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 $$\mathrm{the}\:\mathrm{answer}\:\mathrm{should}\:\mathrm{be}\:\frac{\mathrm{4}}{\mathrm{9}}\:\mathrm{I}\:\mathrm{think}…
Question Number 74383 by aliesam last updated on 23/Nov/19 Commented by mathmax by abdo last updated on 23/Nov/19 $${A}\left({x}\right)=\frac{\mathrm{16}\sqrt{{x}−\sqrt{{x}}}−\mathrm{3}\sqrt{\mathrm{2}}{x}−\mathrm{4}\sqrt{\mathrm{2}}}{\mathrm{16}\left({x}−\mathrm{4}\right)^{\mathrm{2}} }{let}\:{use}\:{hospital}\:{theorem}\:\:{let}\:{f}\left({x}\right)=\mathrm{16}\sqrt{{x}−\sqrt{{x}}}\:−\mathrm{3}\sqrt{\mathrm{2}}{x}−\mathrm{4}\sqrt{\mathrm{2}} \\ $$$${and}\:{g}\left({x}\right)=\mathrm{16}\left({x}−\mathrm{4}\right)^{\mathrm{2}} \:\:{we}\:{have}\:{f}^{'} \left({x}\right)=\mathrm{16}\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}}{\mathrm{2}\sqrt{{x}−\sqrt{{x}}}}\:−\mathrm{3}\sqrt{\mathrm{2}} \\…
Question Number 8839 by tawakalitu last updated on 31/Oct/16 $$\mathrm{Prove}\:\mathrm{that}.\: \\ $$$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{1}\:+\:\mathrm{n}\right)^{\mathrm{1}/\mathrm{n}} \:=\:\mathrm{e} \\ $$ Answered by FilupSmith last updated on 31/Oct/16 $${L}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{1}\:+\:{n}\right)^{\mathrm{1}/{n}}…
Question Number 8628 by sou1618 last updated on 18/Oct/16 $${is}\:{it}\:{always}\:{satisfying}? \\ $$$$\boldsymbol{{A}}=\mathrm{lim}\left[{n}\rightarrow\infty\right]\int{f}\left({n},{x}\right){dx} \\ $$$$\boldsymbol{{B}}=\int\mathrm{lim}\left[{n}\rightarrow\infty\right]{f}\left({n},{x}\right){dx} \\ $$$${A}={B}?? \\ $$$${please}\:{show}\:{counter}\:{example} \\ $$$$ \\ $$$$ \\ $$$${checking} \\…