Question Number 122236 by benjo_mathlover last updated on 15/Nov/20 $$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+{mx}\right)^{{n}} −\left(\mathrm{1}+{nx}\right)^{{m}} }{{x}^{\mathrm{2}} }\:? \\ $$ Answered by Dwaipayan Shikari last updated on 15/Nov/20 $$\underset{{x}\rightarrow\mathrm{0}}…
Question Number 122241 by TANMAY PANACEA last updated on 15/Nov/20 $${limit}… \\ $$ Commented by TANMAY PANACEA last updated on 15/Nov/20 Commented by Dwaipayan Shikari…
Question Number 122235 by benjo_mathlover last updated on 15/Nov/20 $$\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{{x}}{{x}+\mathrm{2}}\right)^{{x}} ?\: \\ $$ Answered by liberty last updated on 15/Nov/20 $$\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{x}}{\mathrm{x}+\mathrm{2}}\right)^{\mathrm{x}} =\:\mathrm{e}^{\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{x}}{\mathrm{x}+\mathrm{2}}\:−\mathrm{1}\right).\mathrm{x}}…
Question Number 122192 by physicstutes last updated on 14/Nov/20 $$\mathrm{Prove}\:\mathrm{using}\:\mathrm{the}\:\mathrm{squeeze}\:\mathrm{law}\:\mathrm{of}\:\mathrm{functions}\:\mathrm{that}\: \\ $$$$\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:\sqrt{{x}}\:=\:\sqrt{{a}}\:. \\ $$$$\:\mathrm{Recall}\:\mathrm{that}\:\mathrm{the}\:\mathrm{squeeze}\:\mathrm{theorem}\:\mathrm{states}\:\mathrm{that}\:\mathrm{if} \\ $$$$\:{f}\left({x}\right)\:\leqslant\:\mathrm{g}\left({x}\right)\:\leqslant\:{h}\left({x}\right)\: \\ $$$$\:\mathrm{and}\:\underset{{x}−{a}} {\mathrm{lim}}\:{f}\left({x}\right)\:=\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:{h}\left({x}\right)\:=\:{L} \\ $$$$\mathrm{then}\:,\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:\mathrm{g}\left({x}\right)\:=\:{L}. \\…
Question Number 122152 by liberty last updated on 14/Nov/20 $$\:\left(\mathrm{1}\right)\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{n}^{\mathrm{2}} +\mathrm{1}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{n}^{\mathrm{2}} +\mathrm{2}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{n}^{\mathrm{2}} +\mathrm{3}}}+…+\frac{\mathrm{1}}{\:\sqrt{\mathrm{n}^{\mathrm{2}} +\mathrm{n}+\mathrm{1}}}\:\right)\:=? \\ $$$$\left(\mathrm{2}\right)\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\left(\mathrm{2}\:\sqrt[{\mathrm{n}}]{\mathrm{n}}\:−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}^{\mathrm{2}} }\:=? \\ $$ Answered by benjo_mathlover…
Question Number 122130 by bemath last updated on 14/Nov/20 $$\:\left(\mathrm{1}\right)\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{3}} \:\mathrm{sin}\:\left(\frac{\mathrm{1}}{{x}−\mathrm{1}}\right)^{\mathrm{3}} ? \\ $$$$\:\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{{x}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{sin}\:^{\mathrm{2}} {x}\right)}{\left({x}+\mathrm{sin}\:{x}\right)^{\mathrm{2}} }\:? \\ $$ Answered by TANMAY…
Question Number 187645 by Mingma last updated on 19/Feb/23 Answered by Ar Brandon last updated on 19/Feb/23 $$\mathscr{L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}^{\mathrm{4}} \left(\pi\mathrm{cos}{x}\right)}{\mathrm{1}−\mathrm{cos}\left(\mathrm{1}−\mathrm{cos}\left(\mathrm{1}−\mathrm{cos}{x}\right)\right)} \\ $$$$\:\:\:\:\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}^{\mathrm{4}} \left(\pi−\frac{\pi{x}^{\mathrm{2}} }{\mathrm{2}}\right)}{\frac{{x}^{\mathrm{8}}…
Question Number 122106 by bemath last updated on 14/Nov/20 $$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{3sin}\:\left(\pi{x}\right)−\mathrm{sin}\:\left(\mathrm{3}\pi{x}\right)}{{x}^{\mathrm{3}} }\:=? \\ $$ Answered by liberty last updated on 14/Nov/20 $$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{3}\left(\pi\mathrm{x}−\frac{\pi^{\mathrm{3}} \mathrm{x}^{\mathrm{3}} }{\mathrm{6}}\right)−\left(\mathrm{3}\pi\mathrm{x}−\frac{\mathrm{27}\pi^{\mathrm{3}}…
Question Number 122071 by bemath last updated on 13/Nov/20 $$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{e}^{\frac{\mathrm{1}}{{x}}} \left(\frac{\mathrm{1}+\mathrm{2}{x}}{\mathrm{1}+\mathrm{3}{x}}\right)^{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} \:? \\ $$ Answered by liberty last updated on 13/Nov/20 $$\:\mathrm{L}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{x}}}…
Question Number 187607 by SonGoku last updated on 19/Feb/23 $$\: \\ $$$$\:\boldsymbol{\mathrm{Help}}!\:\::\left(\right. \\ $$$$\: \\ $$$$\:\boldsymbol{\mathrm{Should}}\:\:\boldsymbol{\mathrm{i}}\:\:\boldsymbol{\mathrm{partially}}\:\:\boldsymbol{\mathrm{differentiate}}\:\:\boldsymbol{\mathrm{the}}\:\:\boldsymbol{\mathrm{numerator}}\:\:\boldsymbol{\mathrm{and}}\:\:\boldsymbol{\mathrm{denominator}}? \\ $$$$\: \\ $$$$\:\underset{\left(\boldsymbol{{x}},\:\boldsymbol{\mathrm{y}}\right)\rightarrow\left(\mathrm{0},\mathrm{1}\right)} {\boldsymbol{\mathrm{lim}}}\left[\frac{\boldsymbol{{x}}^{\boldsymbol{{e}}} \:\:+\:\boldsymbol{{ln}}\left(\frac{\mathrm{1}}{\boldsymbol{\mathrm{y}}}\right)}{\boldsymbol{{x}}^{\boldsymbol{{e}}} \:−\:\boldsymbol{{ln}}\left(\frac{\mathrm{1}}{\boldsymbol{\mathrm{y}}}\right)}\right] \\ $$…