Question Number 178716 by cortano1 last updated on 20/Oct/22 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{3tan}\:\mathrm{4x}−\mathrm{4tan}\:\mathrm{3x}}{\mathrm{3sin}\:\mathrm{4x}−\mathrm{4sin}\:\mathrm{3x}}\:=\:? \\ $$ Answered by a.lgnaoui last updated on 21/Oct/22 $$\:\:{f}\left({x}\right)=\left(\:\:\frac{\mathrm{tan}\:\mathrm{4x}\left(\mathrm{3}−\frac{\mathrm{4tan}\:\mathrm{3x}}{\mathrm{tan}\:\mathrm{4x}\:\:}\right)}{\mathrm{sin}\:\mathrm{4x}\left(\mathrm{3}−\frac{\mathrm{4sin}\:\mathrm{3x}}{\mathrm{sin}\:\mathrm{4x}}\right)}=\frac{\mathrm{1}}{\mathrm{cos}\:\mathrm{4x}}×\left(\frac{\mathrm{3}−\frac{\mathrm{4sin}\:\mathrm{3x}}{\mathrm{cos}\:\mathrm{3x}}×\frac{\mathrm{cos}\:\mathrm{4x}}{\mathrm{sin}\:\mathrm{4x}}}{\mathrm{3}−\:\frac{\mathrm{4sin}\:\mathrm{3x}}{\mathrm{sin}\:\mathrm{4x}}}\right)\:\right. \\ $$$$\frac{\mathrm{1}}{\mathrm{cos}\:\mathrm{4x}}×\left(\frac{\mathrm{3}−\frac{\mathrm{4sin}\:\mathrm{3x}}{\mathrm{sin}\:\mathrm{4x}}×\frac{\mathrm{cos}\:\mathrm{4x}}{\mathrm{cos}\:\mathrm{3x}}}{\mathrm{3}−\frac{\mathrm{4sin}\:\mathrm{3x}}{\mathrm{sin}\:\mathrm{4x}}}\right) \\ $$$$\mathrm{x}\rightarrow\mathrm{0}\:\:\:\:\:{f}\left({x}\right)\rightarrow\frac{\mathrm{3}−\frac{\mathrm{4sin}\:\mathrm{3x}}{\mathrm{sin}\:\mathrm{4x}}}{\mathrm{3}−\frac{\mathrm{4sin}\:\mathrm{3x}}{\mathrm{sin}\:\mathrm{4x}}}=\:\:\frac{\mathrm{3}−\mathrm{4}\frac{\mathrm{sin}\:\mathrm{3x}}{\mathrm{3x}}×\frac{\mathrm{4x}}{\mathrm{sin}\:\mathrm{4x}}}{\mathrm{3}−\mathrm{4}\frac{\mathrm{sin}\:\mathrm{3x}}{\mathrm{3x}}×\frac{\mathrm{3x}}{\mathrm{sin}\:\mathrm{4x}}}\:\:=\mathrm{1}…
Question Number 112974 by bemath last updated on 10/Sep/20 $$\mathrm{solve}\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{2x}^{\mathrm{3}} −\mathrm{x}^{\mathrm{2}} +\mathrm{2}}{\mathrm{2x}^{\mathrm{3}} −\mathrm{4x}^{\mathrm{2}} +\mathrm{1}}\right)^{\mathrm{x}} \\ $$ Answered by john santu last updated on 10/Sep/20…
Question Number 47402 by Aditya789 last updated on 09/Nov/18 Commented by Aditya789 last updated on 10/Nov/18 $$\mathrm{plz}\:\mathrm{explain} \\ $$ Commented by maxmathsup by imad last…
Question Number 47401 by Aditya789 last updated on 09/Nov/18 Commented by maxmathsup by imad last updated on 10/Nov/18 $${let}\:{A}\left({x}\right)=\left({cosx}\right)^{\frac{\mathrm{1}}{{sinx}}} \:\Rightarrow{A}\left({x}\right)={e}^{\frac{\mathrm{1}}{{sinx}}{ln}\left({cosx}\right)} \:\:{but}\:{cosx}\:\sim\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\left({x}\rightarrow\mathrm{0}\right)\:\Rightarrow \\ $$$${ln}\left({cosx}\right)\:\sim\:{ln}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}\:}\right)\:\sim−\frac{{x}^{\mathrm{2}}…
Question Number 112917 by abdullahquwatan last updated on 10/Sep/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left[\mathrm{csc}^{\mathrm{2}} \left(\frac{\mathrm{2}}{\mathrm{x}}\right)−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{x}^{\mathrm{2}} \right] \\ $$ Commented by bobhans last updated on 10/Sep/20 $$\underset{\mathrm{w}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{2w}+\mathrm{sin}\:\left(\mathrm{2w}\right)}{\mathrm{4w}}\right)\left(\frac{\mathrm{2w}−\mathrm{sin}\:\left(\mathrm{2w}\right)}{\mathrm{w}.\mathrm{sin}^{\mathrm{2}} \:\left(\mathrm{2w}\right)}\right)…
Question Number 112900 by bemath last updated on 10/Sep/20 $$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{x}^{\mathrm{3}} −\mathrm{x}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2x}^{\mathrm{3}} +\mathrm{x}^{\mathrm{2}} −\mathrm{2}}\right)^{\mathrm{x}} ? \\ $$ Commented by kaivan.ahmadi last updated on 10/Sep/20…
Question Number 112881 by bemath last updated on 10/Sep/20 Answered by Dwaipayan Shikari last updated on 10/Sep/20 $$\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}+\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}}… \\ $$$$\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{2}}{{n}\left({n}+\mathrm{1}\right)}=\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\mathrm{40}} {\sum}}\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+\mathrm{1}}=\mathrm{2}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{41}}\right)=\frac{\mathrm{80}}{\mathrm{41}} \\…
Question Number 112866 by bemath last updated on 10/Sep/20 Answered by bobhans last updated on 10/Sep/20 $$\left(\mathrm{7}\right)\:\mathrm{L}=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{tan}\:\mathrm{x}} \\ $$$$\:\:\:\mathrm{L}\:=\:\mathrm{e}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}tan}\:\mathrm{x}.\mathrm{ln}\:\left(\frac{\mathrm{1}}{\mathrm{x}}\right)} =\mathrm{e}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}−\mathrm{tan}\:\mathrm{x}.\mathrm{ln}\:\mathrm{x}} \\ $$$$\:\:\mathrm{L}\:=\:\mathrm{e}^{\underset{{x}\rightarrow\mathrm{0}}…
Question Number 112863 by bemath last updated on 10/Sep/20 $$\left(\mathrm{1}\right)\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{2x}+\mathrm{3}}{\mathrm{2x}−\mathrm{1}}\right)^{\mathrm{4x}+\mathrm{2}} \\ $$$$\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{3x}+\mathrm{1}}{\mathrm{3x}−\mathrm{1}}\right)^{\mathrm{4x}−\mathrm{2}} \\ $$ Commented by bobhans last updated on 22/Sep/20 $$\left(\mathrm{1}\right)\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{2}{x}+\mathrm{3}}{\mathrm{2}{x}−\mathrm{1}}\right)^{\mathrm{4}{x}+\mathrm{2}}…
Question Number 112855 by ruwedkabeh last updated on 10/Sep/20 $$ \\ $$$$\mathrm{1}.\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{5}{x}+\mathrm{4}}−\sqrt{\mathrm{3}{x}+\mathrm{9}}}{\mathrm{4}{x}}=… \\ $$$$ \\ $$$$\mathrm{2}.\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{5}−\mathrm{4}{x}+\mathrm{3}{x}^{\mathrm{2}} }−\sqrt{\mathrm{4}−\mathrm{3}{x}+\mathrm{3}{x}^{\mathrm{2}} }}{\mathrm{2}{x}}=… \\ $$$$ \\ $$ Answered…