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Question Number 32291 Answers: 0 Comments: 0
$${let}\:{u}_{{n}} =\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{n}+{k}}\:{prove}\:{that}\:\mathrm{0}\leqslant{u}_{{n}} \leqslant\mathrm{1}\:. \\ $$
Question Number 32290 Answers: 0 Comments: 1
$${let}\:{give}\:{u}_{\mathrm{0}} =\mathrm{1}\:{and}\:{u}_{{n}+\mathrm{1}} =\sqrt{\mathrm{1}+\sqrt{{u}_{{n}} }}\:\:{prove}\:{that}\:{u}_{{n}} \:{is} \\ $$$${increasing}\:. \\ $$
Question Number 32289 Answers: 0 Comments: 0
$${find}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:{ln}\:\left(\:\frac{{e}^{\mathrm{2}{x}} −\mathrm{1}}{{x}}\right)\:. \\ $$
Question Number 32288 Answers: 0 Comments: 0
$${study}\:{the}\:{function}\:{f}\left({x}\right)=\frac{{x}^{\mathrm{2}} }{{x}+\mathrm{1}}\:{e}^{\frac{\mathrm{1}}{{x}}} \:\:. \\ $$
Question Number 32287 Answers: 0 Comments: 0
$$\left.\mathrm{1}\right)\:{for}\:{x}>\mathrm{0}\:{prove}\:{that}\:\frac{\mathrm{1}}{{x}+\mathrm{1}}\:\leqslant{ln}\left({x}+\mathrm{1}\right)−{lnx}\:\leqslant\:\frac{\mathrm{1}}{{x}} \\ $$$$\left.\mathrm{2}\right)\:{let}\:{u}_{{n}} =\:\sum_{{p}=\mathrm{1}} ^{{kn}} \:\frac{\mathrm{1}}{{p}}\:\:\:{find}\:{lim}_{{n}\rightarrow\infty\:} \:{u}_{{n}} . \\ $$
Question Number 32286 Answers: 0 Comments: 0
$${calculate}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\:\frac{\frac{{e}^{{x}} }{\sqrt{\mathrm{1}+{x}}}\:−\mathrm{1}−\frac{{x}}{\mathrm{2}}}{{x}^{\mathrm{2}} }\:. \\ $$
Question Number 32281 Answers: 0 Comments: 0
$${calculate}\:{lim}_{{x}\rightarrow\infty} \sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\:−\sqrt{{x}^{\mathrm{2}} \:−{x}+\mathrm{1}}\:\:. \\ $$
Question Number 32280 Answers: 0 Comments: 0
$${let}\:{u}_{{n}} =\:{e}^{\frac{\mathrm{1}}{{n}^{\mathrm{2}} \:+\mathrm{1}}} \:\:−\mathrm{1} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{equivalent}\:{of}\:{u}_{{n}} \:{and}\:{lim}_{{n}\rightarrow\infty} {u}_{{n}} \\ $$$$\left.\mathrm{2}\right)\:{study}\:{the}\:{convergence}\:{of}\:\:\Sigma{u}_{{n}} \:. \\ $$
Question Number 32279 Answers: 0 Comments: 0
$${find}\:\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\frac{{tan}^{\mathrm{2}} {x}}{\left(\mathrm{1}−{cosx}\right)}\:.\frac{{e}^{{x}} \:−\mathrm{1}}{{x}}\:\:\: \\ $$
Question Number 32282 Answers: 0 Comments: 0
$${let}\:{give}\:{f}\left({x}\right)=\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\:\:{find}\:\:{f}^{\left({n}\right)} \left({o}\right)\:.\: \\ $$
Question Number 32278 Answers: 0 Comments: 0
$${calculate}\:{lim}_{{x}\rightarrow+\infty} \left({x}−\mathrm{1}\right){cos}\left(\frac{\pi}{{x}}\right)\:. \\ $$
Question Number 32277 Answers: 0 Comments: 0
$${for}\:{x}_{{i}} \:\in\left[\mathrm{0},\mathrm{1}\right]\:{prove}\:{that} \\ $$$$\left(\mathrm{1}−{x}_{\mathrm{1}} \right)\left(\mathrm{1}−{x}_{\mathrm{2}} \right)....\left(\mathrm{1}−{x}_{{n}} \right)\:\geqslant\mathrm{1}−\left({x}_{\mathrm{1}} +{x}_{\mathrm{2}} \:+....\:+{x}_{{n}} \right). \\ $$
Question Number 32276 Answers: 0 Comments: 0
$${prove}\:{that}\:\sum_{{i}=\mathrm{1}} ^{{n}} \:\left(\prod_{{j}=\mathrm{0}} ^{{p}} \left({i}+{j}\right)\right)=\frac{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)...\left({n}+{p}+\mathrm{1}\right)}{{p}+\mathrm{2}} \\ $$
Question Number 32285 Answers: 0 Comments: 0
$${let}\:{f}\left({x}\right)=\:{x}^{{n}−\mathrm{1}} {ln}\left(\mathrm{1}+{x}\right)\:{with}\:{n}\:{integr}\:{and}\:{n}\geqslant\mathrm{1} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}^{\left({p}\right)} \left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{find}\:{f}^{\left({n}\right)} \left({x}\right) \\ $$
Question Number 32284 Answers: 0 Comments: 0
$${prove}\:{that}\:\forall\:{x}\in\left[\mathrm{0},\frac{\pi}{\mathrm{4}}\right]\:\:{x}\leqslant{tanx}\leqslant\mathrm{2}{x} \\ $$$$\left.\mathrm{2}\right)\:{find}\:\alpha_{{n}} \:{and}\:\beta_{{n}} \:{from}\:{R}\:/\:\:\alpha_{{n}} \leqslant\:\sum_{{k}=\mathrm{2}} ^{{n}} \:{tan}\left(\frac{\pi}{\mathrm{2}{n}}\right)\leqslant\:\beta_{{n}} \\ $$
Question Number 32283 Answers: 1 Comments: 1
$${let}\:{give}\:{f}\left({x}\right)={x}+\mathrm{2}\:−\sqrt{{x}+\mathrm{1}} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{f}^{−\mathrm{1}} \left({x}\right)\:{inverse}\:{of}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\left({f}^{−\mathrm{1}} \right)^{'} \left({x}\right)\:. \\ $$
Question Number 32274 Answers: 0 Comments: 0
$${prove}\:{that}\:\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\:\:\frac{\mathrm{2}^{{k}} }{{x}^{\mathrm{2}^{{k}} } \:+\mathrm{1}}\:=\:\frac{\mathrm{1}}{{x}−\mathrm{1}}\:−\:\frac{\mathrm{2}^{{n}+\mathrm{1}} }{{x}^{\mathrm{2}^{{n}+\mathrm{1}} \:} −\mathrm{1}}\:\:. \\ $$
Question Number 32273 Answers: 1 Comments: 1
$${let}\:{put}\:{u}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:{k}\left({k}!\right) \\ $$$$\left.\mathrm{1}\right)\:{prove}\:{that}\:{u}_{{n}} =\left({n}+\mathrm{1}\right)!\:−\mathrm{1} \\ $$$$\left.\mathrm{2}\right)\:{study}\:{the}\:{convergence}\:{of}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{{u}_{{n}} }\:. \\ $$
Question Number 32271 Answers: 0 Comments: 0
$${study}\:{the}\:{convergence}\:{of}\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}^{\frac{\mathrm{1}}{\mathrm{3}}} }\:. \\ $$
Question Number 32269 Answers: 1 Comments: 0
$${find}\:\int\:\:\frac{{x}^{\mathrm{3}} }{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:{dx} \\ $$
Question Number 32266 Answers: 0 Comments: 0
Question Number 32264 Answers: 0 Comments: 1
Question Number 32258 Answers: 2 Comments: 0
$${find} \\ $$$$\int\:\frac{\mathrm{1}}{\mathrm{2}−{x}^{\mathrm{2}} }\:{dx} \\ $$
Question Number 32256 Answers: 0 Comments: 0
$$\left.\mathrm{1}\right){let}\:{a}>\mathrm{0}\:{and}\:{x}>\mathrm{0}\:{find}\:{lim}\:_{{x}\rightarrow{a}} \:\frac{{e}^{−{ax}^{\mathrm{2}} } \:−\:{e}^{−{xa}^{\mathrm{2}} } }{{a}^{{x}} \:−{x}^{{a}} }\:. \\ $$$$\left.\mathrm{2}\right){find}\:{lim}_{{x}\rightarrow\mathrm{2}} \:\:\:\frac{{e}^{−\mathrm{2}{x}^{\mathrm{2}} } \:−\:{e}^{−\mathrm{4}{x}} }{\mathrm{2}^{{x}} \:−{x}^{\mathrm{2}} }\:. \\ $$
Question Number 32255 Answers: 0 Comments: 0
$${le}\:{x}>\mathrm{0}\:{and}\:{a}>\mathrm{0}\:{find}\:{lim}_{{x}\rightarrow{a}} \:\frac{{log}_{{a}} \:\left({x}\right)\:−{log}_{{x}} \left({a}\right)}{{sinx}\:−{sina}}\:\:. \\ $$
Question Number 32265 Answers: 0 Comments: 0
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