Question Number 71767 by psyche last updated on 19/Oct/19 $${Let}\:{f}\:{be}\:{continuous}\:{on}\:{a}\:{closed}\:{and}\:{bounded}\:{subset}\:{E},\:{then}\:{show}\:{that}\:{f}\:{is}\:{uniformly}\:{continuous}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 71739 by Tony Lin last updated on 19/Oct/19 $${find}\:{the}\:{asymptote}\:{of}\:{folium}\:{of}\: \\ $$$${Descartes}\:{x}^{\mathrm{3}} +{y}^{\mathrm{3}} =\mathrm{3}{axy},\:{and}\:{a}\:{is}\:{a} \\ $$$${constant}\:>\mathrm{0} \\ $$ Answered by Tony Lin last updated…
Question Number 6189 by enigmeyou last updated on 17/Jun/16 $$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\lfloor{x}^{\lfloor{xlnx}\rfloor} \rfloor=? \\ $$ Commented by FilupSmith last updated on 18/Jun/16 $$\lfloor{x}\rfloor\:\mathrm{is}\:\mathrm{only}\:\mathrm{really}\:\mathrm{applicable}\:\mathrm{for}\:\mathrm{non} \\ $$$$\mathrm{integers}.\:\lfloor{x}\mathrm{ln}{x}\rfloor\:\mathrm{is}\:\mathrm{only}\:\mathrm{applicable}\:\mathrm{if} \\…
Question Number 137255 by Nith last updated on 31/Mar/21 Answered by Dwaipayan Shikari last updated on 31/Mar/21 $${sinx}\sim{x} \\ $$$$\mathrm{2}^{{sinx}} \sim\mathrm{2}^{{x}} \sim\mathrm{1}+{xlog}\left(\mathrm{2}\right) \\ $$$$\mathrm{3}^{{sinx}} \sim\mathrm{1}+{xlog}\left(\mathrm{3}\right)…
Question Number 71649 by TawaTawa last updated on 18/Oct/19 Commented by mathmax by abdo last updated on 18/Oct/19 $$\sum_{{n}=\mathrm{1}} ^{\left[{x}\right]} \sqrt{{n}}=\mathrm{1}+\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}+…+\sqrt{\left[{x}\right]} \\ $$$$\frac{\mathrm{2}}{\mathrm{3}}\sqrt{\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} }=\frac{\mathrm{2}}{\mathrm{3}}\sqrt{{x}^{\mathrm{3}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{x}}\right)^{\mathrm{3}}…
Question Number 137157 by mathlove last updated on 30/Mar/21 Commented by mathlove last updated on 30/Mar/21 $${with}\:{out}\:{macloreen}\:{sirees}\:{and}\:{H}-{pital}\:{rools} \\ $$ Commented by mathlove last updated on…
Question Number 71513 by aliesam last updated on 16/Oct/19 Commented by kaivan.ahmadi last updated on 16/Oct/19 $$×\frac{\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}+\sqrt{{x}−\mathrm{1}}}{\:\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}+\sqrt{{x}−\mathrm{1}}}={lim}_{{x}\rightarrow+\infty} \frac{{x}^{\mathrm{2}} +\mathrm{2}}{\:\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}+\sqrt{{x}−\mathrm{1}}}= \\ $$$$\equiv{lim}_{{x}\rightarrow+\infty} \frac{{x}^{\mathrm{2}}…
Question Number 71362 by malwaan last updated on 14/Oct/19 $$\boldsymbol{\mathrm{please}}\:\boldsymbol{{prove}}\:\boldsymbol{{that}} \\ $$$$\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{lim}}}\frac{\boldsymbol{{x}}−\boldsymbol{{sinx}}}{\boldsymbol{{x}}^{\mathrm{3}} }\:=\frac{\mathrm{1}}{\mathrm{6}}\:\boldsymbol{{by}}\:\boldsymbol{{using}} \\ $$$$\boldsymbol{{x}}=\mathrm{3}\boldsymbol{{y}}\:\boldsymbol{{and}}\: \\ $$$$\boldsymbol{{sin}}\mathrm{3}\boldsymbol{{y}}=\mathrm{3}\boldsymbol{{siny}}−\mathrm{4}\boldsymbol{{sin}}^{\mathrm{3}} \boldsymbol{{y}} \\ $$ Terms of Service Privacy…
Question Number 71360 by 20190927 last updated on 14/Oct/19 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\sqrt{\mathrm{1}+\mathrm{2x}^{\mathrm{4}} }\mathrm{cos}\:\left(\sqrt{\mathrm{2}}\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}^{\mathrm{5}} \mathrm{ln}\:\left(\mathrm{1}−\mathrm{2x}^{\mathrm{3}} \right)} \\ $$ Commented by mathmax by abdo last updated on…
Question Number 5821 by FilupSmith last updated on 30/May/16 $${L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{x}^{{x}} \\ $$ Answered by bahmanfeshki last updated on 27/Feb/17 $$\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:{e}^{{x}\mathrm{ln}\:{x}} =\underset{{x}\rightarrow\mathrm{0}^{+} }…