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 $$\mathrm{8}+\mathrm{88}+\mathrm{888}+…+\mathrm{888}…\mathrm{8}=\mathrm{8}\left(\frac{\mathrm{10}−\mathrm{1}}{\mathrm{9}}+\frac{\mathrm{10}^{\mathrm{2}} −\mathrm{1}}{\mathrm{9}}+…+\frac{\mathrm{10}^{{n}} −\mathrm{1}}{\mathrm{9}}\right) \\ $$$$=\frac{\mathrm{8}}{\mathrm{9}}\left(\mathrm{10}+\mathrm{10}^{\mathrm{2}} +…+\mathrm{10}^{{n}} −{n}\right)=\frac{\mathrm{8}}{\mathrm{9}}\left(\frac{\mathrm{10}^{{n}+\mathrm{1}} −\mathrm{10}}{\mathrm{9}}−{n}\right) \\…
Question Number 59437 by tanmay last updated on 10/May/19 Answered by tanmay last updated on 10/May/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 59393 by Mikael_Marshall last updated on 09/May/19 $$\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…
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Question Number 190182 by SLVR last updated on 29/Mar/23 $${Li}\underset{{x}\rightarrow\pi/\mathrm{2}} {{m}}\frac{{sinx}−{sinx}^{{sinx}} }{\mathrm{1}−{sinx}+{logsinx}} \\ $$$$\left.{a}\left.\right)\left.\mathrm{4}\left.\:\:\:\:\:\:\:\:{b}\right)\mathrm{2}\:\:\:\:\:\:\:\:\:\:{c}\right)\mathrm{1}/\mathrm{2}\:\:\:\:\:\:\:\:{d}\right){none} \\ $$ Commented by SLVR last updated on 29/Mar/23 $${kindly}\:{help}\:{me} \\…