Question Number 107871 by john santu last updated on 13/Aug/20 $$\:\:\:\:\:\:\:\frac{\checkmark\mathcal{JS}\checkmark}{\heartsuit} \\ $$$$\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\sqrt{\frac{{x}\:\mathrm{tan}\:{x}}{\mathrm{sin}\:\mathrm{2}{x}−\mathrm{cos}\:\mathrm{2}{x}\:+\mathrm{1}}}\:?\: \\ $$ Commented by bemath last updated on 13/Aug/20 Answered by…
Question Number 173403 by mathlove last updated on 11/Jul/22 Commented by mr W last updated on 11/Jul/22 $$\frac{\sqrt[{\mathrm{3}}]{{n}^{\mathrm{2}} }\:\mathrm{sin}\:\left({n}!\right)}{{n}+\mathrm{1}}\geqslant−\frac{{n}^{\frac{\mathrm{2}}{\mathrm{3}}} }{{n}+\mathrm{1}}>−\frac{{n}^{\frac{\mathrm{2}}{\mathrm{3}}} }{{n}}=−\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{n}}} \\ $$$$\frac{\sqrt[{\mathrm{3}}]{{n}^{\mathrm{2}} }\:\mathrm{sin}\:\left({n}!\right)}{{n}+\mathrm{1}}\leqslant\frac{{n}^{\frac{\mathrm{2}}{\mathrm{3}}} }{{n}+\mathrm{1}}<\frac{{n}^{\frac{\mathrm{2}}{\mathrm{3}}}…
Question Number 173380 by cortano1 last updated on 10/Jul/22 Commented by kaivan.ahmadi last updated on 10/Jul/22 $$\sim{li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\:\:\frac{{e}^{{x}} −{e}^{−{x}} +\mathrm{2}{sinx}}{\mathrm{1}}=\mathrm{0} \\ $$$$ \\ $$ Answered…
Question Number 173386 by mathlove last updated on 10/Jul/22 Answered by mnjuly1970 last updated on 10/Jul/22 $$\:\:\:\mathrm{lim}_{\:\mathrm{n}\rightarrow\infty} \:\frac{\:\left(\mathrm{n}+\mathrm{1}\right)!−\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)!}\:=\mathrm{1}−\mathrm{lim}_{\:\mathrm{n}\rightarrow\infty} \left(\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)!}\right) \\ $$$$\:\:\:\:\:\:=\:\mathrm{1} \\ $$ Commented by…
Question Number 107828 by Rohit@Thakur last updated on 12/Aug/20 $${lim}\:\:\:\:\:\:\left\{\frac{{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{3}}{{x}^{\mathrm{2}} +{x}+\mathrm{2}}\right\}^{{x}} \\ $$$${x}\rightarrow\mathrm{0} \\ $$$${lim}\:\:\:\:\frac{\mathrm{10}^{{x}} −\mathrm{2}^{{x}} −\mathrm{5}^{{x}} +\mathrm{1}}{{xtanx}} \\ $$$${x}\rightarrow\mathrm{0} \\ $$$$ \\ $$…
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}}…