Question Number 76815 by Rio Michael last updated on 30/Dec/19 $$\mathrm{Which}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following}\:\mathrm{series}\:\mathrm{is}\:\boldsymbol{\mathrm{N}}\mathrm{o}\boldsymbol{\mathrm{t}}\:\mathrm{convergent}? \\ $$$$\mathrm{A}.\:\underset{\mathrm{r}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{r}^{\mathrm{3}} } \\ $$$$\mathrm{B}.\:\underset{\mathrm{r}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{r}^{\mathrm{2}} } \\ $$$$\mathrm{C}.\:\underset{\mathrm{r}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{r}^{\mathrm{2}}…
Question Number 142351 by bramlexs22 last updated on 30/May/21 $$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}\:}]{\mathrm{27}+{x}}−\sqrt[{\mathrm{3}\:}]{\mathrm{27}−{x}}}{\:\sqrt[{\mathrm{3}\:}]{{x}^{\mathrm{2}} }\:+\:\sqrt[{\mathrm{4}\:}]{{x}^{\mathrm{3}} }}\:=? \\ $$ Answered by mathmax by abdo last updated on 30/May/21 $$\mathrm{f}\left(\mathrm{x}\right)=\frac{\left(\mathrm{27}+\mathrm{x}\right)^{\frac{\mathrm{1}}{\mathrm{3}}}…
Question Number 142329 by HarshSahu last updated on 30/May/21 $$\:{lim}_{{x}\rightarrow\infty} \:\left(\frac{{x}!}{{x}^{{x}} }\right)^{\frac{\mathrm{1}}{{x}}} \\ $$ Answered by Dwaipayan Shikari last updated on 30/May/21 $$\left(\frac{{x}!}{{x}^{{x}} }\right)^{\frac{\mathrm{1}}{{x}}} ={y}…
Question Number 142309 by qaz last updated on 29/May/21 $$\underset{\mathrm{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\frac{\mathrm{x}^{\left(\mathrm{sin}\:\mathrm{x}\right)^{\mathrm{x}} } −\left(\mathrm{sin}\:\mathrm{x}\right)^{\mathrm{x}^{\mathrm{sin}\:\mathrm{x}} } }{\mathrm{x}^{\mathrm{3}} }=? \\ $$ Answered by mnjuly1970 last updated on…
Question Number 142308 by qaz last updated on 29/May/21 $$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{lnlnln}\left[\mathrm{x}+\left(\mathrm{1}+\mathrm{x}\right)^{\frac{\left(\mathrm{1}+\mathrm{x}\right)^{\frac{\mathrm{1}}{\mathrm{x}}} }{\mathrm{x}}} \right]+\mathrm{x}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{e}^{\mathrm{e}+\mathrm{1}} }\right)}{\mathrm{x}^{\mathrm{2}} }=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 142307 by qaz last updated on 29/May/21 $$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{tan}\:\left(\mathrm{tan}\:\mathrm{x}\right)−\mathrm{tan}\:\left(\mathrm{tan}\:\left(\mathrm{tan}\:\mathrm{x}\right)\right)}{\mathrm{tan}\:\mathrm{x}\centerdot\mathrm{tan}\:\left(\mathrm{tan}\:\mathrm{x}\right)\centerdot\mathrm{tan}\:\left(\mathrm{tan}\:\left(\mathrm{tan}\:\mathrm{x}\right)\right)}=? \\ $$ Answered by Dwaipayan Shikari last updated on 29/May/21 $${tanx}\approx{x}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\:\:{or}\:\:{x}\:\:\left({sometimes}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}}…
Question Number 142300 by qaz last updated on 29/May/21 $$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{e}^{\mathrm{sin}\:\mathrm{x}} +\mathrm{sin}\:\mathrm{x}\right)^{\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{x}}} −\left(\mathrm{e}^{\mathrm{tan}\:\mathrm{x}} +\mathrm{tan}\:\mathrm{x}\right)^{\frac{\mathrm{1}}{\mathrm{tan}\:\mathrm{x}}} }{\mathrm{x}^{\mathrm{3}} }=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 142299 by mnjuly1970 last updated on 29/May/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 142277 by liberty last updated on 29/May/21 $$\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−{e}^{\mathrm{sin}\:{x}\:\mathrm{ln}\:\left(\mathrm{cos}\:{x}\right)} }{{x}^{\mathrm{3}} }\:=? \\ $$ Answered by Dwaipayan Shikari last updated on 29/May/21 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−{e}^{{sinx}\left(−\frac{{x}^{\mathrm{2}}…
Question Number 142204 by mnjuly1970 last updated on 27/May/21 Answered by MJS_new last updated on 28/May/21 $$\mathrm{what}\:\mathrm{are}\:\mathrm{you}\:\mathrm{waiting}\:\mathrm{for}? \\ $$$${A}^{\mathrm{2}} =\begin{bmatrix}{\mathrm{7}}&{\mathrm{2}{m}−\mathrm{5}}&{\mathrm{5}}\\{\mathrm{2}}&{{m}^{\mathrm{2}} −\mathrm{1}}&{{m}+\mathrm{2}}\\{\mathrm{2}}&{\mathrm{2}−{m}}&{\mathrm{9}}\end{bmatrix} \\ $$$${A}^{\mathrm{4}} =\begin{bmatrix}{\mathrm{4}{m}+\mathrm{49}}&{\mathrm{2}{m}^{\mathrm{3}} −\mathrm{5}{m}^{\mathrm{2}}…