Question Number 125815 by Tanuidesire last updated on 14/Dec/20 $${prove}\:{that}\:{the}\:{function}\:{f}\left({x}\right)=\frac{\mathrm{1}}{{e}^{{x}} −\mathrm{2}}−\mathrm{2}{x},\:{has}\:{a}\:{root}\:{in}\:{the}\:{open}\:{interval}\:\left(\mathrm{0},\mathrm{1}\right).\:{recall}\:{that}\:{e}\approx\mathrm{2}.\mathrm{7} \\ $$ Answered by physicstutes last updated on 14/Dec/20 $${f}\:\left({x}\right)\:=\:\frac{\mathrm{1}}{{e}^{{x}} −\mathrm{2}}−\mathrm{2}{x} \\ $$$${f}\left(\mathrm{0}\right)\:=\:\frac{\mathrm{1}}{−\mathrm{1}}−\mathrm{0}\:=\:−\mathrm{1} \\…
Question Number 125728 by Mammadli last updated on 13/Dec/20 $$\underset{\boldsymbol{{n}}\rightarrow\infty} {\boldsymbol{{lim}}}\left(\mathrm{1}+\frac{\mathrm{3}}{\boldsymbol{{n}}\left(\boldsymbol{{n}}+\mathrm{3}\right)}\right)^{\boldsymbol{{n}}^{\mathrm{2}} } =? \\ $$ Answered by bramlexs22 last updated on 13/Dec/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left[\left(\mathrm{1}+\frac{\mathrm{3}}{{n}^{\mathrm{2}} +\mathrm{3}{n}}\right)^{\frac{{n}^{\mathrm{2}}…
Question Number 125722 by Mammadli last updated on 13/Dec/20 $$\underset{\boldsymbol{{x}}\rightarrow\infty} {\boldsymbol{{lim}}}\left(\frac{\mathrm{3}\boldsymbol{{x}}+\mathrm{6}}{\mathrm{3}\boldsymbol{{x}}+\mathrm{2}}\right)^{\boldsymbol{{x}}+\mathrm{2}} =? \\ $$ Answered by liberty last updated on 13/Dec/20 $${let}\:{us}\:{denote}\:{h}\left({x}\right)=\frac{\mathrm{3}{x}+\mathrm{6}}{\mathrm{3}{x}+\mathrm{2}}\:;\:{r}\left({x}\right)={x}+\mathrm{2} \\ $$$$\:\begin{cases}{\underset{{x}\rightarrow\infty} {\mathrm{lim}}{h}\left({x}\right)=\underset{{x}\rightarrow\infty}…
Question Number 191187 by mathlove last updated on 20/Apr/23 $${prove}\:{that} \\ $$$$\mathrm{1}:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}−{tanx}}{{x}^{\mathrm{3}} }=\frac{−\mathrm{1}}{\mathrm{3}} \\ $$$$\mathrm{2}:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}−{tan}^{−\mathrm{1}} {x}}{{x}^{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{3}} \\ $$ Commented by mathlove last…
Question Number 125649 by bramlexs22 last updated on 12/Dec/20 $$\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{ln}\:\left(\mathrm{1}+{x}\:\mathrm{arc}\:\mathrm{tan}\:{x}\right)+\mathrm{1}−{e}^{{x}} }{\:\sqrt{\mathrm{1}+\mathrm{2}{x}^{\mathrm{4}} }\:−\mathrm{1}}\:?\: \\ $$ Answered by Olaf last updated on 13/Dec/20 $$\mathrm{Let}\:{f}\left({x}\right)\:=\:\frac{\mathrm{ln}\left(\mathrm{1}+{x}\mathrm{arctan}{x}\right)+\mathrm{1}−{e}^{{x}} }{\:\sqrt{\mathrm{1}+\mathrm{2}{x}^{\mathrm{4}} }−\mathrm{1}}…
Question Number 191168 by Mingma last updated on 19/Apr/23 Answered by mehdee42 last updated on 19/Apr/23 $${A}={lim}_{{n}\rightarrow\infty} \frac{\sqrt[{{n}}]{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)…\left(\mathrm{2}{n}\right)}}{{n}}={lim}_{{n}\rightarrow\infty} \sqrt[{{n}}]{\frac{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)…\left(\mathrm{2}{n}\right)}{{n}^{{n}} }} \\ $$$$\Rightarrow{lnA}={lim}_{{n}\rightarrow\infty} \:\frac{\mathrm{1}}{{n}}\left[{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)+{ln}\left(\mathrm{1}+\frac{\mathrm{2}}{{n}}\right)+…+{ln}\left(\mathrm{1}+\frac{{n}}{{n}}\right)\right] \\ $$$$={lim}_{{n}\rightarrow\infty}…
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Question Number 60007 by meme last updated on 17/May/19 $${df}\:{of}\:{f}\left({x}\right)={x}^{{x}} {y}^{{y}} \\ $$ Answered by alex041103 last updated on 18/May/19 $${if}\:{you}\:{mean}\:{f}\left({x},{y}\right)={x}^{{x}} {y}^{{y}} \:{then} \\ $$$${df}=\frac{\partial{f}}{\partial{x}}{dx}+\frac{\partial{f}}{\partial{y}}{dy}=…
Question Number 60006 by meme last updated on 17/May/19 $${li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\frac{{x}^{{x}} }{{x}}=? \\ $$ Commented by maxmathsup by imad last updated on 17/May/19 $${let}\:{A}\left({x}\right)\:=\frac{{x}^{{x}} }{{x}}\:\:\:{for}\:{x}>\mathrm{0}\:\Rightarrow{A}\left({x}\right)\:=\frac{{e}^{{xln}\left({x}\right)}…
Question Number 125523 by Algoritm last updated on 11/Dec/20 Answered by Dwaipayan Shikari last updated on 11/Dec/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{2}{x}+\mathrm{3}\right){log}\left(\frac{{x}+\mathrm{2}}{{x}}\right) \\ $$$$=\left(\mathrm{2}{x}+\mathrm{3}\right)\left(\frac{\mathrm{2}}{{x}}\right)=\mathrm{4}+\frac{\mathrm{6}}{{x}}=\mathrm{4}\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{log}\left(\mathrm{1}+{z}\right)={z} \\ $$ Terms…