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}\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\:\frac{\mathrm{tan}^{−\mathrm{1}}…
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Question Number 169803 by cortano1 last updated on 09/May/22 Answered by greougoury555 last updated on 09/May/22 $$\:{x}−\mathrm{1}=\:{y}\: \\ $$$$\:\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left({y}+\mathrm{1}\right)−\left(\frac{\left({n}+\mathrm{1}\right)\left({y}+\mathrm{1}\right)^{{n}} −\mathrm{1}}{{n}}\right)^{\frac{\mathrm{1}}{{n}+\mathrm{1}}} }{{y}^{\mathrm{2}\:} }\: \\ $$$$=\:\underset{{y}\rightarrow\mathrm{0}}…
Question Number 104218 by bemath last updated on 20/Jul/20 $$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\left[\frac{\mathrm{ln}\:\left(\mathrm{1}+{x}\right)+\underset{{n}\:=\:\mathrm{1}} {\overset{\infty} {\sum}}\left[\frac{\left(\mathrm{1}+{x}^{\mathrm{2}^{{n}} } \right)}{\left(\mathrm{1}−{x}^{\mathrm{2}^{{n}} } \right)}\right]^{\mathrm{2}^{−{n}} } }{\mathrm{ln}\:\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)}\right] \\ $$$$ \\ $$ Commented by…
Question Number 104217 by bemath last updated on 20/Jul/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{tan}\:\left({x}+\mathrm{3}\right)^{\mathrm{2}} −\mathrm{tan}\:\left(\mathrm{9}\right)}{{x}}\:=\:? \\ $$ Answered by bemath last updated on 20/Jul/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}\left({x}+\mathrm{3}\right)^{\mathrm{1}} \mathrm{sec}\:^{\mathrm{2}} \left({x}+\mathrm{3}\right)^{\mathrm{2}}…
Question Number 104215 by bemath last updated on 20/Jul/20 $$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{9}^{\mathrm{3}^{\mathrm{2ln}\:{x}} } −\mathrm{9}^{\mathrm{2}^{\mathrm{ln}\:{x}} } }{\mathrm{ln}\:{x}}\:=? \\ $$ Answered by john santu last updated on 20/Jul/20…
Question Number 169711 by mnjuly1970 last updated on 06/May/22 $$ \\ $$$$\:\:\:\:{prove}\:{that}: \\ $$$$\:\: \\ $$$$\:\:{lim}_{\:{x}\:\rightarrow\:\mathrm{0}} \left(\:\frac{\mathrm{1}}{{x}^{\:\mathrm{2}} }\:\:−\:\frac{{e}^{\:{x}} }{\left({e}^{\:{x}} −\mathrm{1}\:\right)^{\:\mathrm{2}} }\:\right)\:=\:\frac{\mathrm{1}}{\mathrm{12}} \\ $$$$\:\:\:\:\:\: \\ $$…