Question Number 227992 by Lara2440 last updated on 09/Mar/26 $$\mathrm{Q}. \\ $$$$\underset{{x}\rightarrow{a}\:{y}\rightarrow{b}} {\mathrm{lim}\:\mathrm{lim}}\:{A}_{{m},{n}} ={L} \\ $$$$\: \\ $$$$\mathrm{Suppose}\:\mathrm{a}\:\mathrm{double}\:\mathrm{sequence}\:{A}_{{m},{n}} \:\mathrm{converges}\:\mathrm{to}\:\mathrm{L}. \\ $$$$\mathrm{According}\:\mathrm{to}\:\mathrm{the} \\ $$$$\mathrm{Moore}-\mathrm{Osgood}\:\mathrm{Theorem}\:\mathrm{the}\:\mathrm{order}\:\mathrm{of}\:\mathrm{limits}\:\mathrm{can}\:\mathrm{be}\:\mathrm{interchanged} \\ $$$$\mathrm{if}\:\mathrm{at}\:\mathrm{least}\:\mathrm{one}\:\mathrm{direction}\:\mathrm{converges}\:\mathrm{uniformly}.…
Question Number 227956 by mathlove last updated on 05/Mar/26 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{!{x}−\mathrm{1}}{\:\sqrt{{x}−\mathrm{1}}−\mathrm{1}}=? \\ $$ Answered by Lara2440 last updated on 05/Mar/26 $$\mathrm{did}\:\mathrm{you}\:\mathrm{mean}\:!\left({a}\right)\:\mathrm{is}\:\mathrm{subfcatorial}?? \\ $$$$\mathrm{and}\:{x}\:\mathrm{approaching}\:\mathrm{0}\:..\mathrm{but}\:\:\sqrt{{x}−\mathrm{1}}−\mathrm{1}….. \\ $$…
Question Number 227426 by Elie last updated on 25/Jan/26 $${Un}\:=\:\frac{\mathrm{5}{n}−\mathrm{3}}{\mathrm{3}{n}−\mathrm{5}} \\ $$$${Un}\:=\:\frac{{n}\left(\mathrm{5}−\frac{\mathrm{3}}{{n}}\right)}{{n}\left(\mathrm{3}−\frac{\mathrm{5}}{{n}}\right)} \\ $$$${Un}\:=\:\frac{\left(\mathrm{5}−\frac{\mathrm{3}}{{n}}\right)}{\left(\mathrm{3}−\frac{\mathrm{5}}{{n}}\right)} \\ $$$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}{Un}\:=\:\frac{\left(\mathrm{5}−\frac{\mathrm{3}}{\infty}\right)}{\left(\mathrm{3}−\frac{\mathrm{5}}{\infty}\right)}=\:\frac{\mathrm{5}}{\mathrm{3}} \\ $$$${Good}\:{Luck} \\ $$ Terms of Service Privacy…
Question Number 227067 by Spillover last updated on 29/Dec/25 Answered by peace2 last updated on 29/Dec/25 $$\frac{\mathrm{1}}{{x}}\rightarrow{y} \\ $$$$=\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{{y}}.\frac{\left[{y}\right]}{{y}^{\mathrm{2}} }{dy}=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{{k}} ^{{k}+\mathrm{1}}…
Question Number 227051 by mnjuly1970 last updated on 28/Dec/25 $$ \\ $$$$\:\:\:\:\:\underset{{n}\in\mathbb{N}\cup\left\{\mathrm{0}\right\}} {\sum}{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{3}{n}\:+\:\mathrm{2}}\:\right)=?\:\:\:\:\:\:\:\:\:\:\:\:\blacksquare\: \\ $$$$\:\:\:\:\:\:\:\:\:\: \\ $$ Commented by Spillover last updated on…
Question Number 226973 by CrispyXYZ last updated on 23/Dec/25 $$\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\left(\frac{\mathrm{sin}\:\frac{\mathrm{1}}{{n}}}{{n}+\frac{\mathrm{1}}{\mathrm{1}}}\:+\:\frac{\mathrm{sin}\:\frac{\mathrm{2}}{{n}}}{{n}+\frac{\mathrm{1}}{\mathrm{2}}}\:+\:…\:+\:\frac{\mathrm{sin}\:\frac{{n}}{{n}}}{{n}+\frac{\mathrm{1}}{{n}}}\right)=? \\ $$ Terms of Service Privacy Policy Contact: [email protected]
Question Number 226815 by CrispyXYZ last updated on 16/Dec/25 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{cos}\:\frac{{x}}{\mathrm{2}}\:\mathrm{cos}\:\frac{{x}}{\mathrm{2}^{\mathrm{2}} }\:…\:\mathrm{cos}\:\frac{{x}}{\mathrm{2}^{{n}} }\right)\right)\:=\:? \\ $$ Answered by AgniMath last updated on 16/Dec/25 $${sin}\mathrm{2}\theta\:=\:\mathrm{2}{sin}\theta{cos}\theta \\…
Question Number 226550 by gregori last updated on 03/Dec/25 $$\:\:\:\: {a}+{b}+{c}\:=\:{x}\: \\ $$$$\:\:\:\: \underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} }{{abc}}\:=? \\ $$ Answered by peace2 last updated…
Question Number 224713 by efronzo1 last updated on 28/Sep/25 $$\:\:\:\cancel{\underline{\underbrace{\boldsymbol{{x}}}}} \\ $$ Answered by mehdee7396 last updated on 29/Sep/25 $$\mathrm{1} \\ $$ Commented by gregori…
Question Number 224288 by mnjuly1970 last updated on 31/Aug/25 $$ \\ $$$$\:\:\:\:\:\:\mathrm{P}=\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\prod}}\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{k}}}\:\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{k}}\right)}\:=?\:\:\:\: \\ $$$$ \\ $$ Commented by Frix last updated on 31/Aug/25…