Category: Limits

Question Number 140530 by liberty last updated on 09/May/21 $$\underset{x\to 0}{\lim }\frac{(\mathrm{x}+\mathrm{y})\sec (\mathrm{x}+\mathrm{y})-\mathrm{ysec}\mathrm{y}}{\mathrm{x}}=?$$ Answered by EDWIN88 last updated on 09/May/21 $${\mathrm{L}}^{\prime }\mathrm{H}\ddot{\mathrm{o}pital}$$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sec}\:\left(\mathrm{x}+\mathrm{y}\right)+\left(\mathrm{x}+\mathrm{y}\right)\mathrm{sec}\:\left(\mathrm{x}+\mathrm{y}\right)\mathrm{tan}\:\left(\mathrm{x}+\mathrm{y}\right)−\mathrm{0}}{\mathrm{1}}…

Question Number 140471 by EDWIN88 last updated on 08/May/21 $$\mathrm{If}\underset{x\to 0}{\lim }{(\cos \mathrm{x}+\mathrm{a}\sin \mathrm{bx})}^{\frac{1}{\mathrm{x}}}={\mathrm{e}}^2$$$$\{\begin{array}{l}\mathrm{a}=?\\ \mathrm{b}=?\end{array}$$ Answered by benjo_mathlover last updated on 08/May/21 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\mathrm{cos}\:\mathrm{x}+\:\mathrm{a}\:\mathrm{sin}\:\mathrm{bx}\right)^{\frac{\mathrm{1}}{\mathrm{x}}}…

Question Number 140381 by Willson last updated on 07/May/21 $$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{xe}}^{1-\mathrm{x}}$$$$\mathrm{Prove}\mathrm{that}\underset{\mathrm{n}\to +\mathrm{\infty }}{\lim }\sqrt{\mathrm{n}}{\underset{0}{\int }}^1{\left[\mathrm{f}\left(\mathrm{x}\right)\right]}^{\mathrm{n}}\mathrm{dt}=\sqrt{\frac{\pi }{2}}$$ Terms of Service Privacy Policy Contact: info@tinkutara.com

Question Number 74766 by chess1 last updated on 30/Nov/19 Commented by mathmax by abdo last updated on 30/Nov/19 $$changement\frac{1}{x}=tleadyo{lim}_{t\to +\mathrm{\infty }}\sqrt{t+\sqrt{t+\sqrt{t}}}-\sqrt{t-\sqrt{t+\sqrt{t}}}$$$$={lim}_{t\to +\mathrm{\infty }}g\left(t\right)wehave\sqrt{t+\sqrt{t}}=\sqrt{t}\times \sqrt{1+\frac{1}{\sqrt{t}}}\sim \sqrt{t}(1+\frac{1}{2\sqrt{t}})$$$$=\sqrt{{t}}+\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\sqrt{{t}+\sqrt{{t}+\sqrt{{t}}}}\sim\sqrt{{t}+\sqrt{{t}}+\frac{\mathrm{1}}{\mathrm{2}}}\:{also}\:\sqrt{{t}−\sqrt{{t}+\sqrt{{t}}}}\sim\sqrt{{t}−\sqrt{{t}}−\frac{\mathrm{1}}{\mathrm{2}}}…

Question Number 140207 by mathlove last updated on 05/May/21 Answered by liberty last updated on 06/May/21 $$\underset{△x\to 0}{\lim }\frac{{\mathrm{e}}^{\mathrm{x}+\mathrm{m}}.{\mathrm{e}}^{△\mathrm{x}}-{\mathrm{e}}^{\mathrm{x}+\mathrm{m}}}{△\mathrm{x}}=$$$$\mathrm{e}^{\mathrm{x}+\mathrm{m}} \:.\underset{\bigtriangleup{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{e}^{\bigtriangleup\mathrm{x}}…