Question Number 194344 by mathlove last updated on 04/Jul/23 Answered by qaz last updated on 04/Jul/23 $$\mathrm{sin}\:^{\mathrm{2}} {x}=\left({x}−\frac{\mathrm{1}}{\mathrm{6}}{x}^{\mathrm{3}} +…\right)^{\mathrm{2}} ={x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{4}} +… \\ $$$${e}^{{x}} +{e}^{−{x}}…
Question Number 194338 by mathlove last updated on 04/Jul/23 Answered by Peace last updated on 04/Jul/23 $${e}^{{cos}\left({t}\right){ln}\left(\mathrm{2}+{sin}\left({t}\right)\right)} −\mathrm{2}={f}\left({t}\right) \\ $$$$\left.{f}\:{in}\:\left[\mathrm{0},{x}\right];\exists{c}\in\right]\mathrm{0},{x}\left[\right. \\ $$$$\Rightarrow{f}\left({x}\right)−{f}\left(\mathrm{0}\right)={f}'\left({c}\right)\left({x}−\mathrm{0}\right) \\ $$$$\Rightarrow{f}'\left({c}\right)=\frac{\left(\mathrm{2}+{sin}\left({x}\right)\right)^{{cos}\left({x}\right)} −\mathrm{2}}{{x}}…
Question Number 194282 by alcohol last updated on 02/Jul/23 $${f}\left({f}\left({x}\right)\right)\:=\:{ax}\:+\:{b} \\ $$$$\mathrm{1}.\:{show}\:{that}\:{f}\left({ax}+{b}\right)\:=\:{af}\left({x}\right)\:+\:{b} \\ $$$${deduce}\:{f}\:'\left({ax}\:+\:{b}\right) \\ $$$$\mathrm{2}.\:{Show}\:{that}\:{f}\:'\left({x}\right)\:{is}\:{a}\:{constant}\: \\ $$$${hence}\:{deduce}\:{f} \\ $$ Answered by Frix last updated…
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}}…