Question Number 200413 by Mingma last updated on 18/Nov/23 Commented by Mingma last updated on 18/Nov/23 Evaluate! Answered by witcher3 last updated on 18/Nov/23 $$\:\mathrm{x}−\frac{\mathrm{x}^{\mathrm{3}}…
Question Number 200304 by Calculusboy last updated on 16/Nov/23 Answered by Sutrisno last updated on 17/Nov/23 $${lim}_{{n}\rightarrow\infty} \mathrm{0}.\mathrm{2}^{{log}_{\sqrt{\mathrm{5}}} \left(\frac{\frac{\mathrm{1}}{\mathrm{4}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}}\right)} \\ $$$${lim}_{{n}\rightarrow\infty} \mathrm{0}.\mathrm{2}^{{log}_{\sqrt{\mathrm{5}}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)} \\ $$$${lim}_{{n}\rightarrow\infty}…
Question Number 200300 by Calculusboy last updated on 16/Nov/23 Answered by MM42 last updated on 17/Nov/23 $${lnA}={lim}_{{n}\rightarrow\infty} \:\frac{\mathrm{1}}{{n}}\:\left[{ln}\left(\mathrm{1}+\left(\frac{\mathrm{1}}{{n}}\right)^{\mathrm{1}} \right)+{ln}\left(\mathrm{1}+\left(\frac{\mathrm{2}}{{n}}\right)^{\mathrm{2}} +…+{ln}\left(\mathrm{1}+\left(\frac{{n}}{{n}}\right)^{\mathrm{2}} \right)\right.\right. \\ $$$$={lim}_{{n}\rightarrow\infty} \:\frac{\mathrm{1}}{{n}}\:\underset{{i}=\mathrm{1}} {\overset{{n}}…
Question Number 200298 by Calculusboy last updated on 16/Nov/23 Commented by Frix last updated on 17/Nov/23 $$−\frac{\mathrm{1}}{\mathrm{16}} \\ $$ Answered by Rasheed.Sindhi last updated on…
Question Number 200203 by sabiha last updated on 15/Nov/23 $$\frac{{x}}{\mathrm{6}{x}\mathrm{7}{mcnnc}}\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 200105 by cortano12 last updated on 14/Nov/23 Commented by mr W last updated on 14/Nov/23 Commented by mr W last updated on 14/Nov/23…
Question Number 200103 by cortano12 last updated on 14/Nov/23 $$\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{sin}\:\sqrt{\mathrm{x}+\mathrm{1}}−\mathrm{sin}\:\sqrt{\mathrm{x}}\:=? \\ $$ Answered by Frix last updated on 14/Nov/23 $$\mathrm{sin}\:\alpha\:−\mathrm{sin}\:\beta\:=\mathrm{2cos}\:\frac{\alpha+\beta}{\mathrm{2}}\:\mathrm{sin}\:\frac{\alpha−\beta}{\mathrm{2}} \\ $$$$\mathrm{For}\:\mathrm{large}\:{x}:\:\frac{\sqrt{{x}+\mathrm{1}}+\sqrt{{x}}}{\mathrm{2}}\sim\sqrt{{x}}\:\wedge\:\frac{\sqrt{{x}+\mathrm{1}}−\sqrt{{x}}}{\mathrm{2}}\sim\mathrm{0} \\ $$$$\Rightarrow…
Question Number 199983 by cortano12 last updated on 11/Nov/23 $$\:\:\:\mathrm{find}\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{sin}\:\left(\frac{\pi\mathrm{x}}{\mathrm{5}+\mathrm{3x}}\right)\: \\ $$$$\:\:\mathrm{by}\:\mathrm{sequeeze}\:\mathrm{theorem} \\ $$ Answered by tri26112004 last updated on 12/Nov/23 $${We}\:{have}\: \\ $$$$\mid{sin}\left(\:\frac{\pi{x}}{\mathrm{5}+\mathrm{3}{x}}\right)\mid\:\leqslant\:\mathrm{1}…
Question Number 199908 by cortano12 last updated on 11/Nov/23 $$\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{2}^{\mathrm{x}} +\mathrm{3}^{\mathrm{x}} +\mathrm{5}^{\mathrm{x}} }{\mathrm{3}}\right)^{\frac{\mathrm{3}}{\mathrm{x}}} =?\: \\ $$ Answered by cortano12 last updated on 11/Nov/23 $$\:\mathrm{L}\:=\:\underset{{x}\rightarrow\mathrm{0}}…
Question Number 199843 by universe last updated on 10/Nov/23 $$\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{1}}{\mathrm{ln}\left(\mathrm{1}+{x}\right)\:}−\frac{\mathrm{1}}{\mathrm{ln}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\right)}\right)\:=\:?? \\ $$ Answered by witcher3 last updated on 10/Nov/23 $$=\frac{\mathrm{ln}\left(\mathrm{x}+\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\right)−\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)}{\mathrm{ln}\left(\left(\mathrm{1}+\mathrm{x}\right)\mathrm{ln}\left(\mathrm{x}+\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\right)\right.}= \\…