Question Number 60856 by Tony Lin last updated on 26/May/19 $${if}\:\:\mathrm{0}<{x}<\mathrm{1},\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\frac{\left.{x}^{{x}^{{x}^{.^{.^{.^{{x}} } } } } } \right\}{n}}{\left(\left({x}^{{x}} \right)^{{x}} \right)^{\left.{x}…\right\}{n}} }=? \\ $$$$\left(?\:{can}\:{be}\:{expressed}\:{by}\:{x}\right) \\ $$…
Question Number 126376 by benjo_mathlover last updated on 20/Dec/20 $$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\mathrm{1}+\mathrm{3}{x}\right)^{−\frac{\mathrm{5}}{{x}}} \:? \\ $$ Answered by liberty last updated on 20/Dec/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+\mathrm{3}{x}\right)^{−\frac{\mathrm{5}}{{x}}} =\:{e}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+\mathrm{3}{x}−\mathrm{1}\right)\left(−\frac{\mathrm{5}}{{x}}\right)}…
Question Number 126378 by benjo_mathlover last updated on 20/Dec/20 $$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:{x}−\mathrm{cos}\:\mathrm{3}{x}+{x}^{\mathrm{3}} \mathrm{cos}\:\left(\frac{\pi}{{x}}\right)}{{x}^{\mathrm{2}} }? \\ $$ Answered by liberty last updated on 20/Dec/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:{x}−\mathrm{cos}\:\mathrm{3}{x}}{{x}^{\mathrm{2}} }+\underset{{x}\rightarrow\mathrm{0}}…
Question Number 191887 by Rupesh123 last updated on 03/May/23 Answered by mehdee42 last updated on 03/May/23 $${hop}\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} \frac{{sinxcos}\mathrm{2}{xco}\mathrm{3}{x}…{cosnx}+\mathrm{2}{sin}\mathrm{2}{xcosxcos}\left(\mathrm{3}{x}\right)…{cos}\left({nx}\right)+…+{nsin}\left({nx}\right){cosxcos}\left(\mathrm{2}{x}\right)…{cos}\left({n}−\mathrm{1}\right){x}}{\mathrm{2}{x}}= \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:\frac{{x}\left(\mathrm{1}+\mathrm{4}+\mathrm{9}+…+{n}^{\mathrm{2}} \right)}{{x}}=\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}\checkmark \\ $$ Commented…
Question Number 191874 by Subhi last updated on 02/May/23 $${lim}_{{x}\Rightarrow\infty} \frac{{e}^{{x}+\mathrm{1}} +{pi}^{{x}−\mathrm{1}} }{{e}^{{x}−\mathrm{1}} +{pi}^{{x}+\mathrm{1}} } \\ $$$$ \\ $$ Answered by mehdee42 last updated on…
Question Number 126195 by bramlexs22 last updated on 18/Dec/20 $$\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{2}\sqrt{{x}}−\mathrm{sin}\:\left({x}−\mathrm{1}\right)−\mathrm{2cos}\:\left({x}−\mathrm{1}\right)}{\mathrm{arctan}\:\left({x}−\mathrm{1}\right)−\mathrm{ln}\:{x}}\:? \\ $$ Answered by liberty last updated on 18/Dec/20 $${let}\:{w}={x}−\mathrm{1}\: \\ $$$${L}=\underset{{w}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}\sqrt{{w}+\mathrm{1}}−\mathrm{sin}\:{w}−\mathrm{2cos}\:{w}}{\mathrm{arctan}\:{w}−\mathrm{ln}\:\left({w}+\mathrm{1}\right)} \\…
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Question Number 126056 by TITA last updated on 16/Dec/20 Commented by TITA last updated on 16/Dec/20 $$\mathrm{please}\:\mathrm{help} \\ $$ Answered by Olaf last updated on…
Question Number 191359 by amin96 last updated on 23/Apr/23 $$\boldsymbol{\mathrm{Find}}\:\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\boldsymbol{\Gamma}\left(\boldsymbol{\mathrm{x}}\right)\boldsymbol{\psi}^{\left(\mathrm{0}\right)} \left(\boldsymbol{\mathrm{x}}\right)+\pi^{\mathrm{2}} \frac{\boldsymbol{\mathrm{cot}}\left(\pi\boldsymbol{\mathrm{x}}\right)}{\boldsymbol{\mathrm{sin}}\left(\pi\boldsymbol{\mathrm{x}}\right)}\right)=? \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 125814 by Tanuidesire last updated on 14/Dec/20 $${Let}\:{f}:\left[\mathrm{1},\mathrm{5}\right]\rightarrow\mathbb{R}\:{be}\:{defined}\:{by}\:{f}\left({x}\right)=\frac{\mathrm{6}}{{x}+\mathrm{1}}.\:{Show}\:{that}\:{f}\:{has}\:{a}\:{unique}\:{fixed}\:{point}\:{and}\:{find}\:{it}. \\ $$ Answered by Olaf last updated on 14/Dec/20 $${f}\:\mathrm{strictly}\:\mathrm{decreases}\:\mathrm{and}\:: \\ $$$${f}\left(\mathrm{1}\right)\:=\:\frac{\mathrm{6}}{\mathrm{1}+\mathrm{1}}\:=\:\mathrm{3} \\ $$$${f}\left(\mathrm{5}\right)\:=\:\frac{\mathrm{6}}{\mathrm{5}+\mathrm{1}}\:=\:\mathrm{1} \\…