Question Number 204558 by universe last updated on 22/Feb/24 $$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{n}^{−\mathrm{3}/\mathrm{2}} \left[\left(\mathrm{n}+\mathrm{1}\right)^{\left(\mathrm{n}+\mathrm{1}\right)} \left(\mathrm{n}+\mathrm{2}\right)^{\left(\mathrm{n}+\mathrm{2}\right)} …\left(\mathrm{2n}\right)^{\mathrm{2n}} \right]^{\mathrm{1}/\mathrm{n}^{\mathrm{2}} } \:=\:? \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 204477 by universe last updated on 18/Feb/24 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{2n}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{x}^{\mathrm{n}} }{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\right)^{\mathrm{n}} =? \\ $$ Answered by witcher3 last updated on 18/Feb/24…
Question Number 204396 by Thierrybadouana last updated on 15/Feb/24 Answered by Faetmaaa last updated on 27/Feb/24 $$\mathrm{ln}\left(\mathrm{1}+{y}\right)\:\underset{{y}\rightarrow\mathrm{0}} {\sim}\:{y} \\ $$$$\mathrm{sin}\left({y}\right)\:\underset{{y}\rightarrow\mathrm{0}} {\sim}\:{y} \\ $$$$\underset{\begin{matrix}{{x}\rightarrow\mathrm{0}}\\{{x}\neq\mathrm{0}}\end{matrix}} {\mathrm{lim}}\:\frac{\mathrm{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\mathrm{sin}^{\mathrm{2}}…
Question Number 204395 by Thierrybadouana last updated on 15/Feb/24 Answered by Lindemann last updated on 20/Feb/24 $$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} }×\frac{{x}^{\mathrm{2}} }{{sin}^{\mathrm{2}} \left({x}\right)}\right)\:=\:\mathrm{1} \\ $$ Terms…
Question Number 203980 by Davidtim last updated on 03/Feb/24 $$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{{ax}^{\mathrm{2}} +{bx}+\mathrm{6}}{{x}^{\mathrm{2}} −{x}−\mathrm{2}}=\mathrm{10}\:\:;\:\:\:{find}\:\:{a}=?\:\wedge\:{b}=? \\ $$ Answered by AST last updated on 03/Feb/24 $$\frac{{a}\left({x}^{\mathrm{2}} −{x}−\mathrm{2}\right)+\left({b}+{a}\right){x}+\mathrm{6}+\mathrm{2}{a}}{{x}^{\mathrm{2}} −{x}−\mathrm{2}}=\mathrm{10}…
Question Number 203668 by Perelman last updated on 25/Jan/24 Answered by witcher3 last updated on 25/Jan/24 $$\mathrm{x}\overset{\mathrm{f}} {\rightarrow}\frac{\mathrm{1}}{\mathrm{4}−\mathrm{3x}} \\ $$$$\mathrm{f}'\left(\mathrm{x}\right)=\frac{\mathrm{3}}{\left(\mathrm{4}−\mathrm{3x}\right)^{\mathrm{2}} } \\ $$$$\forall\mathrm{x}\in\left[−\mathrm{1},\frac{\mathrm{1}}{\mathrm{3}}\right]\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\in\left[\frac{\mathrm{1}}{\mathrm{7}},\frac{\mathrm{1}}{\mathrm{3}}\right]…..\left(\mathrm{1}\right) \\ $$$$\left.\Rightarrow\mathrm{f}\left[−\mathrm{1},\frac{\mathrm{1}}{\mathrm{3}}\right]\right)\subset\left[\frac{\mathrm{1}}{\mathrm{7}},\frac{\mathrm{1}}{\mathrm{3}}\right]…
Question Number 203508 by Fridunatjan08 last updated on 20/Jan/24 $${Evaluate}\:{the}\:{given}\:{limit}: \\ $$$$\underset{{n}\rightarrow\infty} {{lim}}\frac{\sqrt[{{n}}]{\mathrm{8}}−\mathrm{1}}{\:\sqrt[{{n}}]{\mathrm{16}}−\mathrm{1}} \\ $$ Answered by esmaeil last updated on 20/Jan/24 $$={Y} \\ $$$$\frac{\mathrm{1}}{{n}}={p}\rightarrow\left({n}\rightarrow\infty\rightarrow{p}\rightarrow\mathrm{0}\right)…
Question Number 203330 by Calculusboy last updated on 16/Jan/24 Answered by MM42 last updated on 16/Jan/24 $$={lim}_{{x}\rightarrow\mathrm{1}} \:\frac{{sin}\left({x}−\mathrm{1}\right)}{\left({x}−\mathrm{1}\right)}\:×\frac{\mathrm{1}}{{x}+\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}}\:\:\checkmark \\ $$ Commented by Calculusboy last updated…
Question Number 203247 by mathlove last updated on 13/Jan/24 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{x}\:{tan}\frac{\pi}{\mathrm{2}}\left(\mathrm{1}+{x}\right)=? \\ $$ Answered by MM42 last updated on 13/Jan/24 $$={lim}_{{x}\rightarrow\mathrm{0}} \:−{xcot}\frac{\pi}{\mathrm{2}}{x}={lim}_{{x}\rightarrow\mathrm{0}} \:−\frac{\frac{\pi}{\mathrm{2}}{xcos}\frac{\pi}{\mathrm{2}}{x}}{{sin}\frac{\pi}{\mathrm{2}}{x}}×\frac{\mathrm{2}}{\pi} \\ $$$$=\:−\frac{\mathrm{2}}{\pi}\:\checkmark…
Question Number 203180 by Calculusboy last updated on 11/Jan/24 Answered by Rana_Ranino last updated on 11/Jan/24 $$\mathrm{using}\:\mathrm{arcsin}^{\mathrm{2}} \left(\mathrm{z}\right)=\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{4}^{\mathrm{n}} \mathrm{z}^{\mathrm{2n}} }{\mathrm{n}^{\mathrm{2}} \begin{pmatrix}{\mathrm{2n}}\\{\:\mathrm{n}}\end{pmatrix}}\:\:\mathrm{take}\:\mathrm{z}=\frac{\mathrm{1}}{\mathrm{2}}\: \\ $$$$\underset{\mathrm{n}=\mathrm{1}}…