Question Number 57957 by rahul 19 last updated on 15/Apr/19 Commented by Smail last updated on 15/Apr/19 $${L}_{{n}} =\frac{\mathrm{2}^{{n}} +\left(−\mathrm{2}\right)^{{n}} }{\mathrm{3}^{{n}} }=\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{n}} \left(\mathrm{1}+\left(−\mathrm{1}\right)^{{n}} \right) \\…
Question Number 188985 by normans last updated on 10/Mar/23 $$ \\ $$$$\:\:\:\boldsymbol{\mathrm{Lim}}_{\boldsymbol{{x}}\rightarrow\sim} \:\:\frac{\mathrm{4}\boldsymbol{{x}}^{\mathrm{3}} −\mathrm{2}\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{5}\boldsymbol{{x}}+\mathrm{4}}{\mathrm{9}\boldsymbol{{x}}^{\mathrm{3}} −\mathrm{4}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{9}}\:=\:??\: \\ $$ Commented by MJS_new last updated on…
Question Number 188984 by normans last updated on 10/Mar/23 $$ \\ $$$$\:\:\boldsymbol{\mathrm{Lim}}_{\boldsymbol{{x}}\rightarrow\sim} \:\sqrt{\mathrm{16}\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{{x}}−\mathrm{1}}−\mathrm{4}\boldsymbol{{x}}−\mathrm{5}\:=\:??\:\:\:\: \\ $$$$ \\ $$ Commented by MJS_new last updated on 13/Mar/23…
Question Number 57866 by Mikael_Marshall last updated on 13/Apr/19 $$\underset{{n}\rightarrow\infty} {{lim}}\:\:\frac{\mathrm{3}^{{n}+\mathrm{2}} −\mathrm{2}.\mathrm{5}^{{n}+\mathrm{1}} }{\mathrm{3}^{{n}} −\mathrm{2}.\mathrm{5}^{{n}−\mathrm{1}} } \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 13/Apr/19 $$\underset{{n}\rightarrow\infty}…
Question Number 123398 by bemath last updated on 25/Nov/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}}]{\mathrm{1}+\mathrm{3}{x}}\:−\sqrt{\mathrm{1}+\mathrm{2}{x}}}{{x}^{\mathrm{2}} }\:? \\ $$ Answered by Dwaipayan Shikari last updated on 25/Nov/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}+\frac{\mathrm{3}{x}}{\mathrm{3}}+\frac{\mathrm{9}{x}^{\mathrm{2}} }{\mathrm{2}!}.\left(−\frac{\mathrm{2}}{\mathrm{9}}\right)−\mathrm{1}−\frac{\mathrm{2}{x}}{\mathrm{2}}+\frac{\mathrm{4}{x}^{\mathrm{2}}…
Question Number 188933 by horsebrand11 last updated on 09/Mar/23 $$\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{4}{x}−\mathrm{3}}+\sqrt{\mathrm{2}{x}−\mathrm{1}}−\mathrm{3}{x}+\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}}=? \\ $$ Commented by TUN last updated on 09/Mar/23 $$ \\ $$$$=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\sqrt{\mathrm{4}{x}−\mathrm{3}}−\left(\mathrm{2}{x}−\mathrm{1}\right)+\sqrt{\mathrm{2}{x}−\mathrm{1}}−{x}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}}…
Question Number 57850 by Mikael_Marshall last updated on 13/Apr/19 $$\underset{{n}\rightarrow\infty} {{lim}}\:\:\frac{\mathrm{8}^{{n}} }{\mathrm{2}^{{n}+\mathrm{1}} +\mathrm{3}^{{n}+\mathrm{2}} } \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 13/Apr/19 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{8}^{{n}}…
Question Number 123375 by bemath last updated on 25/Nov/20 $$\:\:\underset{{x}\rightarrow\pi/\mathrm{2}} {\mathrm{lim}}\left(\mathrm{sec}\:^{\mathrm{2}} {x}−\mathrm{sec}\:{x}\:\mathrm{tan}\:{x}\right)\:=? \\ $$ Answered by Dwaipayan Shikari last updated on 25/Nov/20 $$\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\left(\frac{\mathrm{1}−{sinx}}{{cos}^{\mathrm{2}} {x}}\right)=\frac{\mathrm{2}\left({sin}^{\mathrm{2}}…
Question Number 123349 by bemath last updated on 25/Nov/20 $$\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left[\:{x}\:\sqrt{\frac{{x}−\mathrm{1}}{\mathrm{9}{x}+\mathrm{2}}}\:−\frac{{x}}{\mathrm{3}}\:\right]? \\ $$ Answered by liberty last updated on 25/Nov/20 $$\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}\:\left[\sqrt{\frac{{x}−\mathrm{1}}{\mathrm{9}{x}+\mathrm{2}}}\:−\:\frac{\mathrm{1}}{\mathrm{3}}\:\right]= \\ $$$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{{x}\:\left(\mathrm{3}\sqrt{{x}−\mathrm{1}}\:−\sqrt{\mathrm{9}{x}+\mathrm{2}}\right)}{\:\mathrm{3}\sqrt{\mathrm{9}{x}+\mathrm{2}}}\:=…
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