Question Number 115999 by bemath last updated on 30/Sep/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:{x}\:\sqrt{\mathrm{cos}\:\mathrm{2}{x}}\:\sqrt[{\mathrm{3}\:}]{\mathrm{cos}\:\mathrm{3}{x}}\:\sqrt[{\mathrm{4}\:}]{\mathrm{cos}\:\mathrm{4}{x}}}{{x}^{\mathrm{2}} } \\ $$ Answered by bobhans last updated on 30/Sep/20 $${short}\:{cut}\:'{mr}\:{john}\:{santu}\:' \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:{x}\:\sqrt{\mathrm{cos}\:\mathrm{2}{x}}\:\sqrt[{\mathrm{3}\:}]{\mathrm{cos}\:\mathrm{3}{x}}\:\sqrt[{\mathrm{4}\:}]{\mathrm{cos}\:\mathrm{4}{x}}}{{x}^{\mathrm{2}}…
Question Number 115961 by Ar Brandon last updated on 29/Sep/20 Commented by Dwaipayan Shikari last updated on 29/Sep/20 $$\mathrm{Indetermined} \\ $$ Commented by Ar Brandon…
Question Number 181464 by cortano1 last updated on 25/Nov/22 $$\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\left(\mathrm{tan}\:\mathrm{x}\right)−\mathrm{sin}\:\left(\mathrm{x}\right)}{\mathrm{x}−\mathrm{tan}\:\left(\mathrm{x}\right)}\:=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 50395 by Abdo msup. last updated on 16/Dec/18 $${let}\:{U}_{{n}} =\left({e}−\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{{n}} \right)^{\sqrt{{n}^{\mathrm{2}} \:+\mathrm{2}}−\sqrt{{n}^{\mathrm{2}} \:+\mathrm{1}}} \\ $$$${calculate}\:{lim}_{{n}\rightarrow+\infty} \:\:{U}_{{n}} \\ $$ Terms of Service Privacy Policy…
Question Number 181462 by mathlove last updated on 25/Nov/22 Commented by Frix last updated on 26/Nov/22 $$\mathrm{I}\:\mathrm{think}\:\mathrm{it}'\mathrm{s}\:\frac{\mathrm{1}}{\mathrm{2e}}\:\mathrm{but}\:\mathrm{I}'\mathrm{m}\:\mathrm{not}\:\mathrm{100\%}\:\mathrm{sure} \\ $$$$\mathrm{Can}\:\mathrm{someone}\:\mathrm{confirm}\:\mathrm{this}? \\ $$ Answered by SEKRET last…
Question Number 115889 by bemath last updated on 29/Sep/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}\:}]{\mathrm{1}+\mathrm{cos}\:\mathrm{2}{x}}\:−\sqrt[{\mathrm{3}\:}]{\mathrm{2}}}{{x}^{\mathrm{2}} .\mathrm{sin}\:\mathrm{3}{x}} \\ $$ Answered by bemath last updated on 29/Sep/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}\:}]{\mathrm{1}+\left(\mathrm{1}−\frac{\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{2}}\right)}−\sqrt[{\mathrm{3}\:}]{\mathrm{2}}}{{x}^{\mathrm{2}} .\mathrm{sin}\:\mathrm{3}{x}}\:=…
Question Number 50293 by pooja24 last updated on 15/Dec/18 $${li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\:\underset{{r}=\mathrm{1}} {\overset{\mathrm{2}{n}} {\sum}}\left(\frac{\mathrm{1}}{{r}+{n}}\right)=? \\ $$$${please}\:{help} \\ $$ Commented by maxmathsup by imad last updated on…
Question Number 115780 by bemath last updated on 28/Sep/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\mathrm{cos}\:{x}\right)^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} \:=\:? \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left({e}^{\mathrm{3}{x}} −\mathrm{5}{x}\right)^{\frac{\mathrm{1}}{{x}}} \:=? \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{e}^{\mathrm{2}{x}} −\mathrm{2}{e}^{{x}} +\mathrm{1}}{\mathrm{cos}\:\mathrm{3}{x}−\mathrm{2cos}\:\mathrm{2}{x}+\mathrm{cos}\:{x}}=? \\ $$$$\underset{{x}\rightarrow\mathrm{0}}…
Question Number 115777 by bemath last updated on 28/Sep/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:{x}−\mathrm{tan}^{−\mathrm{1}} {x}}{{x}^{\mathrm{2}} \:\mathrm{ln}\:\left(\mathrm{1}+{x}\right)} \\ $$ Answered by bobhans last updated on 28/Sep/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left({x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}+{Px}^{\mathrm{5}}…
Question Number 115775 by bemath last updated on 28/Sep/20 $$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt[{\mathrm{3}\:}]{\frac{\left(\mathrm{3}−\sqrt{{x}}\right)\left(\sqrt{{x}}+\mathrm{2}\right)}{\mathrm{8}{x}−\mathrm{4}}} \\ $$ Answered by bobhans last updated on 28/Sep/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\sqrt[{\mathrm{3}\:}]{\frac{\left(\mathrm{3}−\sqrt{{x}}\right)\left(\sqrt{{x}}+\mathrm{2}\right)}{\mathrm{8}{x}−\mathrm{4}}}\:=\:\sqrt[{\mathrm{3}\:}]{\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{6}+\sqrt{{x}}−{x}}{\mathrm{8}{x}−\mathrm{4}}} \\ $$$$=\sqrt[{\mathrm{3}\:}]{\underset{{x}\rightarrow\infty}…