Question Number 58405 by rahul 19 last updated on 22/Apr/19 $$\left.\mathrm{1}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{e}^{{ax}} −{bx}−\mathrm{1}}{{x}^{\mathrm{2}} }=\mathrm{2}\:. \\ $$$${find}\:{a},{b}\:? \\ $$$$ \\ $$$$\left.\mathrm{2}\right)\:\mathrm{6}\:{balls}\:{are}\:{placed}\:{randomly}\:{into} \\ $$$$\mathrm{6}\:{cells}.\:{Then}\:{the}\:{probability}\:{that}\:{exactly} \\ $$$${one}\:{cell}\:{remains}\:{empty}\:{is}\:? \\…
Question Number 123874 by john_santu last updated on 29/Nov/20 $$\:\left(\mathrm{1}\right)\underset{\theta\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\theta−\mathrm{sin}\:\theta\mathrm{cos}\:\theta}{\mathrm{tan}\:\theta−\theta} \\ $$$$\:\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\mathrm{3}{x}−\mathrm{3}{x}+{x}^{\mathrm{2}} }{\mathrm{sin}\:{x}\mathrm{sin}\:\mathrm{2}{x}} \\ $$ Answered by liberty last updated on 29/Nov/20 $$\:{solve}\:{this}\:{question}\:…
Question Number 189382 by TUN last updated on 15/Mar/23 Answered by Ar Brandon last updated on 15/Mar/23 $$\mathrm{Use}\:\mathrm{integration}\:\mathrm{by}\:\mathrm{parts}\:\mathrm{with}; \\ $$$${a}.\:\mathrm{u}\left({x}\right)={x}^{\mathrm{3}} +\mathrm{1}\:\mathrm{and}\:\mathrm{v}'\left({x}\right)={e}^{\mathrm{3}{x}−\mathrm{2}} \\ $$$${b}.\:\mathrm{u}\left({x}\right)=\mathrm{3}{x}^{\mathrm{2}} −{x}+\mathrm{2}\:\mathrm{and}\:\mathrm{v}'\left({x}\right)={e}^{{x}−\mathrm{2}} \\…
Question Number 58309 by Smail last updated on 21/Apr/19 $${Prove}\:{that}\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\mathrm{1}−{cos}\left({x}\right){cos}\left({x}/\mathrm{2}\right){cos}\left({x}/\mathrm{3}\right)…}{{x}^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$ Commented by maxmathsup by imad last updated on…
Question Number 58267 by salahahmed last updated on 20/Apr/19 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\frac{\mathrm{1}−\mathrm{cos}\:\left({x}\right)\mathrm{cos}\:\left(\mathrm{2}{x}\right)\mathrm{cos}\:\left(\mathrm{3}{x}\right)\mathrm{cos}\:\left(\mathrm{4}{x}\right)}{{x}^{\mathrm{2}} }\right] \\ $$ Commented by Smail last updated on 20/Apr/19 Commented by Smail last…
Question Number 58247 by tanmay last updated on 20/Apr/19 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{{S}_{\mathrm{1}} {S}_{{n}} +{S}_{\mathrm{2}} {S}_{{n}−\mathrm{1}} +{S}_{\mathrm{3}} {S}_{{n}−\mathrm{2}} +…+{S}_{{n}} {S}_{\mathrm{1}} }{{S}_{\mathrm{1}} ^{\mathrm{2}} +{S}_{\mathrm{2}} ^{\mathrm{2}} +…+{S}_{{n}} ^{\mathrm{2}} }…
Question Number 58239 by salahahmed last updated on 20/Apr/19 $$\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\left({x}^{\frac{\mathrm{1}}{{x}}} \right) \\ $$ Answered by salahahmed last updated on 23/Apr/19 $$\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\left({e}^{\frac{\mathrm{ln}\left({x}\right)}{{x}}}…
Question Number 189285 by TUN last updated on 14/Mar/23 $${f}\left({x}\right)\:{is}\:{continous}\:{function}\:{on}\:{R} \\ $$$${and}\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{{f}\left(\frac{{x}+\mathrm{1}}{{x}}\right)−\mathrm{6}}{\left(\frac{{x}−\mathrm{1}}{{x}}\right)^{\mathrm{2}} }=\mathrm{2} \\ $$$${Evalute}\::\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\sqrt{{f}\left({x}\right)+{x}}−{x}}{\left({x}−\mathrm{1}\right)}=¿ \\ $$ Answered by cortano12 last updated on…
Question Number 189145 by mathlove last updated on 12/Mar/23 $${pleas}\:{solve}\:{this} \\ $$$$\left.\mathrm{1}\right)\:\:\:\:\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{{e}^{{x}+\mathrm{2}{x}+\mathrm{3}{x}+\mathrm{4}{x}+\centerdot\centerdot\centerdot\centerdot\centerdot+{nx}} −{e}^{\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}} }{{x}−\mathrm{1}}=? \\ $$$$\left.\mathrm{2}\right)\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{{e}^{\mathrm{2}^{{x}} \centerdot\mathrm{3}^{{x}} \centerdot\mathrm{4}^{{x}} \centerdot\centerdot\centerdot\centerdot{n}^{{x}} } −{e}^{{n}!} }{{x}−\mathrm{1}}=? \\…
Question Number 58067 by mustakim420 last updated on 17/Apr/19 Commented by maxmathsup by imad last updated on 17/Apr/19 $$\left.{let}\:{A}\left({x}\right)\:={x}^{\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}} −{e}} \:\:\:\Rightarrow{A}\left({x}\right)\:=\:{e}^{\left\{\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}} −{e}\right\}{ln}\left({x}\right)} \:\:\Rightarrow{ln}\left({A}\left({x}\right)\right)=\left\{\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}} −{e}\right)\right\}{ln}\left({x}\right) \\…