Question Number 74766 by chess1 last updated on 30/Nov/19 Commented by mathmax by abdo last updated on 30/Nov/19 $${changement}\:\frac{\mathrm{1}}{{x}}={t}\:{lead}\:{yo}\:{lim}_{{t}\rightarrow+\infty} \sqrt{{t}+\sqrt{{t}+\sqrt{{t}}}}\:−\sqrt{{t}−\sqrt{{t}+\sqrt{{t}}}} \\ $$$$={lim}_{{t}\rightarrow+\infty} {g}\left({t}\right)\:\:{we}\:{have}\:\:\sqrt{{t}+\sqrt{{t}}}=\sqrt{{t}}×\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{{t}}}}\sim\sqrt{{t}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{{t}}}\right) \\ $$$$=\sqrt{{t}}+\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\sqrt{{t}+\sqrt{{t}+\sqrt{{t}}}}\sim\sqrt{{t}+\sqrt{{t}}+\frac{\mathrm{1}}{\mathrm{2}}}\:{also}\:\sqrt{{t}−\sqrt{{t}+\sqrt{{t}}}}\sim\sqrt{{t}−\sqrt{{t}}−\frac{\mathrm{1}}{\mathrm{2}}}…
Question Number 9157 by tawakalitu last updated on 21/Nov/16 $$\underset{{x}\rightarrow\mathrm{y}} {\mathrm{lim}}\:\:\frac{\mathrm{x}^{\mathrm{n}} \:−\:\mathrm{y}^{\mathrm{n}} }{\mathrm{x}\:−\:\mathrm{y}} \\ $$ Answered by prakash jain last updated on 21/Nov/16 $${x}^{{n}} −{y}^{{n}}…
Question Number 140207 by mathlove last updated on 05/May/21 Answered by liberty last updated on 06/May/21 $$\underset{\bigtriangleup{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{e}^{\mathrm{x}+\mathrm{m}} .\mathrm{e}^{\bigtriangleup\mathrm{x}} −\mathrm{e}^{\mathrm{x}+\mathrm{m}} }{\bigtriangleup\mathrm{x}}\:= \\ $$$$\mathrm{e}^{\mathrm{x}+\mathrm{m}} \:.\underset{\bigtriangleup{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{e}^{\bigtriangleup\mathrm{x}}…
Question Number 74600 by rajesh4661kumar@gmail.com last updated on 27/Nov/19 Answered by ajfour last updated on 27/Nov/19 $${L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\left\{\left(\frac{{a}^{{mx}} −\mathrm{1}}{{mx}}\right)/\left(\frac{{b}^{{nx}} −\mathrm{1}}{{nx}}\right)\right\}\left(\frac{{mx}}{{nx}}\right)\right] \\ $$$$\:{L}\:=\:\frac{{m}\mathrm{ln}\:{a}}{{n}\mathrm{ln}\:{b}}\:. \\ $$ Terms…
Question Number 74580 by chess1 last updated on 26/Nov/19 Answered by mind is power last updated on 26/Nov/19 $$\sqrt{\mathrm{1}+\mathrm{t}}=\mathrm{1}+\frac{\mathrm{t}}{\mathrm{2}}+\mathrm{o}\left(\mathrm{t}\right)\Rightarrow\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}=\mathrm{1}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{o}\left(\mathrm{x}^{\mathrm{2}} \right) \\ $$$$\sqrt[{\mathrm{3}}]{\left(\mathrm{t}+\mathrm{1}\right)}=\mathrm{1}+\frac{\mathrm{t}}{\mathrm{3}}+\mathrm{o}\left(\mathrm{t}\right)\Rightarrow\sqrt[{\mathrm{3}}]{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }=\mathrm{1}+\frac{\mathrm{x}^{\mathrm{2}}…
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Question Number 74508 by isra.pm07nov@gmail.com last updated on 25/Nov/19 $${li}=\overset{−} {{x}}−{t}_{{v}} ,\frac{\alpha}{\mathrm{2}}.\left(\frac{{s}}{\:\sqrt{{n}}}\right) \\ $$$${ls}=\overset{−} {{x}}+{t}_{{v}} ,\frac{\alpha}{\mathrm{2}}.\left(\frac{{s}}{\:\sqrt{{n}}}\right) \\ $$$$ \\ $$ Terms of Service Privacy Policy…
Question Number 8964 by Joel575 last updated on 08/Nov/16 $$\mathrm{Can}\:\mathrm{you}\:\mathrm{solve}\:\mathrm{this}\:\mathrm{problem}? \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{log}\:\left({x}^{\mathrm{3}} \:+\:\left(\mathrm{log}\:{x}\right)^{\mathrm{3}} \right)}{\mathrm{log}\:\left({x}^{\mathrm{2}} \:+\:\left(\mathrm{log}\:{x}\right)^{\mathrm{2}} \right)}\: \\ $$ Commented by sou1618 last updated on…
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Question Number 140000 by bramlexs22 last updated on 03/May/21 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{1}+\mathrm{x}.\mathrm{2}^{\mathrm{x}} }{\mathrm{1}+\mathrm{x}.\mathrm{3}^{\mathrm{x}} }\right)^{\frac{\mathrm{2}}{\mathrm{x}^{\mathrm{2}} }} =? \\ $$ Commented by MJS_new last updated on 03/May/21 $$\mathrm{the}\:\mathrm{answer}\:\mathrm{should}\:\mathrm{be}\:\frac{\mathrm{4}}{\mathrm{9}}\:\mathrm{I}\:\mathrm{think}…