Question Number 102036 by Study last updated on 06/Jul/20 $${li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\frac{\sqrt[{\mathrm{5}}]{{x}^{\mathrm{2}} −\mathrm{1}}\:+\sqrt[{\mathrm{3}}]{{x}+\mathrm{1}}}{\:\sqrt[{\mathrm{3}}]{{x}−\mathrm{1}}\:+\sqrt{{x}+\mathrm{1}}}=? \\ $$ Answered by john santu last updated on 06/Jul/20 $${L}'{Hopital}\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}}…
Question Number 102034 by Study last updated on 06/Jul/20 $${li}\underset{{x}\rightarrow\mathrm{2}} {{m}}\frac{\sqrt[{\mathrm{3}}]{\mathrm{2}{x}+\mathrm{4}}\:−\mathrm{2}}{{x}^{\mathrm{2}} −{x}−\mathrm{2}}=? \\ $$ Answered by Dwaipayan Shikari last updated on 06/Jul/20 $$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{\sqrt[{\mathrm{3}}]{\mathrm{2}{x}+\mathrm{4}}−\mathrm{2}}{\mathrm{2}{x}+\mathrm{4}−\mathrm{8}}.\frac{\mathrm{2}{x}−\mathrm{4}}{\left({x}−\mathrm{2}\right)\left({x}+\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{8}^{\frac{−\mathrm{2}}{\mathrm{3}}} .\frac{\mathrm{2}}{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{18}}…
Question Number 102020 by Dwaipayan Shikari last updated on 06/Jul/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left({tan}\left(\frac{\pi}{\mathrm{4}}−{x}\right)\right)^{\frac{\mathrm{1}}{{x}}} \\ $$ Answered by john santu last updated on 06/Jul/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+\left(\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}−{x}\right)−\mathrm{1}\right)\right)^{\frac{\mathrm{1}}{{x}}} \\…
Question Number 101981 by Rohit@Thakur last updated on 05/Jul/20 $${li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\left(\frac{{tanx}}{{x}}\right)^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} \\ $$ Answered by Dwaipayan Shikari last updated on 06/Jul/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{{tanx}}{{x}}\right)^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}…
Question Number 101977 by Ar Brandon last updated on 05/Jul/20 $$\mathrm{Given}\:\mathrm{2}\:\mathrm{functions},\:\mathrm{f}\:\mathrm{and}\:\mathrm{g},\:\mathrm{n}-\mathrm{times}\:\mathrm{derivable}\:\mathrm{within} \\ $$$$\mathrm{the}\:\mathrm{open}\:\mathrm{interval},\:\mathbb{R}\:\mathrm{and}\:\mathrm{verify}\:\mathrm{the}\:\mathrm{property} \\ $$$$\mathrm{f}\left(\mathrm{x}_{\mathrm{0}} \right)=\mathrm{f}^{\left(\mathrm{k}\right)} \left(\mathrm{x}_{\mathrm{0}} \right)=\mathrm{0}\:,\:\mathrm{g}\left(\mathrm{x}_{\mathrm{0}} \right)=\mathrm{g}^{\left(\mathrm{k}\right)} \left(\mathrm{x}_{\mathrm{0}} \right)=\mathrm{0}\:,\:\forall\mathrm{k}\in\left\{\mathrm{1},\mathrm{2},…,\mathrm{n}−\mathrm{1}\right\} \\ $$$$\mathrm{Show}\:\mathrm{that}\:\underset{\mathrm{x}\rightarrow\mathrm{x}_{\mathrm{0}} } {\mathrm{lim}}\frac{\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{g}\left(\mathrm{x}\right)}=\frac{\mathrm{f}^{\left(\mathrm{n}\right)}…
Question Number 167434 by greogoury55 last updated on 16/Mar/22 $$\:\:\:\:\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:\frac{{x}^{{n}} −{a}^{{n}} −{na}^{{n}−\mathrm{1}} \left({x}−{a}\right)}{\left({x}−{a}\right)^{\mathrm{2}} }=? \\ $$ Answered by qaz last updated on 16/Mar/22 $$\underset{\mathrm{x}\rightarrow\mathrm{a}}…
Question Number 167421 by infinityaction last updated on 16/Mar/22 Answered by mindispower last updated on 16/Mar/22 $${cot}^{\mathrm{2}} \left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)={cot}^{\mathrm{2}} \left({s}\pi\right)=\left({i}\frac{{e}^{{is}} +{e}^{−{is}} }{{e}^{{is}} −{e}^{−{is}} }\right)^{\mathrm{2}} = \\…
Question Number 101822 by john santu last updated on 04/Jul/20 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\phi^{{n}+\mathrm{1}} −\left(−\phi\right)^{−{n}−\mathrm{1}} }{\phi^{{n}} −\left(−\phi\right)^{−{n}} }\:=\: \\ $$$$\left({JS}\:\circledast\right) \\ $$ Answered by bobhans last updated…
Question Number 167350 by infinityaction last updated on 13/Mar/22 Answered by Jamshidbek last updated on 13/Mar/22 $$\mathrm{Telegram}:@\mathrm{math\_undergraduate} \\ $$$$\mathrm{Problem}\:\mathrm{15}. \\ $$$$\mathrm{This}\:\mathrm{has}\:\mathrm{problem}\:\mathrm{solution}\:\mathrm{in}\:\mathrm{telegram}\:\mathrm{channel} \\ $$ Commented by…
Question Number 101770 by john santu last updated on 04/Jul/20 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}^{\mathrm{13}} +\mathrm{2}^{\mathrm{13}} +\mathrm{3}^{\mathrm{13}} +\mathrm{4}^{\mathrm{13}} +…+{n}^{\mathrm{13}} }{{n}^{\mathrm{14}} }\:? \\ $$ Commented by Dwaipayan Shikari last…