Question Number 112273 by bemath last updated on 07/Sep/20 $$\:\left(\mathrm{1}\right)\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{e}^{\mathrm{x}} +\mathrm{e}^{−\mathrm{x}} \right)^{\frac{\mathrm{2}}{\mathrm{x}}} \:=? \\ $$$$\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\left(\mathrm{cos}\:\mathrm{x}\right)^{−\mathrm{x}+\frac{\pi}{\mathrm{2}}} ? \\ $$$$\left(\mathrm{3}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\sqrt{\frac{\mathrm{1}+\mathrm{tan}\:\mathrm{x}}{\mathrm{x}^{\mathrm{2}} }}−\sqrt{\frac{\mathrm{1}−\mathrm{sin}\:\mathrm{x}}{\mathrm{x}^{\mathrm{2}} }\:}? \\ $$…
Question Number 46563 by Necxx last updated on 28/Oct/18 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{{a}^{{x}} +{b}^{{x}} +{c}^{{x}} }{\mathrm{3}}\right)^{\mathrm{1}/{x}} \\ $$ Answered by ajfour last updated on 28/Oct/18 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+\frac{{a}^{{x}}…
Question Number 112083 by bemath last updated on 06/Sep/20 $$\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{4}^{{n}} \:\left(\mathrm{1}−\mathrm{cos}\:\left(\frac{\alpha}{\mathrm{2}^{{n}} }\right)\right)\:?\: \\ $$$$\:\:\sqrt{{bemath}} \\ $$ Answered by bobhans last updated on 06/Sep/20 $$\:\:\:\underset{\mathrm{n}\rightarrow\infty}…
Question Number 177570 by mathlove last updated on 07/Oct/22 $$\underset{\Delta{x}\rightarrow{cos}\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\:\frac{{sin}^{\mathrm{3}} \left(\Delta{x}+{x}\right)−{sin}^{\mathrm{3}} {x}}{\mathrm{2}^{−\mathrm{1}} \Delta{x}}=? \\ $$ Commented by blackmamba last updated on 07/Oct/22 $$ \\…
Question Number 111960 by john santu last updated on 05/Sep/20 $$\left(\mathrm{1}\right)\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}−{x}\right).\mathrm{tan}\:\left({x}+\frac{\pi}{\mathrm{4}}\right)\:? \\ $$$$\left(\mathrm{2}\right)\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\:\frac{\pi\left(\pi−\mathrm{2}{x}\right)\mathrm{tan}\:\left({x}−\frac{\pi}{\mathrm{2}}\right)}{\mathrm{2}\left({x}−\pi\right)\mathrm{cos}\:^{\mathrm{2}} {x}}\:? \\ $$$$\left(\mathrm{3}\right)\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{3}{x}\left(\mathrm{cos}\:\mathrm{3}{x}−\mathrm{cos}\:\mathrm{7}{x}\right)}{\:\sqrt{\mathrm{sin}\:\mathrm{2}{x}+\mathrm{1}}\:−\sqrt{\mathrm{tan}\:\mathrm{2}{x}+\mathrm{1}}}\:? \\ $$$$\left(\mathrm{4}\right)\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{3}−\sqrt{\mathrm{3}{x}+\mathrm{9}}}\:? \\ $$$$ \\…
Question Number 111923 by bemath last updated on 05/Sep/20 $$\sqrt{{bemath}\:} \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}\:}]{\mathrm{cos}\:\mathrm{4}{x}}\:−\mathrm{1}}{\mathrm{cos}\:\mathrm{3}{x}−\mathrm{cos}\:\mathrm{9}{x}}\:? \\ $$ Answered by bobhans last updated on 05/Sep/20 $$\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}\:}]{\mathrm{cos}\:\mathrm{4x}}\:−\mathrm{1}}{\mathrm{cos}\:\mathrm{3x}\:−\:\mathrm{cos}\:\mathrm{9x}}\:? \\…
Question Number 111914 by bemath last updated on 05/Sep/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{8}{x}^{\mathrm{3}} }{\:\sqrt{\mathrm{4}+\mathrm{sin}\:\mathrm{6}{x}}\:−\sqrt{\mathrm{tan}\:\mathrm{6}{x}+\mathrm{4}}}\:? \\ $$ Answered by bobhans last updated on 05/Sep/20 $$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{8x}^{\mathrm{3}} }{\:\sqrt{\mathrm{4}+\mathrm{sin}\:\mathrm{6x}}\:−\sqrt{\mathrm{4}+\mathrm{tan}\:\mathrm{6x}}}\:=\: \\…
Question Number 111907 by bemath last updated on 05/Sep/20 $$\underset{{x}\rightarrow\frac{\pi}{\mathrm{20}}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{tan}\:\mathrm{5}{x}}{\mathrm{sin}\:\mathrm{5}{x}−\mathrm{cos}\:\mathrm{5}{x}}\:? \\ $$ Answered by john santu last updated on 05/Sep/20 Answered by john santu…
Question Number 177431 by Ar Brandon last updated on 04/Oct/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 111882 by bemath last updated on 05/Sep/20 $$\:\:\sqrt{{bemath}} \\ $$$$\left(\mathrm{1}\right)\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left[\frac{{x}^{\mathrm{2}} }{{x}+\mathrm{1}}\:−\:\frac{{x}^{\mathrm{2}} }{{x}+\mathrm{3}}\:\right]\:? \\ $$$$\left(\mathrm{2}\right)\:{prove}\:{that}\:{n}^{\mathrm{2}} \:\leqslant\:\mathrm{2}^{{n}} \:{for}\:\forall{n}\in\mathbb{N} \\ $$$${by}\:{mathematical}\:{induction} \\ $$ Answered by…