Question Number 131882 by Eric002 last updated on 09/Feb/21 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}}{{x}+\sqrt[{\mathrm{4}}]{{x}+\mathrm{1}}−\mathrm{1}} \\ $$ Answered by liberty last updated on 09/Feb/21 $$\:\mathrm{L}'\mathrm{H}\ddot {\mathrm{o}pital}\:\mathrm{L}=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\:\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}\:\sqrt[{\mathrm{4}}]{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{3}} }}}\:\right]=\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}}=\:\frac{\mathrm{4}}{\mathrm{5}} \\…
Question Number 131836 by liberty last updated on 09/Feb/21 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{x}\left(\mathrm{2}+\mathrm{cos}\:\mathrm{x}\right)−\mathrm{3sin}\:\mathrm{x}}{\mathrm{x}^{\mathrm{5}} } \\ $$ Commented by EDWIN88 last updated on 10/Feb/21 $$\mathrm{another}\:\mathrm{way}\:\mathrm{L}'\mathrm{H}\hat {\mathrm{o}pital} \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}}…
Question Number 131825 by liberty last updated on 09/Feb/21 $$\:\:\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{2n}−\mathrm{1}}{\mathrm{2}^{\mathrm{n}} }\:? \\ $$ Answered by mr W last updated on 09/Feb/21 $${method}\:\mathrm{1} \\…
Question Number 131789 by bemath last updated on 08/Feb/21 $$\mathrm{Let}\:\varphi\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{x}^{\mathrm{4}} +\mathrm{3}\left(\mathrm{a}^{\mathrm{2}} −\sqrt{\mathrm{a}^{\mathrm{4}} +\mathrm{x}^{\mathrm{4}} }\:\right)}{\mathrm{x}^{\mathrm{8}} }\:;\:\mathrm{a}>\mathrm{0} \\ $$$$\mathrm{If}\:\varphi\:\mathrm{is}\:\mathrm{finite}\:\mathrm{then}\: \\ $$$$\left(\mathrm{a}\right)\:\mathrm{a}=\frac{\mathrm{3}}{\mathrm{2}}\:\:\:\:\left(\mathrm{b}\right)\:\mathrm{a}=\sqrt{\frac{\mathrm{3}}{\mathrm{2}}}\:\:\:\:\:\:\left(\mathrm{c}\right)\:\varphi=\frac{\mathrm{1}}{\mathrm{3}}\:\:\:\:\left(\mathrm{d}\right)\:\varphi=\frac{\mathrm{1}}{\mathrm{9}} \\ $$ Answered by Dwaipayan…
Question Number 131769 by liberty last updated on 08/Feb/21 $$\:\:\:\underset{{x}\rightarrow\pi/\mathrm{2}} {\mathrm{lim}}\frac{\mathrm{2}^{−\mathrm{cos}\:\mathrm{x}} \:−\mathrm{1}}{\mathrm{x}\left(\mathrm{x}−\frac{\pi}{\mathrm{2}}\right)}\:=?\: \\ $$ Answered by bemath last updated on 08/Feb/21 $$\:\underset{{x}\rightarrow\pi/\mathrm{2}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{x}}\:.\:\underset{{x}\rightarrow\pi/\mathrm{2}} {\mathrm{lim}}\:\frac{\mathrm{2}^{\mathrm{sin}\:\left(\mathrm{x}−\frac{\pi}{\mathrm{2}}\right)} −\mathrm{1}}{\mathrm{x}−\frac{\pi}{\mathrm{2}}}\:=…
Question Number 676 by 123456 last updated on 22/Feb/15 $${given}\:{two}\:{sequence}\:{a}_{{n}} >\mathrm{0},{b}_{{n}} >\mathrm{0}\:\:{such} \\ $$$$\forall{n}\in\mathbb{N}^{\ast} ,{a}_{{n}} ^{{n}} <{b}_{{n}} <{a}_{{n}} ^{\mathrm{1}/{n}} \\ $$$${a}.\:{if}\:\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}{a}_{{n}} \:{converge}\:{then}\:\underset{{n}=\mathrm{1}} {\overset{+\infty}…
Question Number 585 by 123456 last updated on 02/Feb/15 $${f}_{{n}} \left({x}\right)=\left(\mathrm{1}+\frac{{x}}{{n}}\right)^{{nx}} ,{n}\in\mathbb{N}^{\ast} ,{x}\in\mathbb{R},{x}\geqslant\mathrm{0} \\ $$$$\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\:{f}_{{n}} \left({x}\right)= \\ $$ Answered by prakash jain last updated…
Question Number 561 by 123456 last updated on 27/Jan/15 $${a}_{{n}+\mathrm{2}} =\frac{{a}_{{n}+\mathrm{1}} +{a}_{{n}} +{a}_{{n}+\mathrm{1}} {a}_{{n}} }{\mathrm{3}},{a}_{\mathrm{0}} ={e},{a}_{\mathrm{1}} =\pi \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{a}_{{n}} =? \\ $$ Terms of…
Question Number 66084 by F_Nongue last updated on 09/Aug/19 $${How}\:{to}\:{solve}\:{this}\:{limit}? \\ $$$$\underset{{x}\rightarrow\infty} {{lim}}\left(\mathrm{7}{x}+\frac{\mathrm{2}}{{x}}\right)^{{x}} \\ $$ Commented by mathmax by abdo last updated on 09/Aug/19 $${let}\:{A}\left({x}\right)\:=\left(\mathrm{7}{x}+\frac{\mathrm{2}}{{x}}\right)^{{x}}…
Question Number 494 by 123456 last updated on 15/Jan/15 $${a}_{\mathrm{0}} =\mathrm{0} \\ $$$${a}_{\mathrm{1}} =\mathrm{1} \\ $$$${a}_{{n}} =\frac{{a}_{{n}−\mathrm{1}} +{a}_{{n}−\mathrm{2}} +{a}_{{n}−\mathrm{1}} {a}_{{n}−\mathrm{2}} }{\mathrm{3}} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{a}_{{n}} =…