Question Number 157389 by cortano last updated on 22/Oct/21 $$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{6}{e}^{{x}^{\mathrm{2}} } −\mathrm{20ln}\:\mid{x}\mid}{\mathrm{6}{e}^{{x}^{\mathrm{2}} } −\mathrm{20ln}\:\mid{x}\mid+\mathrm{3}}\right)^{\mathrm{3}{e}^{\mathrm{2}{x}^{\mathrm{2}} } \:\mathrm{sin}\:\left(\frac{\mathrm{6}{k}}{{e}^{{x}^{\mathrm{2}} } }\right)} =\mathrm{26} \\ $$$$\:{k}=? \\ $$$$ \\…
Question Number 26294 by d.monhbayr@gmail.com last updated on 23/Dec/17 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\sqrt{\mathrm{1}+{x}\mathrm{sin}\:{x}\:}−\sqrt{\mathrm{cos}\:\mathrm{2}{x}}}{{tg}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}} \\ $$ Commented by abdo imad last updated on 24/Dec/17 $${for}\:{x}\in{v}\left(\mathrm{0}\right)\:\:\sqrt{\mathrm{1}+{u}}\sim\mathrm{1}+\frac{{u}\:}{\mathrm{2}} \\ $$$${tanu}\sim{u}…
Question Number 26293 by d.monhbayr@gmail.com last updated on 23/Dec/17 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{3}\sqrt{\left.\mathrm{1}−{x}^{\mathrm{3}} +{x}\right)}\right. \\ $$ Answered by Joel578 last updated on 24/Dec/17 $${L}\:=\:\mathrm{3}\:.\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt{\mathrm{1}\:+\:{x}\:−\:{x}^{\mathrm{3}} } \\…
Question Number 26289 by d.monhbayr@gmail.com last updated on 23/Dec/17 $$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\left(\frac{\mathrm{1}}{{x}−\mathrm{2}}−\frac{\mathrm{12}}{{x}^{\mathrm{3}} −\mathrm{8}}\right) \\ $$ Answered by prakash jain last updated on 24/Dec/17 $$\frac{\mathrm{1}}{{x}−\mathrm{2}}−\frac{\mathrm{1}}{\left({x}−\mathrm{2}\right)\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}\right)} \\…
Question Number 91813 by jagoll last updated on 03/May/20 $$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{1}−{x}+\:\mathrm{ln}\:{x}}{\mathrm{1}−\sqrt{\mathrm{2}{x}−{x}^{\mathrm{2}} }}\:? \\ $$ Commented by john santu last updated on 03/May/20 $${L}'{Hopital} \\ $$$$\underset{{x}\rightarrow\mathrm{1}}…
Question Number 26244 by abdo imad last updated on 22/Dec/17 $$\:\left({x}_{{i}} \:\right)_{\mathrm{1}\leqslant{i}\leqslant{n}} \:\:{n}\:{real}\:{number}\:\:{positifs}\:{wish}\:{verfy}\:\:\:\sum_{{i}=\mathrm{1}} ^{{i}={n}} \:{x}_{{i}} =\mathrm{1} \\ $$$${prove}\:{that}\:\:\:\sum_{\mathrm{1}\leqslant{i}\leqslant{n}} {x}_{{i}} ^{\mathrm{2}} \:\:\:\geqslant\:\:\frac{\mathrm{1}}{{n}}\:\:\:. \\ $$ Commented by…
Question Number 26243 by abdo imad last updated on 22/Dec/17 $${let}\:{put}\:{U}_{{n}} \:\:=\:\sum_{\mathrm{1}\leqslant{i}<{j}\leqslant{n}} \:\:\:\frac{\mathrm{1}}{{ij}}\:\:\:\:{find}\:\:{lim}_{{n}−>\propto} \:\:{U}_{{n}} \\ $$ Commented by abdo imad last updated on 28/Dec/17 $$\:{we}\:{have}\:{the}\:{equality}\:\:\:\:\:\left({x}_{\mathrm{1}}…
Question Number 157314 by cortano last updated on 22/Oct/21 $$\:\:\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{3}\:\mathrm{ln}\:\left(\mathrm{1}+\mathrm{5tan}\:\frac{\mathrm{4}}{{x}}\right)}{{x}\:\left(\mathrm{1}−\mathrm{cos}\:\frac{\mathrm{6}}{{x}}\right)}\:=? \\ $$ Commented by john_santu last updated on 22/Oct/21 $$\:{let}\:\frac{\mathrm{1}}{{x}}\:=\:{u}\:{and}\:{u}\rightarrow\mathrm{0} \\ $$$$\:\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{3}{u}\:\mathrm{ln}\:\left(\mathrm{1}+\mathrm{5tan}\:\mathrm{4}{u}\right)}{\mathrm{1}−\mathrm{cos}\:\mathrm{6}{u}} \\…
Question Number 26223 by abdo imad last updated on 22/Dec/17 $${let}\:{put}\:\xi\left({x}\right)=\:\:\sum_{{n}=\mathrm{1}} ^{\propto} \:\:\frac{\mathrm{1}}{{n}^{{x}} }\:\:{with}\:{x}>\mathrm{1} \\ $$$${and}\:\:\delta\left({x}\right)\:\:=\sum_{{n}=\mathrm{1}} ^{\propto} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{{x}} }\:\:\:{find}\:{a}\:{relation} \\ $$$${between}\:\xi\left({x}\right)\:{and}\:\delta\left({x}\right). \\ $$ Commented…
Question Number 26222 by abdo imad last updated on 22/Dec/17 $${find}\:{the}\:{value}\:{of}\:\:\sum_{{n}=\mathrm{1}} ^{\propto} \:\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right){n}^{\mathrm{3}} }\:{in}\:{terms}\:{of}\:\xi\left(\mathrm{3}\right) \\ $$$${we}\:{give}\:\xi\left({x}\right)=\:\sum_{{n}=\mathrm{1}} ^{\propto} \:\:\frac{\mathrm{1}}{{n}^{{x}} }\:\:{and}\:{x}>\mathrm{1} \\ $$$$\left({zeta}\:{function}\:{of}\:{Rieman}\right) \\ $$ Commented by…