Question Number 49326 by Saorey last updated on 05/Dec/18 $$\mathrm{please}\:\mathrm{help}\:\mathrm{me}! \\ $$$$\begin{cases}{\mathrm{u}_{\mathrm{1}} ={a},\:\mathrm{u}_{\mathrm{2}} =\mathrm{b}}\\{\mathrm{u}_{\mathrm{n}+\mathrm{2}} =\mathrm{3}\sqrt[{\mathrm{5}}]{\mathrm{u}_{\mathrm{n}+\mathrm{1}} }+\mathrm{13}\sqrt[{\mathrm{5}}]{\mathrm{u}_{\mathrm{n}} }\:,\mathrm{n}\in\mathbb{N}^{\ast} }\end{cases} \\ $$$$\mathrm{show}\:\mathrm{that}\:\left(\mathrm{u}_{\mathrm{n}} \right)\:\mathrm{have}\:\mathrm{limit}\:\mathrm{and}\:\mathrm{find}\: \\ $$$$\mathrm{its}\:\mathrm{limit}. \\ $$…
Question Number 180359 by mathlove last updated on 11/Nov/22 Answered by Frix last updated on 11/Nov/22 $$\mathrm{certainly}\:\mathrm{you}\:\mathrm{do}\:\mathrm{not}\:\mathrm{mean}\:{x}!\:\mathrm{which}\:\mathrm{is} \\ $$$$\mathrm{defined}\:\mathrm{for}\:{x}\in\mathbb{N}\:\Rightarrow\:\mathrm{no}\:\mathrm{limit}\:\mathrm{exists} \\ $$$$\mathrm{but}\:\mathrm{you}\:\mathrm{mean}\:{x}!=\Gamma\left({x}+\mathrm{1}\right) \\ $$$$\mathrm{we}\:\mathrm{know}\:\Gamma\left(\mathrm{1}\right)=\mathrm{1} \\ $$$$\mathrm{let}\:\Gamma\left({x}+\mathrm{1}\right)={f}\left({x}\right)…
Question Number 114754 by bemath last updated on 21/Sep/20 $${Without}\:{L}'{Hopital}\: \\ $$$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\:\frac{\left(\frac{\mathrm{1}}{\mathrm{cos}\:^{\mathrm{2}} {x}}\:−\mathrm{2tan}\:{x}\:\right)}{\mathrm{cos}\:\mathrm{2}{x}}\:? \\ $$ Commented by bemath last updated on 21/Sep/20 $${gave}\:{kudos}\:{all}\:{sir} \\…
Question Number 114696 by bemath last updated on 20/Sep/20 $$\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sinh}\:\left(\mathrm{2}{x}\right)−\mathrm{sin}\:\mathrm{2}{x}}{{x}^{\mathrm{5}} }\:=? \\ $$ Answered by bobhans last updated on 20/Sep/20 $${e}^{\mathrm{2}{x}} =\mathrm{1}+\mathrm{2}{x}+\mathrm{2}{x}^{\mathrm{2}} +\frac{\mathrm{4}{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{2}{x}^{\mathrm{4}}…
Question Number 114637 by bobhans last updated on 20/Sep/20 $$\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\:\frac{\frac{\pi}{\mathrm{2}}−\mathrm{cos}\:^{−\mathrm{1}} \left(\mathrm{2}{x}−\pi\right)}{\mathrm{1}−\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{2}{x}}{\pi}\right)}\:? \\ $$ Answered by bemath last updated on 20/Sep/20 $${setting}\:{x}=\frac{\pi}{\mathrm{2}}+{p}\rightarrow\mathrm{2}{x}=\pi+\mathrm{2}{p} \\ $$$$\underset{{p}\rightarrow\mathrm{0}}…
Question Number 114543 by bemath last updated on 19/Sep/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{2}} \:\mathrm{arctan}\:\left(\mathrm{2}{x}\right)}{{x}\:\mathrm{cos}\:{x}+\mathrm{tan}\:{x}}\:? \\ $$ Answered by bobhans last updated on 19/Sep/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{2}} \:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}{x}\right)}{{x}\:\mathrm{cos}\:{x}+\mathrm{tan}\:{x}}\:=\:\underset{{x}\rightarrow\mathrm{0}}…
Question Number 114533 by 675480065 last updated on 19/Sep/20 Answered by abdomsup last updated on 19/Sep/20 $${u}_{{n}} =\prod_{{k}=\mathrm{1}} ^{{n}} \left(\mathrm{1}+\frac{{k}}{{n}^{\mathrm{2}} }\right) \\ $$$$\left.{a}\right){let}\:\varphi\left({x}\right)={x}−{ln}\left(\mathrm{1}+{x}\right) \\ $$$${with}\:{x}\geqslant\mathrm{0}\:{we}\:{have}\:\varphi\left(\mathrm{0}\right)=\mathrm{0}…
Question Number 179950 by mathlove last updated on 04/Nov/22 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left[\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)+\left(\mathrm{1}+\frac{\mathrm{2}}{{x}}\right)^{\frac{\mathrm{2}}{\mathrm{2}}} +\left(\mathrm{1}+\frac{\mathrm{3}}{{x}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} +\centerdot\centerdot\centerdot\centerdot+\left(\mathrm{1}+\frac{{x}}{{x}}\right)^{\frac{\mathrm{1}}{{x}}} \right]=? \\ $$ Commented by mahdipoor last updated on 04/Nov/22 $$ \\…
Question Number 114411 by bemath last updated on 19/Sep/20 $$\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+{x}\right)^{−\frac{\mathrm{3}}{\mathrm{4}}} −\left(\mathrm{1}+{x}\right)^{−\frac{\mathrm{1}}{\mathrm{4}}} }{\mathrm{2}{x}}\:=? \\ $$ Answered by john santu last updated on 19/Sep/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+{x}\right)^{−\frac{\mathrm{3}}{\mathrm{4}}}…
Question Number 114405 by bemath last updated on 19/Sep/20 $$\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\:\mathrm{tan}\:{x}}{\:\sqrt{\mathrm{3}}\:\mathrm{cos}\:{x}−\mathrm{sin}\:^{\mathrm{2}} {x}−\sqrt{\mathrm{3}}}\:? \\ $$ Answered by bobhans last updated on 19/Sep/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\:\mathrm{tan}\:{x}}{\:\sqrt{\mathrm{3}}\left(\mathrm{cos}\:{x}−\mathrm{1}\right)−\mathrm{sin}\:^{\mathrm{2}} {x}}\:= \\…