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 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\sqrt[{\mathrm{3}\:}]{\frac{\left(\mathrm{3}−\sqrt{{x}}\right)\left(\sqrt{{x}}+\mathrm{2}\right)}{\mathrm{8}{x}−\mathrm{4}}}\:=\:\sqrt[{\mathrm{3}\:}]{\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{6}+\sqrt{{x}}−{x}}{\mathrm{8}{x}−\mathrm{4}}} \\ $$$$=\sqrt[{\mathrm{3}\:}]{\underset{{x}\rightarrow\infty}…
Question Number 181173 by universe last updated on 22/Nov/22 Commented by CElcedricjunior last updated on 22/Nov/22 $$\left.\boldsymbol{{a}}\right)\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)= \\ $$$$\left.\boldsymbol{{b}}\right)\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)= \\ $$$$\left.\boldsymbol{{c}}\right)\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)=\boldsymbol{{log}}\left(\boldsymbol{{e}}\right) \\ $$ Terms of…
Question Number 115621 by sachin1221 last updated on 27/Sep/20 $$ \\ $$$$\:\:\:\:\:\:\:…\:{advanced}\:\:\:{calculus}…\: \\ $$$$\:\:\:\:\:\:\:{evaluate}\::: \\ $$$${show}\:{that}\:{lim}_{{n}\rightarrow\infty} \frac{\mathrm{1}}{{n}}\left[{cos}^{\mathrm{2}{p}} \frac{\pi}{\mathrm{2}{n}}+{cos}^{\mathrm{2}{p}} \frac{\mathrm{2}\pi}{\mathrm{2}{n}}+{cos}^{\mathrm{2}{p}} \frac{\mathrm{3}\pi}{\mathrm{2}{n}}……{cos}^{\mathrm{2}{p}} \frac{\pi}{\mathrm{2}}\right]\:=\underset{{r}=\mathrm{1}} {\overset{{p}} {\prod}}\frac{{p}+{r}}{\mathrm{4}{r}} \\ $$…
Question Number 115616 by bobhans last updated on 27/Sep/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{8}\:}]{\mathrm{256}+\mathrm{tan}\:\:{x}}\:−\sqrt[{\mathrm{3}\:}]{\mathrm{1}+\mathrm{sin}\:{x}}\:−\mathrm{1}}{\:\sqrt[{\mathrm{6}\:}]{\mathrm{1}+\mathrm{tan}\:{x}}\:−\mathrm{1}}\:? \\ $$ Answered by john santu last updated on 27/Sep/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}\:\sqrt[{\mathrm{8}\:}]{\mathrm{1}+\frac{\mathrm{tan}\:{x}}{\mathrm{256}}}−\sqrt[{\mathrm{3}\:}]{\mathrm{1}+\mathrm{sin}\:{x}}\:−\mathrm{1}}{\:\sqrt[{\mathrm{6}\:}]{\mathrm{1}+\mathrm{tan}\:{x}}\:−\mathrm{1}}\:= \\ $$$$\underset{{x}\rightarrow\mathrm{0}}…
Question Number 115604 by bemath last updated on 27/Sep/20 $$\left(\mathrm{1}\right)\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{tan}\:{x}}\:=? \\ $$$$\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\pi} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{2}\:+\mathrm{cos}\:{x}}\:−\mathrm{1}}{\left(\pi−{x}\right)^{\mathrm{2}} }\:=\:? \\ $$ Commented by Dwaipayan Shikari last updated on 27/Sep/20…
Question Number 181104 by cortano1 last updated on 21/Nov/22 $$\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{ln}\:\left(\mathrm{1}+\frac{\mathrm{4}}{\mathrm{x}}\right)}{\pi−\mathrm{arctan}\:\left(\mathrm{2x}\right)}\:=?\: \\ $$ Commented by Frix last updated on 21/Nov/22 $$\mathrm{wrong}. \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{ln}\:\left(\mathrm{1}+\frac{\mathrm{4}}{{x}}\right)}{\pi−\mathrm{arctan}\:\mathrm{2}{x}}\:=\frac{\mathrm{0}}{\frac{\pi}{\mathrm{2}}}=\mathrm{0} \\…