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
$${L}'{Hopital}\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{\frac{\mathrm{2}{x}}{\mathrm{5}\:\sqrt[{\mathrm{5}}]{\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{4}} }}+\:\frac{\mathrm{1}}{\mathrm{3}\:\sqrt[{\mathrm{3}}]{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }}}{\frac{\mathrm{1}}{\mathrm{3}\:\sqrt[{\mathrm{3}}]{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }}\:+\:\frac{\mathrm{1}}{\mathrm{2}\:\sqrt{\left({x}+\mathrm{1}\right)}}}\:=\: \\ $$$$\frac{\frac{\mathrm{1}}{\mathrm{3}}}{\frac{\mathrm{1}}{\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{2}}}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\:×\:\frac{\mathrm{6}}{\mathrm{5}}\:=\:\frac{\mathrm{2}}{\mathrm{5}} \\ $$$$\left({JS}\:\circledast\right) \\ $$
Answered by bemath last updated on 06/Jul/20
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\sqrt[{\mathrm{3}}]{{x}+\mathrm{1}}−\sqrt[{\mathrm{5}}]{\mathrm{1}−{x}^{\mathrm{2}} }}{\:\sqrt{{x}+\mathrm{1}}−\sqrt[{\mathrm{3}}]{\mathrm{1}−{x}}}\:= \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\frac{{x}}{\mathrm{3}}+\mathrm{1}\right)−\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{5}}\right)}{\left(\frac{{x}}{\mathrm{2}}+\mathrm{1}\right)−\left(\mathrm{1}−\frac{{x}}{\mathrm{3}}\right)}\:= \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}\left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{{x}}{\mathrm{5}}\right)}{{x}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{3}}×\frac{\mathrm{6}}{\mathrm{5}}\:=\:\frac{\mathrm{2}}{\mathrm{5}} \\ $$
Commented by 1549442205 last updated on 06/Jul/20
$$\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{m}} =\mathrm{1}+\frac{\mathrm{m}}{\mathrm{1}!}\mathrm{x}+\frac{\mathrm{m}\left(\mathrm{m}−\mathrm{1}\right)}{\mathrm{2}!}\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{m}\left(\mathrm{m}−\mathrm{1}\right)\left(\mathrm{m}−\mathrm{2}\right)}{\mathrm{3}!}\mathrm{x}^{\mathrm{3}} +… \\ $$
Commented by Study last updated on 06/Jul/20
$${write}\:{the}\:{furmolla} \\ $$
Commented by bemath last updated on 06/Jul/20
$${maclaurin}\:{series} \\ $$
Commented by Dwaipayan Shikari last updated on 06/Jul/20
Answered by Dwaipayan Shikari last updated on 06/Jul/20
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\mathrm{2}{x}}{\mathrm{5}\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\frac{\mathrm{4}}{\mathrm{5}}} }+\frac{\mathrm{1}}{\mathrm{3}\left({x}+\mathrm{1}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} }}{\frac{\mathrm{1}}{\mathrm{3}\left({x}−\mathrm{1}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} }+\frac{\mathrm{1}}{\mathrm{2}\left({x}+\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} }}=\frac{\frac{\mathrm{2}{x}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{3}}}{\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{2}}}=\frac{\mathrm{2}}{\mathrm{5}} \\ $$