sinx}}{{x}}\:=\:\mathrm{0}\:{by}\:{Squeez}\:{theorem} \\ $$$${is}\:{there}\:{something}\:{wrong}? \\ $$
Question Number 73451 Answers: 0 Comments: 3
Question Number 73441 Answers: 1 Comments: 0
Question Number 73428 Answers: 1 Comments: 2
$${Evaluate}\:: \\ $$$$\left.\mathrm{1}\right)\:\int_{−\mathrm{2}} ^{\:\mathrm{2}} \int_{−\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }} ^{\:\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }} \:\left(\mathrm{3}−{x}\right){dydx}\:. \\ $$$$\left({after}\:{changing}\:{the}\:{integral}\:{to}\:{polar}\:{form}\right). \\ $$$$ \\ $$$$\left.\mathrm{2}\right)\:\int_{\mathrm{0}} ^{\mathrm{4}} \int_{\mathrm{0}} ^{\mathrm{4}−{x}} \int_{\mathrm{0}} ^{\:\mathrm{4}−\frac{{y}^{\mathrm{2}} }{\mathrm{4}}} \:{dzdydx}\:. \\ $$
Question Number 73429 Answers: 1 Comments: 3
$$\:\:\:{Solve}\::\:\int\frac{\left[{cos}^{−\mathrm{1}} {x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right]^{−\mathrm{1}} }{{log}_{{e}} \left[\mathrm{2}+\frac{{sin}\left(\mathrm{2}{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)}{\pi}\right]}{dx} \\ $$$$\:\:{Evaluate}\:\:\int_{−\pi/\mathrm{2}} ^{\:\pi/\mathrm{2}} {sin}^{\mathrm{2}} {xcos}^{\mathrm{2}} {x}\left({cosx}+{sinx}\right){dx} \\ $$
Question Number 73411 Answers: 2 Comments: 2
$${calculate}\:\: \\ $$$$\left.\mathrm{1}\right){cos}\left(\mathrm{1}+{i}\right)\:,\:{sin}\left(\mathrm{1}+\mathrm{3}{i}\right) \\ $$$$\left.\mathrm{2}\right)\:{arctan}\left({i}\right),\:{arctan}\left(\mathrm{2}{i}\right)\:,\:{arctan}\left(\mathrm{1}+{i}\right)\:,{arctan}\left(\mathrm{1}−{i}\right)\:, \\ $$$${arctan}\left(\mathrm{1}+\mathrm{2}{i}\right). \\ $$$$\left.\mathrm{3}\right)\:{have}\:{us}\:\:{conj}\left({arctanz}\right)={arctan}\left(\overset{−} {{z}}\right)? \\ $$
Question Number 73406 Answers: 0 Comments: 0
$${Use}\:{the}\:{Sandwich}\left(\:{Pinchin}\:{or}\:{Squeez}\:\right)\:{theorem}\:{to}\:{prove} \\ $$$${that}\: \\ $$$$\:\underset{{x}\rightarrow{a}} {\mathrm{Lim}}\:\sqrt{{x}}\:=\:\sqrt{{a}}\: \\ $$
Question Number 73405 Answers: 0 Comments: 0
$${can}\:{someone}\:{please}\:{prove}\:{the}\: \\ $$$${Chinese}\:{Remainder}\:{theorem},\:{for}\: \\ $$$${modula}\:{arithmetic}? \\ $$
Question Number 73399 Answers: 3 Comments: 1
$$\begin{cases}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{65}}\\{\left({x}−\mathrm{1}\right)\left({y}−\mathrm{1}\right)=\mathrm{17}}\end{cases} \\ $$$$ \\ $$$${please}\:{help}\:{me}\:{to}\:{solve}\:{it}... \\ $$
Question Number 73397 Answers: 1 Comments: 1
$${find}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−{t}} {ln}\left(\mathrm{1}−{xt}^{\mathrm{2}} \right){dt}\:\:{with}\:\mid{x}\mid<\mathrm{1} \\ $$$$\left.\mathrm{2}\right){calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−{t}} {ln}\left(\mathrm{1}−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\right){dt} \\ $$
Question Number 73396 Answers: 1 Comments: 1
$${find}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−{t}^{\mathrm{2}} } {ln}\left(\mathrm{1}−{t}\right){dt} \\ $$
Question Number 73378 Answers: 0 Comments: 3
$${Hello}\:,{i}\:{shar}\:{withe}\:{you}\:{nice}\:{problem}\: \\ $$$${show}\:{that}\:\forall{k}\in\mathbb{N}^{\ast} \:\exists{n}\in\mathbb{N}\:{such}\:{that} \\ $$$${k}\leqslant\underset{{j}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{j}}<{k}+\mathrm{1} \\ $$$${have}\:{a}\:{very}\:{Nice}\:{day} \\ $$$$ \\ $$
Question Number 73358 Answers: 0 Comments: 1
$$\mathrm{1}/\mathrm{4}{x}\mathrm{2}−\mathrm{1}/\mathrm{2}{x}−\mathrm{13}=\mathrm{0} \\ $$
Question Number 73356 Answers: 0 Comments: 2
Question Number 73347 Answers: 0 Comments: 0
Question Number 73346 Answers: 3 Comments: 5
Question Number 73340 Answers: 0 Comments: 2
$$\mathrm{Is}\:\mathrm{there}\:\mathrm{any}\:\mathrm{pi}/\mathrm{product}\:\mathrm{notation}\:\mathrm{rules} \\ $$$$\mathrm{I}\:\mathrm{discovered}\:\mathrm{some}\:\mathrm{such}\:\mathrm{as}: \\ $$$$\underset{{k}={a}} {\overset{{b}} {\prod}}\left[{k}\right]=\frac{{b}!}{\left({a}−\mathrm{1}\right)!} \\ $$$$\underset{{k}={a}} {\overset{{b}} {\prod}}\left[{c}\right]={c}^{{b}−{a}+\mathrm{1}} \\ $$$$\underset{{k}={a}} {\overset{{b}} {\prod}}\left[{c}\centerdot{k}\right]=\underset{{k}={a}} {\overset{{b}} {\prod}}\left[{c}\right]\centerdot\underset{{k}={a}} {\overset{{b}} {\prod}}\left[{k}\right] \\ $$$$\underset{{k}={a}} {\overset{{b}} {\prod}}\left[{k}+{c}\right]=\underset{{k}={a}+{c}} {\overset{{b}+{c}} {\prod}}\left[{k}\right] \\ $$$$\mathrm{But}\:\mathrm{what}\:\mathrm{about}\:\mathrm{this} \\ $$$$\underset{{k}={a}} {\overset{{b}} {\prod}}\left[{c}\centerdot{k}+{d}\right] \\ $$
Question Number 73338 Answers: 0 Comments: 1
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left(\mathrm{2}{cosx}\right)}{\mathrm{3}+{x}^{\mathrm{2}} }{dx} \\ $$
Question Number 73337 Answers: 0 Comments: 3
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{cos}\left({artan}\left(\mathrm{2}{x}\right)\right)}{\left(\mathrm{3}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx} \\ $$
Question Number 73336 Answers: 1 Comments: 1
$${find}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}} {ln}\left(\mathrm{1}+{e}^{{t}} \right){dt} \\ $$
Question Number 73335 Answers: 1 Comments: 2
$${eplcit}\:\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({x}+{t}+{t}^{\mathrm{2}} \right){dt}\:\:\:\:\:\:{with}\:{x}>\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\left.\mathrm{2}\right){calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({t}^{\mathrm{2}} \:+{t}\:+\sqrt{\mathrm{2}}\right){dt} \\ $$
Question Number 73334 Answers: 1 Comments: 0
$${calculate}\:{lim}_{{n}\rightarrow+\infty} \:\:\:{n}^{\mathrm{2}} \left(\:{e}^{{sin}\left(\frac{\pi}{{n}^{\mathrm{2}} }\right)} −{cos}\left(\frac{\pi}{{n}}\right)\right) \\ $$
Question Number 73333 Answers: 0 Comments: 1
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{cos}\left(\pi\:+\mathrm{2}{x}^{\mathrm{2}} \right)}{\left({x}^{\mathrm{2}} \:+\mathrm{4}\right)^{\mathrm{2}} }{dx} \\ $$
Question Number 73332 Answers: 0 Comments: 0
$${let}\:{U}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{\sqrt{\mathrm{2}{k}+\mathrm{1}}}\:\:{determine}\:{a}\:{equivalent}\:{of}\:{n}\:{when}\:{n}\rightarrow+\infty \\ $$
Question Number 73331 Answers: 0 Comments: 1
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{ln}\left(\mathrm{1}+{e}^{−\mathrm{3}{x}^{\mathrm{2}} } \right)}{\mathrm{3}+{x}^{\mathrm{2}} }{dx} \\ $$
Question Number 73330 Answers: 1 Comments: 1
$${find}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\frac{{ln}\left(\mathrm{2}−{cos}\left(\mathrm{2}{x}\right)\right)}{{ln}\left(\mathrm{1}+{xsin}\left(\mathrm{3}{x}\right)\right)} \\ $$
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