Question Number 170130 by sciencestudent last updated on 17/May/22 $${li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\left(\frac{\mathrm{1}}{{sin}^{\mathrm{2}} {x}}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)=? \\ $$ Commented by cortano1 last updated on 17/May/22 $$\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left({x}−\mathrm{sin}\:{x}\right)\left({x}+\mathrm{sin}\:{x}\right)}{{x}^{\mathrm{2}} \:\mathrm{sin}\:^{\mathrm{2}}…
Question Number 170057 by nimnim last updated on 15/May/22 $$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\frac{{x}}{{a}}\lfloor\frac{{b}}{{x}}\rfloor\:\:{where}\:{a},{b}>\mathrm{0} \\ $$$${please}…. \\ $$ Answered by mahdipoor last updated on 15/May/22 $${get}\:\frac{{b}}{{x}}={t}\:\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}}…
Question Number 170018 by cortano1 last updated on 14/May/22 $$\:\:\:\:\:\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\left(\mathrm{1}+\mathrm{sin}\:\left(\pi−\mathrm{2}{x}\right)\right)^{\frac{\mathrm{5}}{\mathrm{sin}\:\left({x}−\frac{\pi}{\mathrm{2}}\right)}} .\mathrm{ln}\:\left(\mathrm{4}.\frac{\mathrm{cos}\:\left(\pi−\mathrm{2}{x}\right)−\mathrm{1}}{\left(\frac{\pi}{\mathrm{2}}−{x}\right)^{\mathrm{2}} }\right)=? \\ $$ Answered by Mathspace last updated on 14/May/22 $${u}\left({x}\right)=\left(\mathrm{1}+{sin}\left(\pi−\mathrm{2}{x}\right)\right)^{\frac{\mathrm{5}}{{sin}\left({x}−\frac{\pi}{\mathrm{2}}\right)}} \\ $$$${and}\:{v}\left({x}\right)={ln}\left(\mathrm{4}.\frac{{cos}\left(\pi−\mathrm{2}{x}\right)−\mathrm{1}}{\left(\frac{\pi}{\mathrm{2}}−{x}\right)^{\mathrm{2}}…
Question Number 38893 by gunawan last updated on 01/Jul/18 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\left({a}+{b}\:\mathrm{cos}\:{x}\right)−{c}\:\mathrm{sin}\:{x}}{{x}^{\mathrm{5}} }=\mathrm{1} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:{a},{b},\:\mathrm{and}\:{c} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 01/Jul/18 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{ax}+{bx}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}}…
Question Number 38892 by gunawan last updated on 01/Jul/18 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\left(\mathrm{1}+\:{a}\:\mathrm{cos}\:{x}\right)−{b}\:\mathrm{sin}\:{x}}{{x}^{\mathrm{5}} }=\mathrm{1} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:{a}\:\mathrm{and}\:{b} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 01/Jul/18 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}+{ax}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}}…
Question Number 169966 by cortano1 last updated on 13/May/22 $$\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\left(\mathrm{tan}\:{x}\right)−\mathrm{tan}\:\left(\mathrm{sin}\:{x}\right)}{\mathrm{2}{x}\:\mathrm{cos}\:\left(\mathrm{tan}\:{x}\right)−\mathrm{2}{x}\:\mathrm{cos}\:\left(\mathrm{sin}\:{x}\right)+{x}^{\mathrm{5}} }\:=? \\ $$ Answered by qaz last updated on 13/May/22 $$\mathrm{sin}\:\mathrm{tan}\:\mathrm{x}=\mathrm{x}+\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{6}}−\frac{\mathrm{x}^{\mathrm{5}} }{\mathrm{40}}−\frac{\mathrm{55x}^{\mathrm{7}} }{\mathrm{1008}}+……
Question Number 169899 by mathlove last updated on 12/May/22 Commented by infinityaction last updated on 12/May/22 $${i}\:{think}\:\:{x}\rightarrow\mathrm{0}^{+\:} {should}\:{be} \\ $$ Commented by mathlove last updated…
Question Number 104350 by bemath last updated on 21/Jul/20 $$\underset{\bigtriangleup{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\left(\left(\alpha+\bigtriangleup{x}\right)^{{n}} \right)−\mathrm{sin}\:\left(\alpha^{{n}} \right)}{\mathrm{cos}\:\left(\left(\alpha+\bigtriangleup{x}\right)^{{n}} \right)\mathrm{sin}\:\left(\alpha+\bigtriangleup{x}\right)−\mathrm{cos}\:\left(\alpha^{{n}} \right)\mathrm{sin}\:\left(\alpha\right)} \\ $$ Answered by john santu last updated on 21/Jul/20…
Question Number 104348 by bemath last updated on 21/Jul/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{arc}\:\mathrm{tan}\:\left({x}\right)−\mathrm{arc}\:\mathrm{sin}\:\left({x}\right)}{{x}\left(\mathrm{1}−\mathrm{cos}\:\left({x}\right)\right)}\right) \\ $$ Answered by john santu last updated on 21/Jul/20 $${L}'{Hopital}\:{rule}\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\:\frac{\mathrm{tan}^{−\mathrm{1}}…
Question Number 169859 by mathlove last updated on 11/May/22 Terms of Service Privacy Policy Contact: info@tinkutara.com