Question Number 184794 by Spillover last updated on 11/Jan/23 $${Evaluate}\: \\ $$$$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{5}} −\mathrm{32}}{{x}^{\mathrm{3}} −\mathrm{8}} \\ $$ Commented by MJS_new last updated on 11/Jan/23 $$\frac{\mathrm{20}}{\mathrm{3}}…
Question Number 119253 by bemath last updated on 23/Oct/20 $${Given}\:{three}\:{function}\:{f}\left({x}\right)\:,{g}\left({x}\right)\:{and}\:{h}\left({x}\right). \\ $$$${where}\:{f}\left({x}\right)={x}^{\mathrm{2}} +{x}−\mathrm{2}\:{and}\:{h}\left({x}\right)={x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{1}. \\ $$$${If}\:\frac{{f}\left({x}\right)}{{x}+\mathrm{3}}\:\leqslant\:\frac{{g}\left({x}\right)}{{x}+\mathrm{3}}\:\leqslant\:\frac{{h}\left({x}\right)}{{x}+\mathrm{3}}\:,\:{then}\:{the}\:{value}\:{of} \\ $$$$\underset{{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}\:{g}\left({x}\right)\:=\:? \\ $$ Answered by benjo_mathlover last…
Question Number 184790 by Spillover last updated on 11/Jan/23 $${Evaluate} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+{x}\right)^{\mathrm{6}} −\mathrm{1}}{\left(\mathrm{1}+{x}\right)^{\mathrm{5}} −\mathrm{1}} \\ $$ Commented by MJS_new last updated on 11/Jan/23 $$\mathrm{l}'\mathrm{H}\hat…
Question Number 184791 by Spillover last updated on 11/Jan/23 $${Evaluate}\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{{e}^{{x}} +{e}^{−{x}} −\mathrm{2}}{{x}^{\mathrm{2}} } \\ $$ Commented by MJS_new last updated on 11/Jan/23…
Question Number 119216 by benjo_mathlover last updated on 23/Oct/20 $$\:\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{3}} −\mathrm{3}{x}+\mathrm{2}}{\:\sqrt[{\mathrm{3}\:}]{{x}^{\mathrm{2}} }\:−\mathrm{2}\:\sqrt[{\mathrm{3}\:}]{{x}}\:+\mathrm{1}}\:?\: \\ $$ Commented by MJS_new last updated on 23/Oct/20 $$\mathrm{we}\:\mathrm{can}\:“\mathrm{hospitalize}''\:\mathrm{it}\:\mathrm{2}\:\mathrm{times},\:\mathrm{answer}\:\mathrm{is}\:\mathrm{27} \\ $$…
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}}…