Question Number 47960 by Meritguide1234 last updated on 17/Nov/18 Answered by ajfour last updated on 17/Nov/18 $${L}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{n}}\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\underset{{j}=\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{{i}+{j}}{{i}^{\mathrm{2}} +{j}^{\mathrm{2}} } \\…
Question Number 113438 by bemath last updated on 13/Sep/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}\left(\mathrm{5}^{\frac{\mathrm{1}}{{x}}} −\mathrm{1}\right)\:? \\ $$ Answered by bemath last updated on 13/Sep/20 Answered by JDamian last…
Question Number 178942 by cortano1 last updated on 23/Oct/22 Answered by mr W last updated on 23/Oct/22 Commented by mr W last updated on 23/Oct/22…
Question Number 113274 by bemath last updated on 12/Sep/20 $$\:\left(\mathrm{1}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\mathrm{x}+\mathrm{4tan}\:\mathrm{2x}−\mathrm{3tan}\:\mathrm{3x}}{\mathrm{x}^{\mathrm{2}} \:\mathrm{tan}\:\mathrm{x}} \\ $$$$\:\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{x}}−\sqrt{\mathrm{sin}\:\mathrm{x}}}{\mathrm{x}^{\frac{\mathrm{3}}{\mathrm{2}}} }\: \\ $$$$\:\left(\mathrm{3}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{x}}+\sqrt{\mathrm{sin}\:\mathrm{x}}}{\mathrm{x}^{\frac{\mathrm{5}}{\mathrm{2}}} } \\ $$ Commented by bemath…
Question Number 178751 by mathlove last updated on 21/Oct/22 Answered by cortano1 last updated on 21/Oct/22 $$\:\mathrm{With}\:\mathrm{L}'\mathrm{Hopital}\:\mathrm{rule} \\ $$$$\:\mathrm{L}=\underset{{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{2020x}^{\mathrm{2019}} −\mathrm{2021}}{\mathrm{3x}^{\mathrm{2}} } \\ $$$$\:=\frac{−\mathrm{2020}−\mathrm{2021}}{\mathrm{3}}=\:−\frac{\mathrm{4041}}{\mathrm{3}}=−\mathrm{1347} \\…
Question Number 178716 by cortano1 last updated on 20/Oct/22 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{3tan}\:\mathrm{4x}−\mathrm{4tan}\:\mathrm{3x}}{\mathrm{3sin}\:\mathrm{4x}−\mathrm{4sin}\:\mathrm{3x}}\:=\:? \\ $$ Answered by a.lgnaoui last updated on 21/Oct/22 $$\:\:{f}\left({x}\right)=\left(\:\:\frac{\mathrm{tan}\:\mathrm{4x}\left(\mathrm{3}−\frac{\mathrm{4tan}\:\mathrm{3x}}{\mathrm{tan}\:\mathrm{4x}\:\:}\right)}{\mathrm{sin}\:\mathrm{4x}\left(\mathrm{3}−\frac{\mathrm{4sin}\:\mathrm{3x}}{\mathrm{sin}\:\mathrm{4x}}\right)}=\frac{\mathrm{1}}{\mathrm{cos}\:\mathrm{4x}}×\left(\frac{\mathrm{3}−\frac{\mathrm{4sin}\:\mathrm{3x}}{\mathrm{cos}\:\mathrm{3x}}×\frac{\mathrm{cos}\:\mathrm{4x}}{\mathrm{sin}\:\mathrm{4x}}}{\mathrm{3}−\:\frac{\mathrm{4sin}\:\mathrm{3x}}{\mathrm{sin}\:\mathrm{4x}}}\right)\:\right. \\ $$$$\frac{\mathrm{1}}{\mathrm{cos}\:\mathrm{4x}}×\left(\frac{\mathrm{3}−\frac{\mathrm{4sin}\:\mathrm{3x}}{\mathrm{sin}\:\mathrm{4x}}×\frac{\mathrm{cos}\:\mathrm{4x}}{\mathrm{cos}\:\mathrm{3x}}}{\mathrm{3}−\frac{\mathrm{4sin}\:\mathrm{3x}}{\mathrm{sin}\:\mathrm{4x}}}\right) \\ $$$$\mathrm{x}\rightarrow\mathrm{0}\:\:\:\:\:{f}\left({x}\right)\rightarrow\frac{\mathrm{3}−\frac{\mathrm{4sin}\:\mathrm{3x}}{\mathrm{sin}\:\mathrm{4x}}}{\mathrm{3}−\frac{\mathrm{4sin}\:\mathrm{3x}}{\mathrm{sin}\:\mathrm{4x}}}=\:\:\frac{\mathrm{3}−\mathrm{4}\frac{\mathrm{sin}\:\mathrm{3x}}{\mathrm{3x}}×\frac{\mathrm{4x}}{\mathrm{sin}\:\mathrm{4x}}}{\mathrm{3}−\mathrm{4}\frac{\mathrm{sin}\:\mathrm{3x}}{\mathrm{3x}}×\frac{\mathrm{3x}}{\mathrm{sin}\:\mathrm{4x}}}\:\:=\mathrm{1}…
Question Number 112974 by bemath last updated on 10/Sep/20 $$\mathrm{solve}\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{2x}^{\mathrm{3}} −\mathrm{x}^{\mathrm{2}} +\mathrm{2}}{\mathrm{2x}^{\mathrm{3}} −\mathrm{4x}^{\mathrm{2}} +\mathrm{1}}\right)^{\mathrm{x}} \\ $$ Answered by john santu last updated on 10/Sep/20…
Question Number 47402 by Aditya789 last updated on 09/Nov/18 Commented by Aditya789 last updated on 10/Nov/18 $$\mathrm{plz}\:\mathrm{explain} \\ $$ Commented by maxmathsup by imad last…
Question Number 47401 by Aditya789 last updated on 09/Nov/18 Commented by maxmathsup by imad last updated on 10/Nov/18 $${let}\:{A}\left({x}\right)=\left({cosx}\right)^{\frac{\mathrm{1}}{{sinx}}} \:\Rightarrow{A}\left({x}\right)={e}^{\frac{\mathrm{1}}{{sinx}}{ln}\left({cosx}\right)} \:\:{but}\:{cosx}\:\sim\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\left({x}\rightarrow\mathrm{0}\right)\:\Rightarrow \\ $$$${ln}\left({cosx}\right)\:\sim\:{ln}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}\:}\right)\:\sim−\frac{{x}^{\mathrm{2}}…
Question Number 112917 by abdullahquwatan last updated on 10/Sep/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left[\mathrm{csc}^{\mathrm{2}} \left(\frac{\mathrm{2}}{\mathrm{x}}\right)−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{x}^{\mathrm{2}} \right] \\ $$ Commented by bobhans last updated on 10/Sep/20 $$\underset{\mathrm{w}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{2w}+\mathrm{sin}\:\left(\mathrm{2w}\right)}{\mathrm{4w}}\right)\left(\frac{\mathrm{2w}−\mathrm{sin}\:\left(\mathrm{2w}\right)}{\mathrm{w}.\mathrm{sin}^{\mathrm{2}} \:\left(\mathrm{2w}\right)}\right)…