Question Number 215667 by mathlove last updated on 14/Jan/25 $$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}{sec}\left(\frac{\pi}{\mathrm{2}}{x}\right)\left({arctanx}−\frac{\pi}{\mathrm{4}}\right)=? \\ $$ Answered by issac last updated on 14/Jan/25 $$\mathrm{sec}\left(\frac{\pi}{\mathrm{2}}\right)\left(\mathrm{arctan}\left(\mathrm{1}\right)−\frac{\pi}{\mathrm{4}}\right) \\ $$ Commented by…
Question Number 215343 by RoseAli last updated on 03/Jan/25 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:\mathrm{3}{x}\mathrm{cos5}{x}\:}{{x}^{\mathrm{2}} } \\ $$ Commented by Frix last updated on 04/Jan/25 $$\mathrm{Does}\:\mathrm{not}\:\mathrm{exist}. \\ $$ Answered…
Question Number 215051 by MrGaster last updated on 27/Dec/24 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\frac{{k}^{\mathrm{2}} +\mathrm{1}}{\:\sqrt{{k}^{\mathrm{2}} +\mathrm{4}}}=? \\ $$$$ \\ $$ Answered by MathematicalUser2357 last…
Question Number 214719 by RoseAli last updated on 18/Dec/24 Answered by som(math1967) last updated on 18/Dec/24 $$\:{let}\:{x}−\mathrm{1}={a} \\ $$$$\:{x}\rightarrow\mathrm{1}\Rightarrow\left({x}−\mathrm{1}\right)\rightarrow\mathrm{0}\Rightarrow{a}\rightarrow\mathrm{0} \\ $$$$\:\underset{{a}\rightarrow\mathrm{0}} {{lim}}\frac{{sina}}{{a}}\:=\mathrm{1} \\ $$ Answered…
Question Number 214712 by efronzo1 last updated on 17/Dec/24 $$\:\:\:\cancel{\underline{\underbrace{\boldsymbol{{x}}}}} \\ $$ Answered by mr W last updated on 17/Dec/24 $${r}^{\mathrm{2}} −\mathrm{3}{r}+\mathrm{2}=\mathrm{0} \\ $$$$\left({r}−\mathrm{1}\right)\left({r}−\mathrm{2}\right)=\mathrm{0} \\…
Question Number 214645 by efronzo1 last updated on 14/Dec/24 Answered by golsendro last updated on 15/Dec/24 $$\:=\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{4}−\mathrm{3x}}−\mathrm{1}}{\:\sqrt{\frac{\mathrm{1}}{\mathrm{3}−\mathrm{2x}}}−\mathrm{1}}\: \\ $$$$\:\:=\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{4}−\mathrm{3x}}−\mathrm{1}}{\mathrm{1}−\sqrt{\mathrm{3}−\mathrm{2x}}}\:.\:\sqrt{\mathrm{3}−\mathrm{2x}} \\ $$$$\:\:=\:\mathrm{1}.\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{3}−\mathrm{3x}}{\mathrm{2x}−\mathrm{2}}\:.\:\frac{\mathrm{1}+\sqrt{\mathrm{3}−\mathrm{2x}}}{\:\sqrt{\mathrm{4}−\mathrm{3x}}+\mathrm{1}} \\…
Question Number 214485 by depressiveshrek last updated on 09/Dec/24 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{1}}{\mathrm{2}\centerdot\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{5}\centerdot\mathrm{7}}+…+\frac{\mathrm{1}}{\left(\mathrm{3}{n}−\mathrm{1}\right)\left(\mathrm{3}{n}+\mathrm{1}\right)}\right) \\ $$ Answered by mr W last updated on 10/Dec/24 $$\psi\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\right)=−\gamma+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}−\frac{\mathrm{1}}{\mathrm{3}}}\right) \\…
Question Number 214326 by mathlove last updated on 05/Dec/24 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{4}}]{{x}}−\sqrt[{\mathrm{6}}]{{x}}}{\:\sqrt[{\mathrm{4}}]{{x}}+\sqrt[{\mathrm{6}}]{{x}}}=? \\ $$ Answered by malwan last updated on 05/Dec/24 $${put}\:{y}={x}^{\mathrm{12}} \\ $$$$\underset{{y}\rightarrow\infty} {{lim}}\:\frac{{y}^{\mathrm{3}} −{y}^{\mathrm{2}}…
Question Number 214258 by efronzo1 last updated on 03/Dec/24 $$\:\:\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt[{\mathrm{6}}]{\mathrm{x}^{\mathrm{6}} −\mathrm{x}^{\mathrm{5}} }−\sqrt[{\mathrm{6}}]{\mathrm{x}^{\mathrm{6}} +\mathrm{5x}^{\mathrm{5}} }\:=? \\ $$ Answered by issac last updated on 03/Dec/24 $$−\mathrm{1}…
Question Number 214216 by MathematicalUser2357 last updated on 01/Dec/24 $${a}_{{n}} ={a}_{{n}−\mathrm{1}} +\frac{{x}−{a}_{{n}−\mathrm{1}} }{{t}} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{a}_{{n}} =? \\ $$ Answered by mr W last updated…