Question Number 194279 by cortano12 last updated on 02/Jul/23 $$\:\underbrace{ } \\ $$ Answered by horsebrand11 last updated on 02/Jul/23 $$\:\:\underline{\underbrace{ }\cancel{ }} \\ $$$$…
Question Number 194139 by cortano12 last updated on 28/Jun/23 Commented by MM42 last updated on 28/Jun/23 $${for} \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:\frac{{x}^{\mathrm{2}} +\mathrm{2}{cosx}−\mathrm{2}}{{x}^{\mathrm{4}} }\:\rightarrow{hop} \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:\frac{\mathrm{2}\left({x}−{sinx}\right)}{\mathrm{4}{x}^{\mathrm{3}}…
Question Number 194256 by horsebrand11 last updated on 01/Jul/23 $$\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{\mathrm{1}+\mathrm{tan}\:\mathrm{x}}{\mathrm{1}−\mathrm{tan}\:\mathrm{x}}\:−\mathrm{1}}{\mathrm{x}}\:=? \\ $$ Answered by tri26112004 last updated on 01/Jul/23 $$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}{tan}\:{x}}{{x}\left(\mathrm{1}−{tan}\:{x}\right)} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{tan}\:{x}}{{x}}\:.\:\underset{{x}\rightarrow\mathrm{0}}…
Question Number 194250 by horsebrand11 last updated on 01/Jul/23 $$\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{a}\:\mathrm{for}\:\mathrm{which}\:\mathrm{the}\:\mathrm{limit} \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\left(\mathrm{ax}\right)−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}\right)−\mathrm{x}}{\mathrm{x}^{\mathrm{3}} +\mathrm{x}^{\mathrm{4}} }\:\mathrm{is}\:\mathrm{finite}\: \\ $$$$\:\mathrm{and}\:\mathrm{then}\:\mathrm{evaluate}\:\mathrm{the}\:\mathrm{limit}\: \\ $$ Answered by qaz last updated…
Question Number 194185 by mnjuly1970 last updated on 29/Jun/23 $$ \\ $$$$\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\mathrm{lim}_{\:{x}\rightarrow\:\mathrm{0}^{\:−} } \:\left\{\:\frac{\:{x}^{\:\mathrm{2}} \:+\mathrm{2}{cos}\left({x}\right)\:+\:\lfloor−\frac{{tan}\left({x}\right)}{{x}}\:\rfloor}{{ax}^{\:\mathrm{4}} }\:\right\}\:=\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a}\:=\:? \\ $$$$\:\:\:\:\:\:\:\:{a}:\:\:\:\frac{\mathrm{1}}{\mathrm{12}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{b}:\:\:−\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{c}:\:\:\:\mathrm{12}\:\:\:\:\:\:\:\:\:\:\:\:\:\:{d}:\:\:−\mathrm{12}\:\:\: \\…
Question Number 194020 by mnjuly1970 last updated on 26/Jun/23 $$\:\:\:\:\:\:{F}_{{n}} =\:{F}_{{n}} \:_{−\mathrm{1}} +{F}_{{n}−\mathrm{2}} \:\:\:\:\:\:{F}_{\mathrm{2}} =\:{F}_{\mathrm{1}} =\mathrm{1}\:\:\:\:\:\:\: \\ $$$$\:\:\:\:{F}_{{n}} \::\:\:\:\:\mathrm{1}\:,\:\mathrm{1}\:,\:\mathrm{2}\:,\:\mathrm{3}\:,\mathrm{5}…\:\: \\ $$$$\:\:\:\:\:\:\:{f}\left({x}\right)=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:{F}_{{n}} \:{x}^{\:{n}} \:=\:{x}\:+\:{x}^{\:\mathrm{2}}…
Question Number 193976 by Risandu last updated on 25/Jun/23 Answered by Subhi last updated on 25/Jun/23 $$ \\ $$$${lim}_{{x}\rightarrow\mathrm{1}} \frac{\left(^{\mathrm{3}} \sqrt{{x}}−\mathrm{1}\right)^{\mathrm{2}} }{\left({x}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${lim}_{{x}\rightarrow\mathrm{1}}…
Question Number 193967 by Risandu last updated on 24/Jun/23 Answered by MM42 last updated on 24/Jun/23 $${lim}_{\theta\rightarrow\pi} \:\:\frac{\mathrm{4}{sin}\theta{cos}\theta\left({cos}\theta−{sin}\mathrm{3}\theta\right)}{\left(\pi−\theta\right)\left(\pi+\theta\right)} \\ $$$${let}\:\:\:\:\theta=\pi−{u} \\ $$$$\Rightarrow{lim}_{{u}\rightarrow\mathrm{0}} \:\frac{−\mathrm{4}{sinu}×{cosu}×\left({sin}\mathrm{3}{u}−{cosu}\right)}{{u}×\left(\mathrm{2}\pi−{u}\right)} \\ $$$$\:=\frac{\mathrm{2}}{\pi}\:\checkmark…
Question Number 193874 by cortano12 last updated on 22/Jun/23 Answered by Subhi last updated on 22/Jun/23 $$ \\ $$$${cos}\left({y}\right)=\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{y}}{\mathrm{2}}\right) \\ $$$$\mathrm{1}−{cos}\left({cx}^{\mathrm{2}} +{bx}+{a}\right)=\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{cx}^{\mathrm{2}} +{bx}+{a}}{\mathrm{2}}\right)…
Question Number 193863 by horsebrand11 last updated on 21/Jun/23 $$\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\mathrm{2}} −\mathrm{cos}\:\left(\frac{\mathrm{x}}{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\right)}{\mathrm{x}^{\mathrm{4}} }\:=? \\ $$ Answered by MM42 last updated on 21/Jun/23 $${u}\rightarrow\mathrm{0}\Rightarrow\mathrm{1}−{cosu}\:\sim\:\frac{\mathrm{1}}{\mathrm{2}}{u}^{\mathrm{2}} \\…