Question Number 223082 by DarthMath last updated on 14/Jul/25 $${if}\:\underset{{x}\rightarrow+\infty} {{lim}x}−{f}\left({x}\right)=+\infty\:{and}\:\underset{{x}\rightarrow+\infty} {{lim}x}+{f}\left({x}\right)=+\infty \\ $$$${can}\:{we}\:{determine}\:\underset{{x}\rightarrow+\infty} {{lim}}\frac{{x}−{f}\left({x}\right)}{{x}+{f}\left({x}\right)} \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 222756 by Osefavour last updated on 06/Jul/25 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\sqrt{\boldsymbol{\mathrm{cos}}\left(\boldsymbol{\mathrm{x}}\right)}}{\:\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{xcos}}\left(\sqrt{\boldsymbol{\mathrm{x}}}\right)} \\ $$ Answered by gregori last updated on 07/Jul/25 $$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:{x}}{{x}\left(\mathrm{1}−\mathrm{cos}\:\sqrt{{x}}\:\right)\left(\mathrm{1}+\sqrt{\mathrm{cos}\:{x}}\right)} \\ $$$$\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2sin}\:^{\mathrm{2}}…
Question Number 222478 by mnjuly1970 last updated on 28/Jun/25 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{S}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \frac{{H}_{{n}} }{{n}^{\mathrm{2}} }\:=\:? \\ $$$$\:{note}:\:\:\:{H}_{{n}} =\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{3}}\:+…+\frac{\mathrm{1}}{{n}}\: \\ $$ Answered by MrGaster…
Question Number 222261 by MathematicalUser2357 last updated on 21/Jun/25 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{tan}\left({x}^{\mathrm{2}} +\mathrm{4}{x}\right)}{\mathrm{sin}\left(\mathrm{9}{x}^{\mathrm{2}} +{x}\right)} \\ $$$$\mathrm{No}\:\mathrm{L}'\mathrm{h}\hat {\mathrm{o}pital}'\mathrm{s}\:\mathrm{rule}\:\mathrm{allowed}! \\ $$ Answered by gregori last updated on 21/Jun/25…
Question Number 221392 by Davidtim last updated on 02/Jun/25 $$\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\sqrt{{x}−\mathrm{3}}=? \\ $$$$\left.\mathrm{1}\right)\:\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{3} \\ $$$$\left.\mathrm{3}\right)\:{Does}\:{not}\:{exist} \\ $$$$\left.\mathrm{4}\right)\:{Undefined} \\ $$ Answered by Frix last…
Question Number 221348 by RoseAli last updated on 31/May/25 $$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\mathrm{4}−\mathrm{2}^{{x}} }{{x}−\mathrm{2}} \\ $$ Answered by mr W last updated on 31/May/25 $$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{4}\left(\mathrm{1}−\mathrm{2}^{{x}} \right)}{{x}}…
Question Number 221347 by RoseAli last updated on 31/May/25 $$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\mathrm{4}−{x}^{\mathrm{2}} }{{x}−\mathrm{2}} \\ $$ Answered by AntonCWX8 last updated on 31/May/25 $$\frac{\mathrm{4}−{x}^{\mathrm{2}} }{{x}−\mathrm{2}} \\ $$$$=\frac{\left(\mathrm{2}+{x}\right)\left(\mathrm{2}−{x}\right)}{{x}−\mathrm{2}}…
Question Number 221260 by mnjuly1970 last updated on 28/May/25 Answered by maths2 last updated on 30/May/25 $$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\sqrt{{n}}.\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\frac{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!} \\ $$$${n}!\sim\sqrt{\mathrm{2}\pi{n}}.\left(\frac{{n}}{{e}}\right)^{{n}} ..…
Question Number 221047 by universe last updated on 23/May/25 Answered by breniam last updated on 24/May/25 $$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\underset{{k}=\mathrm{0}} {\overset{\mathrm{2}{r}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}+\mathrm{1}} \left({k}−\mathrm{1}\right)}{\left({k}+\mathrm{1}\right)!}−\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\underset{{k}=\mathrm{0}}…
Question Number 221034 by Nicholas666 last updated on 23/May/25 Answered by vnm last updated on 23/May/25 $${let}\:{lim}={a} \\ $$$$\mathrm{sin}\frac{{a}}{\:\sqrt{{n}}}=\frac{{a}}{\:\sqrt{{n}}}−\frac{\mathrm{1}}{\mathrm{6}}\frac{{a}^{\mathrm{3}} }{{n}\sqrt{{n}}}+{o}\left(\frac{\mathrm{1}}{{n}\sqrt{{n}}}\right)=\frac{{a}}{\:\sqrt{{n}}}\left(\mathrm{1}−\frac{{a}^{\mathrm{2}} }{\mathrm{6}{n}}+{o}\left(\frac{\mathrm{1}}{{n}}\right)\right) \\ $$$$\sqrt{{n}+\mathrm{1}}=\sqrt{{n}}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{n}}}=\sqrt{{n}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{n}}+{o}\left(\frac{\mathrm{1}}{{n}}\right)\right) \\ $$$$\mathrm{sin}\frac{{a}}{\:\sqrt{{n}}}\centerdot\sqrt{{n}+\mathrm{1}}=\frac{{a}}{\:\sqrt{{n}}}\sqrt{{n}}\left(\left(\mathrm{1}−\frac{{a}^{\mathrm{2}}…