Question Number 18854 by 3+4 last updated on 30/Jul/17 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}sin}\:\underset{\mathrm{x}} {\mathrm{x}}\:\:\:\:=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 84377 by john santu last updated on 12/Mar/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{1}+\mathrm{tan}\:\mathrm{x}}−\sqrt{\mathrm{1}+\mathrm{sin}\:\mathrm{x}}}{\mathrm{x}^{\mathrm{2}} \mathrm{sin}\:\mathrm{x}} \\ $$ Answered by john santu last updated on 12/Mar/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{tan}\:\mathrm{x}}+\sqrt{\mathrm{1}+\mathrm{sin}\:\mathrm{x}}}\:×\:\underset{{x}\rightarrow\mathrm{0}}…
Question Number 84367 by bagjamath last updated on 12/Mar/20 $$ \\ $$$$ \\ $$$$ \\ $$$$\mathrm{Given}\:\mathrm{the}\:\mathrm{squence}\:\mathrm{of}\:\mathbb{R} \\ $$$$\left({x}_{{n}} \right).\:{If}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{x}_{{k}} <\infty, \\ $$$${Find} \\…
Question Number 84364 by jagoll last updated on 12/Mar/20 $$\underset{{x}\rightarrow\pi} {\mathrm{lim}}\:\frac{\sqrt{\pi}−\sqrt{\pi+\mathrm{4x}}}{\mathrm{cos}\:\left(\frac{\pi\left(\mathrm{x}+\mathrm{1}\right)}{\mathrm{2}}\right)}\:=\:? \\ $$ Answered by john santu last updated on 12/Mar/20 $$\sqrt{\pi}\:×\:\underset{{x}\rightarrow\pi} {\mathrm{lim}}\:\frac{\mathrm{1}−\sqrt{\mathrm{1}+\frac{\mathrm{4x}}{\pi}}}{−\mathrm{sin}\:\left(\frac{\pi\mathrm{x}}{\mathrm{2}}\right)}\:= \\ $$$$−\sqrt{\pi}\:×\:\underset{{x}\rightarrow\pi}…
Question Number 149886 by liberty last updated on 08/Aug/21 $$\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\frac{\mathrm{ln}\:^{\mathrm{2}} \left(\mathrm{x}\right)}{\mathrm{x}^{\mathrm{2}} }\left(\frac{\mathrm{ln}\:\left(\mathrm{sin}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\right)}{\mathrm{ln}\:\left(\mathrm{sin}\:\left(\mathrm{x}\right)\right)}\:+\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{ln}\:\left(\mathrm{x}\right)}\right)\:=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 18803 by daffa22 last updated on 30/Jul/17 Commented by daffa22 last updated on 30/Jul/17 $$\mathrm{help} \\ $$ Answered by behi.8.3.4.1.7@gmail.com last updated on…
Question Number 149871 by ArielVyny last updated on 07/Aug/21 $${lim}_{{x}\rightarrow\mathrm{2}} \frac{\mathrm{3}^{{x}!} −\mathrm{9}}{{x}−\mathrm{2}} \\ $$ Answered by Ar Brandon last updated on 08/Aug/21 $$\mathscr{L}=\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\frac{\mathrm{3}^{{x}!} −\mathrm{9}}{{x}−\mathrm{2}}…
Question Number 84330 by M±th+et£s last updated on 11/Mar/20 $$\underset{{x}\rightarrow\mathrm{1}} {{lim}}\frac{\left(\mathrm{2}{x}^{\mathrm{3}} +{x}+\mathrm{1}\right)−\mathrm{64}}{{x}^{\mathrm{3}} −\mathrm{1}} \\ $$ Commented by M±th+et£s last updated on 11/Mar/20 $${typo}\:\left(\mathrm{2}{x}^{\mathrm{3}} +{x}+\mathrm{1}\right)^{\mathrm{3}} \\…
Question Number 84309 by jagoll last updated on 11/Mar/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}−\left(\mathrm{ln}\:\mid\mathrm{x}\mid\right)^{\mathrm{n}+\mathrm{1}} }{\mathrm{1}−\mathrm{ln}\mid\mathrm{x}\mid} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 84283 by jagoll last updated on 11/Mar/20 $$\underset{{x}\rightarrow\mathrm{a}} {\mathrm{lim}}\:\frac{\left(\mid\mathrm{x}\mid−\mathrm{1}\right)^{\mathrm{2}} −\left(\mid\mathrm{a}\mid−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{x}−\mathrm{a}}\:=\:\mathrm{k}\:,\:\mathrm{a}>\mathrm{0} \\ $$$$\underset{{x}\rightarrow\mathrm{a}} {\mathrm{lim}}\:\frac{\left(\mid\mathrm{x}\mid−\mathrm{1}\right)^{\mathrm{3}} −\left(\mid\mathrm{a}\mid−\mathrm{1}\right)^{\mathrm{3}} }{\mathrm{x}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} }\:=\: \\ $$$$ \\ $$ Answered…