Question Number 93648 by i jagooll last updated on 14/May/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{sin}\:\mathrm{2x}}{\mathrm{x}^{\mathrm{3}} }\:? \\ $$ Answered by 675480065 last updated on 14/May/20 Answered by john…
Question Number 28098 by tawa tawa last updated on 20/Jan/18 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\frac{\mathrm{log}_{\mathrm{e}} \mathrm{x}}{\mathrm{x}^{\mathrm{h}} }\:,\:\:\:\:\:\:\:\:\:\:\:\mathrm{h}\:>\:\mathrm{0} \\ $$ Answered by ajfour last updated on 20/Jan/18 $$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\left(\mathrm{1}/{x}\right)}{{hx}^{{h}−\mathrm{1}}…
Question Number 28097 by tawa tawa last updated on 20/Jan/18 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\frac{\mathrm{3}^{\mathrm{x}} \:−\:\mathrm{2}^{\mathrm{x}} }{\mathrm{x}^{\mathrm{2}} } \\ $$ Commented by abdo imad last updated on 20/Jan/18…
Question Number 28096 by tawa tawa last updated on 20/Jan/18 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{x}\:−\:\mathrm{sinx}} \\ $$ Commented by abdo imad last updated on 20/Jan/18 $$=\:{lim}_{{x}\rightarrow\propto\:\:} \:\:\frac{{x}}{\mathrm{1}−\frac{{sinx}}{{x}}}\:\:\:{but}\:\:\:/\frac{{sinx}}{{x}}/\leqslant\:\:\:\frac{\mathrm{1}}{/{x}/}\:_{{x}\rightarrow\propto}…
Question Number 28093 by tawa tawa last updated on 20/Jan/18 $$\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}\:\:\left(\mathrm{1}\:+\:\mathrm{tanx}\right)^{\mathrm{cotx}} \\ $$ Commented by abdo imad last updated on 20/Jan/18 $$={lim}_{{x}\rightarrow\mathrm{0}^{−} }…
Question Number 28095 by tawa tawa last updated on 20/Jan/18 $$\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}\:\:\left(\mathrm{1}\:+\:\mathrm{tanx}\right)^{−\mathrm{cotx}} \\ $$ Commented by abdo imad last updated on 20/Jan/18 $$=\:{lim}_{{x}\rightarrow\mathrm{0}^{−} }…
Question Number 28088 by tawa tawa last updated on 20/Jan/18 $$\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\:\left(\mathrm{sinx}\right)^{\left(\mathrm{tanx}\right)} \\ $$ Commented by çhëý böý last updated on 20/Jan/18 $$\underset{{x}\rightarrow\mathrm{0}^{+} }…
Question Number 28084 by tawa tawa last updated on 20/Jan/18 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\left(\mathrm{x}\:−\:\mathrm{log}_{\mathrm{e}} \mathrm{x}\right) \\ $$ Commented by çhëý böý last updated on 20/Jan/18 $$\underset{{x}\rightarrow\propto} {\mathrm{lim}}\:\:\left({x}−{lnx}\right)…
Question Number 159142 by tounghoungko last updated on 13/Nov/21 $$\:\underset{{x}\rightarrow−\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\:\frac{\frac{\pi}{\:\sqrt{\mathrm{8}}}\:−\sqrt{\mathrm{2}}\:{x}.\:\mathrm{tan}\:{x}}{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}\:=? \\ $$ Commented by cortano last updated on 14/Nov/21 $$\:\underset{{x}\rightarrow−\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\frac{\frac{\pi\:\mathrm{cos}\:{x}−\mathrm{4}{x}\:\mathrm{sin}\:{x}}{\:\sqrt{\mathrm{8}}}}{\:\sqrt{\mathrm{2}}\:\mathrm{sin}\:\left({x}+\frac{\pi}{\mathrm{4}}\right)\mathrm{cos}\:{x}} \\ $$$$=\:\frac{\mathrm{1}}{\:\mathrm{2}\sqrt{\mathrm{2}}}\:\underset{{x}\rightarrow−\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\:\frac{\pi\mathrm{cos}\:{x}−\mathrm{4}{x}\:\mathrm{sin}\:{x}}{\mathrm{sin}\:\left({x}+\frac{\pi}{\mathrm{4}}\right)}…
Question Number 159130 by metamorfose last updated on 13/Nov/21 $${f}\:\in\:\mathcal{C}^{\mathrm{0}} \left(\mathbb{R},\mathbb{R}\right)\:,\:\frac{{f}\left({x}\right)}{{x}}\underset{{x}\rightarrow+\infty} {\rightarrow}\mathrm{0} \\ $$$${Does}\:\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\:{f}\left({x}+\mathrm{1}\right)\:−\:{f}\left({x}\right)\:{exist}? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com