Question Number 108956 by bobhans last updated on 20/Aug/20 $$\:\:\:\:\:\:\:\:\:\overset{\bigstar} {{b}}\overset{\bigstar} {{o}}\overset{\bigstar} {{b}}\overset{\Box} {{h}}\overset{\Box} {{a}}\overset{\Box} {{n}}\overset{\spadesuit} {{s}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\mathrm{8}{x}−\mathrm{sin}\:\mathrm{4}{x}.\mathrm{cos}\:\mathrm{4}{x}}{{x}.\mathrm{sin}\:\mathrm{4}{x}.\mathrm{tan}\:\mathrm{8}{x}}\:? \\ $$ Answered by john…
Question Number 174407 by infinityaction last updated on 31/Jul/22 $$\:{evaluate} \\ $$$$\:\:\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\frac{\mathrm{3}^{\mathrm{sin}{x}\:} +\mathrm{2}{x}\:+\mathrm{1}}{\mathrm{sin}{x}−\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:} \\ $$ Answered by CElcedricjunior last updated on 31/Jul/22 $$\underset{{x}\rightarrow−\infty}…
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Question Number 108842 by Study last updated on 19/Aug/20 Commented by Study last updated on 19/Aug/20 $${i}\:{need}\:{u}\:{help} \\ $$ Commented by 150505R last updated on…
Question Number 174331 by cortano1 last updated on 30/Jul/22 Commented by kaivan.ahmadi last updated on 30/Jul/22 $$\overset{{hop}} {\sim}\:{li}\underset{{x}\rightarrow\infty} {{m}}\:\frac{\frac{\mathrm{3}{e}^{\mathrm{3}{x}} }{\mathrm{2}+{e}^{\mathrm{3}{x}} }}{\frac{\mathrm{2}{e}^{\mathrm{2}{x}} }{\mathrm{3}+{e}^{\mathrm{2}{x}} }}={li}\underset{{x}\rightarrow\infty} {{m}}\frac{\mathrm{3}{e}^{\mathrm{3}{x}} \left(\mathrm{3}+{e}^{\mathrm{2}{x}}…
Question Number 174327 by aminee last updated on 30/Jul/22 Answered by Mathspace last updated on 31/Jul/22 $${we}\:{have}\:{sin}\left({a}+{b}\right)={sina}.{cosb}+{cosasinb} \\ $$$${sin}\left({a}−{b}\right)={sinacosb}−{cosasinb}\:\Rightarrow \\ $$$$\mathrm{2}{sina}.{cosb}={sin}\left({a}+{b}\right)+{sin}\left({a}−{b}\right)\:\Rightarrow \\ $$$${sina}.{cosb}=\frac{\mathrm{1}}{\mathrm{2}}\left\{{sin}\left({a}+{b}\right)+{sin}\left({a}−{b}\right)\right\} \\ $$$$\Rightarrow\sum_{{n}=\mathrm{1}}…
Question Number 108757 by bemath last updated on 19/Aug/20 $$\:\:\frac{\vdots\mathcal{B}{e}\mathcal{M}{ath}\vdots}{\triangleright\heartsuit\triangleleft} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\sqrt{\mathrm{cos}\:{x}\mathrm{cos}\:\mathrm{2}{x}\mathrm{cos}\:\mathrm{3}{x}\mathrm{cos}\:\mathrm{4}{x}…\mathrm{cos}\:{nx}}}{{x}^{\mathrm{2}} }\:=? \\ $$ Answered by bemath last updated on 19/Aug/20 Terms of…
Question Number 174276 by cortano1 last updated on 28/Jul/22 Answered by blackmamba last updated on 28/Jul/22 $$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\sqrt[{\mathrm{2022}}]{\mathrm{cos}\:\left(\mathrm{2021}{x}\right)}}{{x}^{\mathrm{2}} }\:= \\ $$$$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\sqrt[{\mathrm{2022}}]{\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{\mathrm{2021}{x}}{\mathrm{2}}\right)}}{{x}^{\mathrm{2}} } \\…
Question Number 174258 by blackmamba last updated on 28/Jul/22 $$\:\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt{\sqrt{{x}^{\mathrm{4}} −{x}^{\mathrm{3}} }}\:−\:{x}\:=? \\ $$ Answered by cortano1 last updated on 28/Jul/22 $$\:\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{{x}^{\mathrm{4}} −{x}^{\mathrm{3}}…
Question Number 108623 by bemath last updated on 18/Aug/20 $$\:\:\frac{\supset\mathcal{B}{e}\mathcal{M}{ath}\supset}{\bigstar} \\ $$$$\:\left(\mathrm{1}\right)\underset{{b}\rightarrow{a}} {\mathrm{lim}}\:\frac{{b}\sqrt{{a}}−{a}\sqrt{{b}}}{{a}\sqrt{{a}}+{b}\sqrt{{a}}−\mathrm{2}{a}\sqrt{{a}}} \\ $$$$\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{3}{e}^{\mathrm{2}{x}} +{e}^{{x}} −\mathrm{4}}{{x}} \\ $$ Answered by ajfour last updated…