Question Number 132138 by benjo_mathlover last updated on 11/Feb/21 $$\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}}\right)^{\mathrm{n}} }{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{n}}\right)^{\mathrm{n}} }\:−\:\mathrm{e}^{\mathrm{2}} \:\right)\mathrm{n}^{\mathrm{2}} =? \\ $$ Answered by EDWIN88 last updated on 11/Feb/21 $$\mathrm{L}=\mathrm{e}^{\mathrm{2}}…
Question Number 132079 by liberty last updated on 11/Feb/21 $$\mathrm{If}\:\mathrm{L}\:=\:\underset{{x}\rightarrow\pi/\mathrm{4}} {\mathrm{lim}}\frac{\mathrm{tan}\:^{\mathrm{3}} \mathrm{x}−\mathrm{tan}\:\mathrm{x}}{\mathrm{cos}\:\left(\mathrm{x}+\frac{\pi}{\mathrm{4}}\right)} \\ $$$$\mathrm{then}\:\frac{\mathrm{L}}{\mathrm{4}}\:=? \\ $$ Answered by EDWIN88 last updated on 11/Feb/21 $$\:\mathrm{L}=\underset{{x}\rightarrow\pi/\mathrm{4}} {\mathrm{lim}}\frac{\mathrm{tan}\:\mathrm{x}\left(\mathrm{tan}\:\mathrm{x}+\mathrm{1}\right)\left(\mathrm{tan}\:\mathrm{x}−\mathrm{1}\right)}{\mathrm{cos}\:\left(\mathrm{x}+\frac{\pi}{\mathrm{4}}\right)}…
Question Number 66518 by Masumsiddiqui399@gmail.com last updated on 16/Aug/19 Commented by mathmax by abdo last updated on 16/Aug/19 $${let}\:{f}\left({x}\right)=\frac{{x}\sqrt{{x}}−{a}\sqrt{{a}}}{{x}−{a}}\:\:\:{cha}\mathrm{7}{gement}\:{x}−{a}={t}\:{give} \\ $$$${lim}_{{x}\rightarrow{a}} {f}\left({x}\right)\:={lim}_{{t}\rightarrow\mathrm{0}} \:\:\:\frac{\left({t}+{a}\right)\sqrt{{t}+{a}}−{a}\sqrt{{a}}}{{t}} \\ $$$$={lim}_{{t}\rightarrow\mathrm{0}}…
Question Number 132007 by abdullahquwatan last updated on 10/Feb/21 Answered by bramlexs22 last updated on 10/Feb/21 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\mathrm{2x}+\mathrm{sin}\:\mathrm{3x}\left(\mathrm{cos}\:\mathrm{x}−\mathrm{cos}\:\mathrm{2x}\right)−\mathrm{cos}\:\mathrm{x}}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}+\mathrm{cos}\:\mathrm{2x}−\mathrm{1}} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{cos}\:\mathrm{2x}−\mathrm{cos}\:\mathrm{x}\right)\left(\mathrm{1}−\mathrm{sin}\:\mathrm{3x}\right)}{\mathrm{cos}\:\mathrm{2x}−\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}}…
Question Number 66444 by ~ À ® @ 237 ~ last updated on 15/Aug/19 $$\:{calculate}\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\left({x}!\right)^{\frac{\mathrm{1}}{{x}}} \:\:\:\:\:\:\:{if}\:\:\:\:\:{x}!=\Pi\left({x}\right)=\int_{\mathrm{0}} ^{\infty} {t}^{{x}} \:{e}^{−{t}} {dt} \\ $$ Commented by…
Question Number 66434 by iklima_0412 last updated on 15/Aug/19 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{cos}^{\mathrm{2}} \mathrm{x}−\mathrm{x}}{\mathrm{1}−\mathrm{2x}} \\ $$ Commented by mathmax by abdo last updated on 15/Aug/19 $${let}\:{f}\left({x}\right)=\frac{{cos}^{\mathrm{2}} {x}−{x}}{\mathrm{1}−\mathrm{2}{x}}\:\Rightarrow\:{for}\:{x}\:\neq\mathrm{0}\:\:{we}\:{have}\:{f}\left({x}\right)=\frac{{x}−{cos}^{\mathrm{2}}…
Question Number 131962 by EDWIN88 last updated on 10/Feb/21 $$\:\underset{{x}\rightarrow\mathrm{1}^{+} } {\mathrm{lim}}\left(\frac{\mathrm{1}}{\mathrm{ln}\:\mathrm{x}}\:−\:\frac{\mathrm{1}}{\mathrm{x}−\mathrm{1}}\right)=? \\ $$ Answered by liberty last updated on 10/Feb/21 $$\:\mathrm{let}\:\mathrm{u}=\mathrm{ln}\:\mathrm{x}\:\Rightarrow\mathrm{x}=\mathrm{e}^{\mathrm{u}} \\ $$$$\:\underset{\mathrm{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}}{\mathrm{u}}−\frac{\mathrm{1}}{\mathrm{e}^{\mathrm{u}}…
Question Number 66406 by ~ À ® @ 237 ~ last updated on 14/Aug/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 131938 by Raxreedoroid last updated on 09/Feb/21 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\:\sqrt{{n}!}}=? \\ $$ Answered by Faetma last updated on 09/Feb/21 $$\left.\begin{matrix}{\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\:{n}!=+\infty}\\{\underset{\mathrm{N}\rightarrow+\infty} {\mathrm{lim}}\:\sqrt{\mathrm{N}}=+\infty}\\{\underset{\mathrm{N}'\rightarrow+\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{N}'}=\mathrm{0}^{+} }\end{matrix}\right\}\underset{{n}\rightarrow+\infty}…
Question Number 131882 by Eric002 last updated on 09/Feb/21 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}}{{x}+\sqrt[{\mathrm{4}}]{{x}+\mathrm{1}}−\mathrm{1}} \\ $$ Answered by liberty last updated on 09/Feb/21 $$\:\mathrm{L}'\mathrm{H}\ddot {\mathrm{o}pital}\:\mathrm{L}=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\:\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}\:\sqrt[{\mathrm{4}}]{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{3}} }}}\:\right]=\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}}=\:\frac{\mathrm{4}}{\mathrm{5}} \\…