Category: Limits

Question Number 126761 by bemath last updated on 24/Dec/20 Answered by liberty last updated on 24/Dec/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{3}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}{x}−\frac{\mathrm{1}}{\mathrm{8}}{x}^{\mathrm{2}} +{O}\left({x}^{\mathrm{3}} \right)\right)−\left({x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}+{O}\left({x}^{\mathrm{5}} \right)\right)^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{3}} \left({x}+\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{3}}…

Question Number 126744 by bemath last updated on 24/Dec/20 $$Givenf\left(x\right)=\frac{2-\sqrt{3}\cos x-\sin x}{{(6x-\pi )}^2}$$$$Find\underset{x\to \frac{\pi }{6}}{\lim }f\left(x\right).$$ Answered by liberty last updated on 24/Dec/20 $$\:\underset{{x}\rightarrow\pi/\mathrm{6}} {\mathrm{lim}}\frac{\sqrt{\mathrm{3}}\:\mathrm{sin}\:{x}−\mathrm{cos}\:{x}}{\mathrm{12}\left(\mathrm{6}{x}−\pi\right)}\:=\:\underset{{x}\rightarrow\pi/\mathrm{6}}…

Question Number 126599 by liberty last updated on 22/Dec/20 $$\underset{x\to \mathrm{\infty }}{\lim }(x+\sqrt{\frac{x^3}{x-1}})$$ Answered by bramlexs22 last updated on 22/Dec/20 $$\underset{x\to \mathrm{\infty }}{\lim }(x+x\sqrt{\frac{x}{x-1}})=$$$$\underset{{x}\rightarrow\infty}…

Question Number 60981 by necx1 last updated on 28/May/19 Commented by Prithwish sen last updated on 28/May/19 $$\mathrm{Let}$$$$\mathrm{A}={\lim }_{\mathrm{x}\to \mathrm{\infty }}{\left(\frac{\mathrm{x}!}{{\mathrm{x}}^{\mathrm{x}}}\right)}^{\frac{1}{\mathrm{x}}}$$$$\therefore\mathrm{lnA}=\mathrm{lim}_{\mathrm{x}\rightarrow\infty} \frac{\mathrm{1}}{\mathrm{x}}\mathrm{ln}\left(\frac{\mathrm{1}.\mathrm{2}.\mathrm{3}………..\mathrm{x}}{\mathrm{x}.\mathrm{x}.\mathrm{x}…………\mathrm{x}}\right)…

Question Number 60856 by Tony Lin last updated on 26/May/19 $$if0<x<1,\underset{n\to +\mathrm{\infty }}{\lim }\frac{x^{x^{x^{{.}^{{.}^{{.}^x}}}}}\}n}{{\left({\left(x^x\right)}^x\right)}^{x\dots \}n}}=?$$$$(?canbeexpressedbyx)$$…