Question Number 125056 by benjo_mathlover last updated on 08/Dec/20 $$\:\left(\mathrm{1}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{5}\:\sqrt[{\mathrm{3}}]{{x}}\:−\mathrm{3}\:\sqrt[{\mathrm{4}}]{{x}}\:−\sqrt[{\mathrm{6}}]{{x}}\:}{\mathrm{4}\:\sqrt[{\mathrm{3}}]{{x}}\:−\sqrt[{\mathrm{24}}]{{x}}}\:? \\ $$$$\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{5}\:\sqrt[{\mathrm{3}}]{{x}}\:−\mathrm{3}\:\sqrt[{\mathrm{4}}]{{x}}\:−\sqrt[{\mathrm{6}}]{{x}}}{\mathrm{4}\:\sqrt[{\mathrm{3}}]{{x}}\:−\:\sqrt[{\mathrm{24}}]{{x}}}\:?\: \\ $$ Answered by bramlexs22 last updated on 08/Dec/20 $$\left(\mathrm{1}\right)\:{let}\:{x}\:=\:{t}^{\mathrm{24}} \\…
Question Number 125049 by bramlexs22 last updated on 08/Dec/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{2}+\mathrm{cos}\:\left(\mathrm{3}{x}\right)−\mathrm{3cosh}\:\left({x}\right)\right)^{\mathrm{4}} }{\mathrm{ln}\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\:? \\ $$ Answered by Olaf last updated on 08/Dec/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{2}+\mathrm{cos3}{x}−\mathrm{3cosh}{x}\right)^{\mathrm{4}} }{\mathrm{ln}\left(\mathrm{1}+{x}^{\mathrm{2}}…
Question Number 124995 by liberty last updated on 07/Dec/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{2}+\mathrm{sin}\:{x}}\:−\sqrt{\mathrm{1}+\mathrm{cos}\:{x}}}{\mathrm{tan}\:{x}}\:=? \\ $$ Answered by bemath last updated on 07/Dec/20 Answered by Dwaipayan Shikari last…
Question Number 190521 by Rupesh123 last updated on 04/Apr/23 Answered by mehdee42 last updated on 04/Apr/23 $$\mathrm{8}+\mathrm{88}+\mathrm{888}+…+\mathrm{888}…\mathrm{8}=\mathrm{8}\left(\frac{\mathrm{10}−\mathrm{1}}{\mathrm{9}}+\frac{\mathrm{10}^{\mathrm{2}} −\mathrm{1}}{\mathrm{9}}+…+\frac{\mathrm{10}^{{n}} −\mathrm{1}}{\mathrm{9}}\right) \\ $$$$=\frac{\mathrm{8}}{\mathrm{9}}\left(\mathrm{10}+\mathrm{10}^{\mathrm{2}} +…+\mathrm{10}^{{n}} −{n}\right)=\frac{\mathrm{8}}{\mathrm{9}}\left(\frac{\mathrm{10}^{{n}+\mathrm{1}} −\mathrm{10}}{\mathrm{9}}−{n}\right) \\…
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Question Number 59393 by Mikael_Marshall last updated on 09/May/19 $$\underset{{x}\rightarrow+\infty} {{lim}}\:\left(\frac{{x}^{\mathrm{2}} −\mathrm{4}}{{x}^{\mathrm{2}} +\mathrm{2}}\overset{\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}+\mathrm{1}}} {\right)} \\ $$$${pls}. \\ $$ Commented by maxmathsup by imad last…
Question Number 124915 by liberty last updated on 06/Dec/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2tan}\:{x}−\mathrm{5sin}\:{x}+\mathrm{3}{x}−{x}^{\mathrm{3}} }{\mathrm{1}−\sqrt[{\mathrm{5}}]{\mathrm{1}−\mathrm{2}{x}^{\mathrm{3}} }}\:=?\: \\ $$ Answered by bemath last updated on 07/Dec/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}\left({x}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\right)−\mathrm{5}\left({x}−\frac{{x}^{\mathrm{3}}…
Question Number 124802 by john_santu last updated on 06/Dec/20 $$\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{4}}]{{x}+\mathrm{1}}\:−\:\sqrt[{\mathrm{4}}]{{x}}}{\:\sqrt[{\mathrm{3}}]{{x}+\mathrm{1}}\:−\:\sqrt[{\mathrm{3}}]{{x}}}\:=?\: \\ $$ Answered by bramlexs22 last updated on 06/Dec/20 $$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{4}}]{{x}}\:\left(\sqrt[{\mathrm{4}}]{\mathrm{1}+\frac{\mathrm{1}}{{x}}}−\mathrm{1}\right)}{\:\sqrt[{\mathrm{3}}]{{x}}\:\left(\sqrt[{\mathrm{3}}]{\mathrm{1}+\frac{\mathrm{1}}{{x}}}−\mathrm{1}\right)}\:= \\ $$$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{\:\sqrt[{\mathrm{12}}]{{x}}}\:\left(\frac{\sqrt[{\mathrm{4}\:}]{\mathrm{1}+\frac{\mathrm{1}}{{x}}}\:−\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{1}+\frac{\mathrm{1}}{{x}}}−\mathrm{1}}\right)\:=\:\mathrm{0}…
Question Number 190317 by krishvasoya last updated on 31/Mar/23 Answered by senestro last updated on 31/Mar/23 $$\frac{\mathrm{1}}{{e}+\mathrm{1}}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{2}}−\mathrm{4}\right) \\ $$ Terms of Service Privacy Policy…
Question Number 190300 by mnjuly1970 last updated on 31/Mar/23 $$ \\ $$$$\:\:\:\mathrm{lim}_{\:{x}\rightarrow\mathrm{0}} \frac{\:{sin}\left({x}\right)\:−{tan}\left({x}\right)}{{x}^{\:\mathrm{3}} } \\ $$$$\:\:\:\:\:{a}:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:{b}:\:\:\:\:\frac{−\mathrm{1}}{\mathrm{4}}\:\:\:\:\:\:\:\:\:{c}:\:\:\:\:\:\frac{−\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:{d}:\:\:\:\:\frac{−\mathrm{1}}{\mathrm{4}} \\ $$$$ \\ $$ Answered by som(math1967) last updated…