Category: Limits

Question Number 107882 by Ar Brandon last updated on 13/Aug/20 $$\underset{\mathrm{x}\to \mathrm{\infty }}{\lim }{\left(\frac{{\mathrm{a}}^{1/\mathrm{x}}+{\mathrm{b}}^{1/\mathrm{x}}}{2}\right)}^{\mathrm{x}}$$ Commented by bemath last updated on 13/Aug/20 $${coolll} \…

Question Number 107883 by Ar Brandon last updated on 13/Aug/20 $$\mathrm{Expand}{\mathrm{e}}^{1/\mathrm{x}}\sqrt{\mathrm{x}(\mathrm{x}+2)}$$ 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 173403 by mathlove last updated on 11/Jul/22 Commented by mr W last updated on 11/Jul/22 $$\frac{\sqrt[3]{n^2}\sin (n!)}{n+1}⩾-\frac{n^{\frac{2}{3}}}{n+1}>-\frac{n^{\frac{2}{3}}}{n}=-\frac{1}{\sqrt[3]{n}}$$$$\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 173386 by mathlove last updated on 10/Jul/22 Answered by mnjuly1970 last updated on 10/Jul/22 $${\lim }_{\mathrm{n}\to \mathrm{\infty }}\frac{(\mathrm{n}+1)!-1}{(\mathrm{n}+1)!}=1-{\lim }_{\mathrm{n}\to \mathrm{\infty }}\left(\frac{1}{(\mathrm{n}+1)!}\right)$$$$=1$$ Commented by…

Question Number 173351 by cortano1 last updated on 10/Jul/22 Answered by aleks041103 last updated on 10/Jul/22 $$lny=\frac{ln\left(tgx\right)}{1+\sqrt[3]{1+{ln}^{2}x}}$$$$\Rightarrow L=\underset{x\to 0}{lim}ln\left(y\right)=\left[\frac{-\mathrm{\infty }}{\mathrm{\infty }}\right]$$$$\Rightarrow L^{\prime }Hopital$$$${L}=\underset{{x}\rightarrow\mathrm{0}}…