Question Number 61834 by aliesam last updated on 09/Jun/19 Commented by maxmathsup by imad last updated on 09/Jun/19 $${let}\:{A}_{{n}} =\:\left\{\frac{\mathrm{1}}{{p}}\:\sum_{{k}=\mathrm{1}} ^{{p}} \:\left(\mathrm{1}+\frac{{k}}{{p}}\right)^{\frac{\mathrm{1}}{{n}}} \right\}^{{n}} \:\Rightarrow{ln}\left({A}_{{n}} \right)\:={n}\:{ln}\left(\frac{\mathrm{1}}{{p}}\sum_{{k}=\mathrm{1}}…
Question Number 192867 by beto last updated on 30/May/23 $${derivate}\:{of}\:\:{csc}\left(\mathrm{2}{x}\right)\:{by}\:\:{definition} \\ $$ Answered by cortano12 last updated on 02/Jun/23 $$\:\frac{\mathrm{d}\left(\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{2x}}\right)}{\mathrm{dx}}=\underset{\mathrm{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{\mathrm{1}}{\mathrm{sin}\:\left(\mathrm{2x}+\mathrm{2h}\right)}−\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{2x}}}{\mathrm{h}} \\ $$$$\:=\:\underset{\mathrm{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\mathrm{2x}−\mathrm{sin}\:\left(\mathrm{2x}+\mathrm{2h}\right)}{\mathrm{h}\:\mathrm{sin}\:\mathrm{2x}\:\mathrm{sin}\:\left(\mathrm{2x}+\mathrm{2h}\right)} \\…
Question Number 192846 by mathlove last updated on 29/May/23 $$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{3}{h}}{\:\sqrt[{\mathrm{5}}]{\mathrm{3}{h}+{x}}−\sqrt[{\mathrm{5}}]{{x}}}=? \\ $$ Answered by MM42 last updated on 29/May/23 $${for}\:\:{f}\left({x}\right)=\sqrt[{\mathrm{5}}]{{x}}\:\Rightarrow{f}'\left({x}\right)=\frac{\mathrm{1}}{\mathrm{5}\sqrt[{\mathrm{5}}]{{x}^{\mathrm{4}} }} \\ $$$${lim}_{{h}\rightarrow\mathrm{0}} \frac{\mathrm{3}{h}}{\:\sqrt[{\mathrm{5}}]{\mathrm{3}{h}+{x}}−\sqrt[{\mathrm{5}}]{{x}}}=\frac{\mathrm{1}}{{f}'\left({x}\right)}=\mathrm{5}\sqrt[{\mathrm{5}}]{{x}^{\mathrm{4}}…
Question Number 192841 by TUN last updated on 29/May/23 Answered by witcher3 last updated on 02/Jun/23 $$\mathrm{f}\left(\mathrm{a}\right)=\int_{−\mathrm{a}} ^{\mathrm{a}} \frac{\mathrm{dx}}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{e}^{\mathrm{bx}} \right)}\leqslant\int_{−\mathrm{a}} ^{\mathrm{a}} \frac{\mathrm{dx}}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}=\mathrm{2tan}^{−\mathrm{1}} \left(\mathrm{a}\right)…
Question Number 127213 by Study last updated on 27/Dec/20 $${li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\left({cosx}\right)^{{logx}} =???\:\:\:\:{by}\:{sandiwich}\:{rule} \\ $$ Commented by Study last updated on 27/Dec/20 $${help}\:{me} \\ $$ Commented…
Question Number 192652 by beto last updated on 24/May/23 $${lim}_{{x}\rightarrow\mathrm{0}} \frac{\mathrm{1}−\sqrt{{cos}\left({x}\right)}}{\mathrm{1}+{cos}\left(\sqrt{{x}}\right)} \\ $$ Answered by MM42 last updated on 24/May/23 $$\mathrm{0} \\ $$ Answered by…
Question Number 192651 by beto last updated on 24/May/23 $$ \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \frac{\mathrm{2}−\sqrt{{cos}\left({x}\right)}−{cos}\left({x}\right)}{{x}^{\mathrm{2}} } \\ $$ Answered by cortano12 last updated on 24/May/23 $$\:\mathrm{L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}−\sqrt{\mathrm{cos}\:\mathrm{x}}\right)+\left(\mathrm{1}−\mathrm{cos}\:\mathrm{x}\right)}{\mathrm{x}^{\mathrm{2}}…
Question Number 192650 by beto last updated on 24/May/23 $$ \\ $$$$\boldsymbol{\mathrm{lim}}_{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{0}} \frac{{x}−{sen}\left({x}\right)}{{tan}^{\mathrm{3}} \left({x}\right)} \\ $$$${without}\:{lhopital}\:{rule} \\ $$ Answered by MM42 last updated on 24/May/23…
Question Number 192646 by pascal889 last updated on 24/May/23 $$\boldsymbol{\mathrm{lim}}_{\boldsymbol{\mathrm{x}}\Rightarrow\mathrm{3}} \sqrt{\frac{\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{4}\boldsymbol{\mathrm{x}}+\mathrm{3}}{\boldsymbol{\mathrm{x}}−\mathrm{3}}} \\ $$$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{limit}} \\ $$ Answered by Subhi last updated on 24/May/23 $${lim}_{{x}\rightarrow\mathrm{3}} \sqrt{\frac{\left({x}−\mathrm{3}\right)\left({x}−\mathrm{1}\right)}{{x}−\mathrm{3}}}…
Question Number 192639 by pascal889 last updated on 24/May/23 $$ \\ $$lim_(x⇒∝ ) ((3x^3 +3x^2 +1)/(5x^3 +4x^2 +x)) find the limit Answered by Subhi last…