Category: Limits

Question Number 96685 by Ar Brandon last updated on 03/Jun/20 $$\underset{\omega \to \mathrm{\infty }}{\lim 20log}\sqrt{1+{\left(\frac{\omega }{100}\right)}^2}$$ Answered by Rio Michael last updated on 04/Jun/20 $$\mathrm{By}\:\mathrm{l}'\mathrm{hopitals}\:\mathrm{rule}\:\mathrm{i}\:\mathrm{get}\:+\infty\:\mathrm{as}\:\mathrm{answer} \…

Question Number 96602 by Ar Brandon last updated on 03/Jun/20 $$\underset{\mathrm{n}\to +\mathrm{\infty }}{\lim }\frac{1}{\sqrt{\mathrm{n}}}\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\sqrt{\mathrm{k}}$$ Answered by Sourav mridha last updated on 03/Jun/20 $$\boldsymbol{{by}}\:\boldsymbol{{inspection}}\:\underset{\boldsymbol{{k}}=\mathrm{1}}…

Question Number 96489 by Ar Brandon last updated on 01/Jun/20 $$\mathcal{G}\mathrm{iven}\{\begin{array}{l}{\mathrm{u}}_0=\mathrm{a}\\ {\mathrm{u}}_{\mathrm{n}+1}=\mathrm{a}+\frac{1}{2}(1-\frac{1}{{\mathrm{b}}^{\mathrm{n}}}){\mathrm{u}}_{\mathrm{n}}\end{array}\mathrm{a}>0\mathrm{b}>1$$$$\mathrm{a}\mathrm{\setminus }\mathcal{C}\mathrm{alculate}{\mathrm{u}}_1,{\mathrm{u}}_2,\mathrm{and}{\mathrm{u}}_3.$$$$\mathrm{b}\backslash\:\:\mathcal{S}\mathrm{how}\:\:\mathrm{that}\:\:\mathrm{the}\:\:\mathrm{sequence}\:\:\left(\mathrm{u}_{\mathrm{n}} \right)_{\mathrm{n}\in\mathbb{N}} \:\:\mathrm{is}\:\:\mathrm{increasing}. \…

Question Number 96319 by  M±th+et+s last updated on 31/May/20 $$\underset{x\to 0}{lim}{\left(\frac{{(1+x)}^{\frac{1}{x}}}{e}\right)}^{\frac{1}{x}}$$ Answered by abdomathmax last updated on 31/May/20 $$\mathrm{f}\left(\mathrm{x}\right)=\left(\frac{\left(\mathrm{1}+\mathrm{x}\right)^{\frac{\mathrm{1}}{\mathrm{x}}} }{\mathrm{e}}\right)^{\frac{\mathrm{1}}{\mathrm{x}}} \:\Rightarrow\mathrm{ln}\left(\mathrm{f}\left(\mathrm{x}\right)\right)=\frac{\mathrm{1}}{\mathrm{x}}\mathrm{ln}\left(\frac{\left(\mathrm{1}+\mathrm{x}\right)^{\frac{\mathrm{1}}{\mathrm{x}}} }{\mathrm{e}}\right)…

Question Number 30759 by abdo imad last updated on 25/Feb/18 $$find{lim}_{n\to \mathrm{\infty }}{\left(\frac{n!}{n^n}\right)}^{\frac{1}{n}}.$$ Commented by abdo imad last updated on 27/Feb/18 $${let}\:{use}\:{the}\:{stirling}\:{formula}\:{we}\:{have}…