Question Number 53647 by Tawa1 last updated on 24/Jan/19 $$\underset{\boldsymbol{\phi}\rightarrow\infty} {\mathrm{lim}}\:\:\left(\mathrm{1}\:+\:\frac{\mathrm{1}}{\phi}\right)^{\phi} \:\:=\:\:\mathrm{e} \\ $$ Commented by maxmathsup by imad last updated on 24/Jan/19 $${we}\:{have}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}} \:={e}^{{xln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)}…
Question Number 184683 by alcohol last updated on 10/Jan/23 $${x}''\:−\:\frac{\mathrm{5}}{{t}}\:{x}'\:+\:\frac{\mathrm{8}}{{t}^{\mathrm{2}} }\:{x}\:=\:\frac{\mathrm{2}\:{ln}\:{t}}{{t}^{\mathrm{2}} } \\ $$$${solve}\:{the}\:{differential}\:{eqn} \\ $$ Answered by qaz last updated on 10/Jan/23 $${t}^{\mathrm{2}} {x}''−\mathrm{5}{tx}'+\mathrm{8}{x}=\mathrm{2}{lnt}…
Question Number 119074 by benjo_mathlover last updated on 22/Oct/20 $$\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\left({x}−\mathrm{1}\right)\:\mathrm{tan}\:\left(\frac{\pi{x}}{\mathrm{2}}\right)\: \\ $$ Answered by Dwaipayan Shikari last updated on 22/Oct/20 $$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\left({x}−\mathrm{1}\right){cot}\left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{2}}{x}\right) \\ $$$$=\underset{{x}\rightarrow\mathrm{1}}…
Question Number 119000 by bramlexs22 last updated on 21/Oct/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}^{\frac{\mathrm{3}}{\mathrm{2}}} \:\mathrm{sin}\:\mathrm{2}{x}\:=? \\ $$ Commented by MJS_new last updated on 21/Oct/20 $$\mathrm{not}\:\mathrm{defined} \\ $$$${x}^{\frac{\mathrm{3}}{\mathrm{2}}} \rightarrow\infty\wedge−\mathrm{1}\leqslant\mathrm{sin}\:\mathrm{2}{x}\:\leqslant\mathrm{1}…
Question Number 118966 by bramlexs22 last updated on 21/Oct/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{{x}+{bx}^{\mathrm{2}} }−\sqrt{{x}}}{{bx}\sqrt{{x}}}\:=? \\ $$ Answered by benjo_mathlover last updated on 21/Oct/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{{x}}\:\left(\sqrt{\mathrm{1}+{bx}}\:−\mathrm{1}\:\right)}{{bx}\:\sqrt{{x}}}\:=\: \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}}…
Question Number 118950 by benjo_mathlover last updated on 21/Oct/20 $$\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}^{\mathrm{2}} } .{e}^{−{x}} \:=\:?. \\ $$ Answered by john santu last updated on 21/Oct/20 $$\:\mathcal{L}\:=\:\underset{{x}\rightarrow\infty}…
Question Number 53376 by rajeshghorai130@gmail.com last updated on 21/Jan/19 Commented by Abdo msup. last updated on 21/Jan/19 $${let}\:{A}\left({x}\right)=\frac{\left({sin}\alpha\right)^{{x}} −\left({cos}\alpha\right)^{{x}} \:−{cos}\left(\mathrm{2}\alpha\right)}{{x}−\mathrm{4}} \\ $$$${A}\left({x}\right)=_{{x}−\mathrm{4}={t}} \:\:\:\:\frac{\left({sin}\alpha\right)^{{t}+\mathrm{4}} \:−\left({cos}\alpha\right)^{{t}+\mathrm{4}} −{cos}\left(\mathrm{2}\alpha\right)}{{t}}={B}\left({t}\right)…
Question Number 184435 by moh777 last updated on 06/Jan/23 Answered by CElcedricjunior last updated on 06/Jan/23 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\boldsymbol{{arsinx}}−\boldsymbol{{arctanx}}}{\boldsymbol{{x}}}=\mathrm{0} \\ $$$$ \\ $$ Terms of Service…
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Question Number 184329 by greougoury555 last updated on 05/Jan/23 $$\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}^{−\mathrm{1}} \left({p}+{x}\right)−\mathrm{tan}^{−\mathrm{1}} \left({p}−{x}\right)}{{x}}=? \\ $$ Answered by SEKRET last updated on 05/Jan/23 $$\:\:\frac{\mathrm{2}}{\mathrm{1}+\boldsymbol{\mathrm{p}}^{\mathrm{2}} } \\…