Question Number 184797 by Spillover last updated on 11/Jan/23 $${Show}\:{that}\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{\frac{\mathrm{1}}{{x}}} −\mathrm{1}}{{e}^{\frac{\mathrm{1}}{{x}}} +\mathrm{1}}\:\:\:{does}\:{not}\:{exist} \\ $$ Commented by MJS_new last updated on 11/Jan/23 $$\mathrm{let}\:{x}=\frac{\mathrm{1}}{{t}}…
Question Number 184796 by Spillover last updated on 11/Jan/23 $${Evaluate}\: \\ $$$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{6}}} {\mathrm{lim}}\frac{\sqrt{\mathrm{3}}\mathrm{sin}\:{x}−\mathrm{cos}\:{x}}{{x}−\frac{\pi}{\mathrm{6}}} \\ $$ Commented by MJS_new last updated on 11/Jan/23 $$\mathrm{2} \\ $$…
Question Number 184793 by Spillover last updated on 11/Jan/23 $${Evaluate}\: \\ $$$$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\:\frac{{x}^{\mathrm{2}} −\mathrm{4}}{\:\sqrt{\mathrm{3}{x}−\mathrm{2}}−\sqrt{{x}+\mathrm{2}}} \\ $$$$ \\ $$ Answered by MJS_new last updated on 11/Jan/23…
Question Number 184798 by Spillover last updated on 11/Jan/23 $${Show}\:{that}\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}}{\mid{x}\mid}\:\:{does}\:{not}\:{exist} \\ $$ Commented by MJS_new last updated on 11/Jan/23 $${x}\rightarrow\mathrm{0}^{−} :\:\frac{{x}}{\mid{x}\mid}=\frac{{x}}{−{x}}=−\mathrm{1} \\…
Question Number 184795 by Spillover last updated on 11/Jan/23 $${Evaluate}\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{cos}\:{x}\sqrt{\mathrm{cos}\:\mathrm{2}{x}}\:}{{x}^{\mathrm{2}} } \\ $$ Commented by MJS_new last updated on 11/Jan/23 $$\frac{\mathrm{3}}{\mathrm{2}} \\…
Question Number 184792 by Spillover last updated on 11/Jan/23 $${Evaluate}\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:{x}−\mathrm{sin}\:{x}}{\mathrm{sin}\:^{\mathrm{3}} {x}} \\ $$$$ \\ $$ Commented by MJS_new last updated on 11/Jan/23…
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…