Question Number 55635 by gunawan last updated on 01/Mar/19 $$\mathrm{Series}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\Sigma}}\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }=.. \\ $$ Commented by Joel578 last updated on 01/Mar/19 $${Ans}\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\…
Question Number 55634 by gunawan last updated on 01/Mar/19 $$\mathrm{If}\:\underset{{x}\rightarrow{c}} {\mathrm{lim}}\:\frac{{a}_{\mathrm{0}} +{a}_{\mathrm{1}} \left({x}−{c}\right)+{a}_{\mathrm{2}} \left({x}−{c}\right)^{\mathrm{2}} +…+{a}_{{n}} \left({x}−{c}\right)^{{n}} }{\left({x}−{c}\right)^{{n}} }=\mathrm{0} \\ $$$$\mathrm{then}\:{a}_{\mathrm{0}} +{a}_{\mathrm{1}} +{a}_{\mathrm{2}} +..+{a}_{{n}} =.. \\…
Question Number 55592 by gunawan last updated on 27/Feb/19 $$\underset{{x}\rightarrow\pi/\mathrm{3}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:{x}−\mathrm{sin}\:\frac{\pi}{\mathrm{6}}}{\frac{\pi}{\mathrm{6}}−\frac{{x}}{\mathrm{2}}}=.. \\ $$ Commented by maxmathsup by imad last updated on 27/Feb/19 $${let}\:{A}\left({x}\right)=\frac{{cosx}−{sin}\left(\frac{\pi}{\mathrm{6}}\right)}{\frac{\pi}{\mathrm{6}}−\frac{{x}}{\mathrm{2}}}\:\Rightarrow\:{A}\left({x}\right)=\mathrm{6}\frac{{cosx}−\frac{\mathrm{1}}{\mathrm{2}}}{\pi−\mathrm{3}{x}}\:=\mathrm{3}\:\:\frac{\mathrm{2}{cosx}−\mathrm{1}}{\mathrm{3}\left(\frac{\pi}{\mathrm{3}}−{x}\right)} \\ $$$$=\frac{\mathrm{1}−\mathrm{2}{cosx}}{{x}−\frac{\pi}{\mathrm{3}}}\:\:\:{changement}\:{x}−\frac{\pi}{\mathrm{3}}\:={t}\:{give}\:{lim}_{{x}\rightarrow\frac{\pi}{\mathrm{3}}}…
Question Number 186659 by cortano12 last updated on 08/Feb/23 $${find}\:{the}\:{value}\:{a}\:{and}\:{b}\:{if}\: \\ $$$$\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{sin}\:\mathrm{2}{x}}{{x}^{\mathrm{2}} }\:+{a}+\frac{{b}}{{x}^{\mathrm{2}} }\right)=\mathrm{0} \\ $$ Commented by mr W last updated on 08/Feb/23…
Question Number 121074 by bramlexs22 last updated on 05/Nov/20 $$\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{x}−\mathrm{1}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}}? \\ $$ Answered by liberty last updated on 05/Nov/20 $$\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{x}−\mathrm{1}}{\:\sqrt{\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}+\mathrm{1}\right)}}\:=\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\sqrt{\frac{\mathrm{x}−\mathrm{1}}{\mathrm{x}+\mathrm{1}}}\:=\:\mathrm{0} \\…
Question Number 186580 by qaz last updated on 06/Feb/23 $$\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{{x}^{\mathrm{2}} } {dx}={A}\:\:,{Find}\:\underset{{n}\rightarrow\infty} {{lim}n}\left[\frac{\mathrm{1}}{{n}}\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\int_{\mathrm{1}} ^{{i}/{n}} {e}^{{x}^{\mathrm{2}} } {dx}+\frac{{e}−\mathrm{1}}{\mathrm{2}}\right]=? \\ $$ Terms of…
Question Number 121042 by bramlexs22 last updated on 05/Nov/20 $$\mathrm{prove}\:\mathrm{that}\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{x}} =\:\mathrm{e}\: \\ $$ Answered by bobhans last updated on 05/Nov/20 $${Let}\:{w}\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}} \\ $$$${then}\:\mathrm{ln}\:\left({w}\right)=\:\mathrm{ln}\:\left(\underset{{x}\rightarrow\infty}…
Question Number 121031 by bramlexs22 last updated on 05/Nov/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{x}+\:\mid\mathrm{x}\mid\:\left(\mathrm{1}+\mathrm{x}\right)}{\mathrm{x}}.\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{x}}\right)? \\ $$ Commented by liberty last updated on 05/Nov/20 $$\:\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}\:\frac{\mathrm{x}+\mid\mathrm{x}\mid\left(\mathrm{1}+\mathrm{x}\right)}{\mathrm{x}}.\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{x}}\right)\:=\: \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}^{−}…
Question Number 55491 by Tinkutara last updated on 25/Feb/19 Answered by tanmay.chaudhury50@gmail.com last updated on 26/Feb/19 $${trying}\:{to}\:\:{understand}\: \\ $$$${restricting}\:{under}\:{following}\:{condition} \\ $$$$\left.\mathrm{1}\right){radius}\left({r}\right)\:{taken}\:{as}\:{integer} \\ $$$$\left.\mathrm{2}\right){centre}\:{and}\:{otherpoints}\:\boldsymbol{{lie}}\:\boldsymbol{{on}}\:\boldsymbol{{the}}\:\boldsymbol{{circle}} \\ $$$$\boldsymbol{{not}}\:\boldsymbol{{counted}}\:\boldsymbol{{in}}\:\boldsymbol{{f}}\left(\boldsymbol{{r}}\right)…
Question Number 121010 by mathocean1 last updated on 04/Nov/20 $$\mathrm{calculate}: \\ $$$$\underset{\mathrm{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\:\:\:\frac{\sqrt{\mathrm{x}+\mathrm{1}}−\mathrm{2}}{\mathrm{3}−\sqrt{\mathrm{x}+\mathrm{6}}} \\ $$ Answered by 675480065 last updated on 04/Nov/20 $$=\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\frac{\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{x}+\mathrm{1}}}}{−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{x}+\mathrm{6}}}}=−\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\sqrt{\frac{\mathrm{x}+\mathrm{6}}{\mathrm{x}+\mathrm{1}}}=−\mathrm{3}/\mathrm{2}…