Question Number 173499 by cortano1 last updated on 12/Jul/22 Commented by blackmamba last updated on 12/Jul/22 $$\:{x}=\mathrm{1}+{r}\: \\ $$$$\:{L}\:=\:\underset{{r}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}}]{\mathrm{7}+\left(\mathrm{1}+{r}\right)^{\mathrm{3}} }−\sqrt{\mathrm{3}+\left(\mathrm{1}+{r}\right)^{\mathrm{2}} }}{{r}} \\ $$$$\:\:\:\:\:=\:\underset{{r}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}}]{\mathrm{8}+\left(\mathrm{3}{r}+\mathrm{3}{r}^{\mathrm{2}}…
Question Number 107930 by bemath last updated on 13/Aug/20 $$\:\:\frac{\mathbb{B}{e}\mathbb{M}{ath}}{\bullet\cap\bullet} \\ $$$$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\mathrm{sin}\:{x}\right)^{\frac{\mathrm{1}}{\mathrm{ln}\:\sqrt{{x}}}} \:? \\ $$ Answered by bemath last updated on 13/Aug/20 $$\Rightarrow{L}=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{sin}\:\right)^{\frac{\mathrm{1}}{\mathrm{ln}\:\sqrt{{x}}}}…
Question Number 42367 by Joel578 last updated on 24/Aug/18 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\sqrt[{{n}^{\mathrm{2}} +{n}}]{\:\begin{pmatrix}{{n}}\\{\mathrm{0}}\end{pmatrix}\begin{pmatrix}{{n}}\\{\mathrm{1}}\end{pmatrix}\begin{pmatrix}{{n}}\\{\mathrm{2}}\end{pmatrix}…\begin{pmatrix}{{n}}\\{{n}}\end{pmatrix}}\right) \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 24/Aug/18 $${T}_{{n}} ={nc}_{\mathrm{0}} ×{nc}_{\mathrm{1}} ×{nc}_{\mathrm{2}}…
Question Number 107882 by Ar Brandon last updated on 13/Aug/20 $$\underset{\mathrm{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{a}^{\mathrm{1}/\mathrm{x}} +\mathrm{b}^{\mathrm{1}/\mathrm{x}} }{\mathrm{2}}\right)^{\mathrm{x}} \\ $$ Commented by bemath last updated on 13/Aug/20 $${coolll} \\…
Question Number 173417 by mathlove last updated on 11/Jul/22 Commented by kaivan.ahmadi last updated on 11/Jul/22 $${Q}_{\mathrm{4}} . \\ $$$$=\underset{\mathrm{1}} {\overset{\infty} {\sum}}{a}^{\mathrm{2}{n}−\mathrm{1}} −\underset{\mathrm{1}} {\overset{\infty} {\sum}}{a}^{\mathrm{2}{n}}…
Question Number 107883 by Ar Brandon last updated on 13/Aug/20 $$\mathrm{Expand}\:\mathrm{e}^{\mathrm{1}/\mathrm{x}} \sqrt{\mathrm{x}\left(\mathrm{x}+\mathrm{2}\right)} \\ $$ Answered by Dwaipayan Shikari last updated on 13/Aug/20 $${e}^{\frac{\mathrm{1}}{{x}}} =\mathrm{1}+\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} \mathrm{2}!}+\frac{\mathrm{1}}{{x}^{\mathrm{3}}…
Question Number 107871 by john santu last updated on 13/Aug/20 $$\:\:\:\:\:\:\:\frac{\checkmark\mathcal{JS}\checkmark}{\heartsuit} \\ $$$$\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\sqrt{\frac{{x}\:\mathrm{tan}\:{x}}{\mathrm{sin}\:\mathrm{2}{x}−\mathrm{cos}\:\mathrm{2}{x}\:+\mathrm{1}}}\:?\: \\ $$ Commented by bemath last updated on 13/Aug/20 Answered by…
Question Number 173403 by mathlove last updated on 11/Jul/22 Commented by mr W last updated on 11/Jul/22 $$\frac{\sqrt[{\mathrm{3}}]{{n}^{\mathrm{2}} }\:\mathrm{sin}\:\left({n}!\right)}{{n}+\mathrm{1}}\geqslant−\frac{{n}^{\frac{\mathrm{2}}{\mathrm{3}}} }{{n}+\mathrm{1}}>−\frac{{n}^{\frac{\mathrm{2}}{\mathrm{3}}} }{{n}}=−\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{n}}} \\ $$$$\frac{\sqrt[{\mathrm{3}}]{{n}^{\mathrm{2}} }\:\mathrm{sin}\:\left({n}!\right)}{{n}+\mathrm{1}}\leqslant\frac{{n}^{\frac{\mathrm{2}}{\mathrm{3}}} }{{n}+\mathrm{1}}<\frac{{n}^{\frac{\mathrm{2}}{\mathrm{3}}}…
Question Number 173380 by cortano1 last updated on 10/Jul/22 Commented by kaivan.ahmadi last updated on 10/Jul/22 $$\sim{li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\:\:\frac{{e}^{{x}} −{e}^{−{x}} +\mathrm{2}{sinx}}{\mathrm{1}}=\mathrm{0} \\ $$$$ \\ $$ Answered…
Question Number 173386 by mathlove last updated on 10/Jul/22 Answered by mnjuly1970 last updated on 10/Jul/22 $$\:\:\:\mathrm{lim}_{\:\mathrm{n}\rightarrow\infty} \:\frac{\:\left(\mathrm{n}+\mathrm{1}\right)!−\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)!}\:=\mathrm{1}−\mathrm{lim}_{\:\mathrm{n}\rightarrow\infty} \left(\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)!}\right) \\ $$$$\:\:\:\:\:\:=\:\mathrm{1} \\ $$ Commented by…