Question Number 173351 by cortano1 last updated on 10/Jul/22 Answered by aleks041103 last updated on 10/Jul/22 $${ln}\:{y}\:=\:\frac{{ln}\left({tgx}\right)}{\mathrm{1}+\sqrt[{\mathrm{3}}]{\mathrm{1}+{ln}^{\mathrm{2}} {x}}} \\ $$$$\Rightarrow{L}=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:{ln}\left({y}\right)=\left[\frac{−\infty}{\infty}\right] \\ $$$$\Rightarrow{L}'{Hopital} \\ $$$${L}=\underset{{x}\rightarrow\mathrm{0}}…
Question Number 173334 by Raxreedoroid last updated on 09/Jul/22 $$\mathrm{Let}\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{3}^{{x}} }{\mathrm{2}^{{f}\left({x}\right)} }=\mathrm{9} \\ $$$$\mathrm{what}\:\mathrm{is}\:{f}\left({x}\right)? \\ $$ Answered by mahdipoor last updated on 09/Jul/22 $$\mathrm{ln9}=\underset{{x}\rightarrow\infty}…
Question Number 173309 by mathlove last updated on 09/Jul/22 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}} {x}−{ex}\right)=? \\ $$ Answered by CElcedricjunior last updated on 12/Jul/22 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left[\left(\mathrm{1}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\right)^{\boldsymbol{\mathrm{x}}} \boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{ex}}\right]=+\infty−\infty=\boldsymbol{\mathrm{FI}} \\…
Question Number 107745 by ajfour last updated on 12/Aug/20 $$\left({i}\right)\:\:\:{L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\frac{\left(\mathrm{1}+{x}\right)^{\mathrm{1}/{x}} }{{e}}\right]^{\mathrm{1}/{x}} \:\:=\:? \\ $$$$\left({ii}\right)\:\:{L}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left[\frac{{x}}{{e}}−{x}\left(\frac{{x}}{{x}+\mathrm{1}}\right)^{{x}} \right]\:=\:? \\ $$ Commented by ajfour last updated on…
Question Number 107747 by ajfour last updated on 12/Aug/20 $${L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}−\mathrm{cos}\:{x}\mathrm{cos}\:\mathrm{2}{x}\mathrm{cos}\:\mathrm{3}{x}}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{2}{x}}\right)\:=\:? \\ $$ Answered by bobhans last updated on 12/Aug/20 $$\:\:\:\:\:\:\:\:\frac{\mathcal{B}\mathrm{obhans}}{\Pi} \\ $$$$\mathrm{L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}−\left(\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}}…
Question Number 107714 by bemath last updated on 12/Aug/20 $$\:\:\:\frac{\checkmark\mathcal{B}{e}\mathcal{M}{ath}\checkmark}{{cooll}} \\ $$$$\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}−\mathrm{tan}\:{x}\right)^{\frac{\mathrm{4}}{\mathrm{sin}\:^{\mathrm{2}} {x}}} \:=? \\ $$ Answered by bemath last updated on 12/Aug/20 Answered…
Question Number 107710 by bemath last updated on 12/Aug/20 $$\:\:\:\checkmark\frac{\mathcal{B}{e}\mathcal{M}{ath}\checkmark}{{cooll}} \\ $$$$\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{{x}+\mathrm{5}}{{x}−\mathrm{3}}\right)^{\frac{\mathrm{3}{x}+\mathrm{1}}{\mathrm{cos}\:\mathrm{2}{x}}} \:? \\ $$ Answered by Dwaipayan Shikari last updated on 12/Aug/20 $$\underset{{x}\rightarrow\infty}…
Question Number 173236 by mathlove last updated on 08/Jul/22 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{{x}}{\mathrm{1}+{x}}\right)^{{x}} =? \\ $$ Answered by floor(10²Eta[1]) last updated on 08/Jul/22 $$\mathrm{L}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{x}}{\mathrm{1}+\mathrm{x}}\right)^{\mathrm{x}} \\ $$$$\mathrm{lnL}=\underset{{x}\rightarrow\infty}…
Question Number 173225 by cortano1 last updated on 08/Jul/22 $$\:\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:{x}+\mathrm{4tan}\:\mathrm{2}{x}−\mathrm{3tan}\:\mathrm{3}{x}}{{x}^{\mathrm{2}} \:\mathrm{tan}\:{x}}\:=?\: \\ $$ Answered by floor(10²Eta[1]) last updated on 08/Jul/22 $$\mathrm{tg}\left(\mathrm{2x}\right)=\frac{\mathrm{2tgx}}{\mathrm{1}−\mathrm{tg}^{\mathrm{2}} \mathrm{x}} \\ $$$$\mathrm{tg3x}=\frac{\frac{\mathrm{2tgx}}{\mathrm{1}−\mathrm{tg}^{\mathrm{2}}…
Question Number 173192 by mathlove last updated on 08/Jul/22 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{n}^{\mathrm{2}} \sqrt{\left(\mathrm{1}−{cos}\frac{\mathrm{1}}{{n}}\right)\sqrt{\left(\mathrm{1}−{cos}\frac{\mathrm{1}}{{n}}\right)\sqrt{\left(\mathrm{1}−{cos}\frac{\mathrm{1}}{{n}}\right)……\infty}}}=? \\ $$ Answered by Frix last updated on 08/Jul/22 $${u}=\sqrt{{t}\sqrt{{t}\sqrt{{t}\sqrt{…}}}}\:\Leftrightarrow\:{u}=\sqrt{{tu}}\:\Leftrightarrow\:{u}={t} \\ $$$$ \\…