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Question Number 95903 by 675480065 last updated on 28/May/20

Commented by PRITHWISH SEN 2 last updated on 28/May/20

$$\because \mathrm{x}>0$$$$\therefore {\mathrm{x}}^2>01+\mathrm{x}>1$$$$1-{\mathrm{x}}^2<1\therefore \frac{1}{1+\mathrm{x}}<1$$$$1-\mathrm{x}<\frac{1}{1+\mathrm{x}}$$$$\therefore 1-\mathrm{x}<\frac{1}{1+\mathrm{x}}<1$$$$\mathbf{as}\mathbf{we}\mathbf{know}\mathbf{that}\mathbf{if}\mathbf{f}\left(\mathbf{x}\right)<>\mathbf{g}\left(\mathbf{x}\right)\mathbf{then}{\int }_{\mathbf{a}}^{\mathbf{b}}\mathbf{f}\left(\mathbf{x}\right)\mathbf{dx}<>{\int }_{\mathbf{a}}^{\mathbf{b}}\mathbf{g}\left(\mathbf{x}\right)\mathbf{dx}$$$$\therefore {\int }_0^1(1-\mathrm{x})\mathrm{dx}<{\int }_0^1\frac{\mathrm{dx}}{1+\mathrm{x}}<{\int }_0^1\mathrm{dx}$$$$=\mathbf{x}-\frac{{\mathbf{x}}^2}{2}<\mathbf{ln}\mid 1+\mathbf{x}\mid <\mathbf{x}\mathbf{proved}$$$$$$$$$$$$$$

Commented by 675480065 last updated on 28/May/20

$$\mathrm{thanks}.$$$$\mathrm{what}\mathrm{about}\mathrm{the}\mathrm{second}\mathrm{part}$$

Commented by PRITHWISH SEN 2 last updated on 28/May/20

$$\text{Double subscripts: use braces to clarify}$$$$=\underset{\mathrm{t}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{t}}\underset{1}{\sum ^{\sqrt{\mathrm{t}}}}\mathrm{tln}(1+\frac{\mathrm{k}}{\mathrm{t}})\mathrm{let}{\mathrm{n}}^2=\mathrm{t}\mathrm{n}\to \mathrm{\infty }\Rightarrow \mathrm{t}\to \mathrm{\infty }$$$$=\underset{\mathrm{t}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{t}}\underset{1}{\sum ^{\sqrt{\mathrm{t}}}}\ln {(1+\frac{\mathrm{k}}{\mathrm{t}})}^{\mathrm{t}}=\underset{\mathrm{t}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{t}}\underset{1}{\sum ^{\sqrt{\mathrm{t}}}}\mathrm{k}\{\because \underset{\mathrm{t}\to \mathrm{\infty }}{\lim }{(1+\frac{\mathrm{k}}{\mathrm{t}})}^{\mathrm{t}}={\mathbf{e}}^{\mathbf{k}}\}$$$$=\underset{\mathrm{t}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{t}}.\frac{\sqrt{\mathrm{t}}(\sqrt{\mathrm{t}}+1)}{2}=\underset{\mathrm{t}\to \mathrm{\infty }}{\lim }\frac{(1+\frac{1}{\sqrt{\mathrm{t}}})}{2}=\frac{1}{2}$$$$\therefore {\mathbf{U}}_{\mathbf{n}}={\mathbf{e}}^{\frac{1}{2}}=\sqrt{\mathbf{e}}$$$$$$

Answered by Rio Michael last updated on 28/May/20

$$\mathrm{Alternatively},$$$$\mathrm{for}x>0,x-\frac{x^2}{2}\mathrm{is}\mathrm{decreasing},\mathrm{hence}\mathrm{if}f\left(x\right)=x-\frac{x^2}{2}\Rightarrow f{}^{\prime }\left(x\right)<0$$$$\mathrm{for}x>0,\ln (1+x)\mathrm{is}\mathrm{increasing},\mathrm{hence}\mathrm{if}\mathrm{g}\left(x\right)=\ln (1+x)\Rightarrow {\mathrm{g}}^{\prime }\left(x\right)>0$$$$\mathrm{for}x>0,x\mathrm{is}\mathrm{increasing}\mathrm{strictly},\mathrm{hence}\mathrm{if}h\left(x\right)=x\Rightarrow h^{\prime }\left(x\right)>>0$$$$\mathrm{thus}x-\frac{x^2}{2}<\ln (1+x)$$$$\mathrm{but}\ln (1+x)<x\mathrm{for}x>0$$$$\Rightarrow x-\frac{x^2}{2}<\ln (1+x)<x\mathrm{for}x>0\mathrm{proved}!$$

Answered by mathmax by abdo last updated on 28/May/20

$$1)\mathrm{for}\mathrm{x}>0\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}-\ln (1+\mathrm{x})\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\mathrm{x}\right)=1-\frac{1}{1+\mathrm{x}}=\frac{\mathrm{x}}{1+\mathrm{x}}>0\mathrm{f}\mathrm{is}\mathrm{increasing}$$$$\mathrm{and}\mathrm{f}\left(0\right)=0\Rightarrow \mathrm{f}\left(\mathrm{x}\right)>0\Rightarrow \ln (1+\mathrm{x})<\mathrm{x}\mathrm{let}\mathrm{g}\left(\mathrm{x}\right)=\ln (1+\mathrm{x})-\mathrm{x}+\frac{{\mathrm{x}}^2}{2}$$$${\mathrm{g}}^{{}^{\prime }}\left(\mathrm{x}\right)=\frac{1}{1+\mathrm{x}}-1+\mathrm{x}=\frac{1+{\mathrm{x}}^2-1}{1+\mathrm{x}}=\frac{{\mathrm{x}}^2}{1+\mathrm{x}}>0\Rightarrow \mathrm{g}\mathrm{is}\mathrm{increazing}\mathrm{and}\mathrm{g}\left(0\right)=0\Rightarrow$$$$\mathrm{g}\left(\mathrm{x}\right)>0\Rightarrow \mathrm{x}-\frac{{\mathrm{x}}^2}{2}<\ln (1+\mathrm{x})\Rightarrow \mathrm{x}-\frac{{\mathrm{x}}^2}{2}<\ln (1+\mathrm{x})<\mathrm{x}$$$$2){\mathrm{U}}_{\mathrm{n}}>0\Rightarrow \ln \left({\mathrm{U}}_{\mathrm{n}}\right)=\sum _{\mathrm{k}=1}^{\mathrm{n}}\ln (1+\frac{\mathrm{k}}{{\mathrm{n}}^2})\mathrm{we}\mathrm{have}\frac{\mathrm{k}}{{\mathrm{n}}^2}-\frac{{\mathrm{k}}^2}{{2\mathrm{n}}^4}<\ln (1+\frac{\mathrm{k}}{{\mathrm{n}}^2})<\frac{\mathrm{k}}{{\mathrm{n}}^2}\Rightarrow$$$$\sum _{\mathrm{k}=1}^{\mathrm{n}}\left(\frac{\mathrm{k}}{{\mathrm{n}}^2}\right)-\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{{\mathrm{k}}^2}{{2\mathrm{n}}^4}<\sum _{\mathrm{k}=1}^{\mathrm{n}}\ln (1+\frac{\mathrm{k}}{{\mathrm{n}}^2})<\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{\mathrm{k}}{{\mathrm{n}}^2}\Rightarrow$$$$\frac{1}{{\mathrm{n}}^2}\times \frac{\mathrm{n}(\mathrm{n}+1)}{2}-\frac{1}{{2\mathrm{n}}^4}\times \frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{6}<\ln \left({\mathrm{U}}_{\mathrm{n}}\right)<\frac{\mathrm{n}(\mathrm{n}+1)}{{2\mathrm{n}}^2}\Rightarrow$$$${\lim }_{\mathrm{n}\to +\mathrm{\infty }}\ln \left({\mathrm{U}}_{\mathrm{n}}\right)=\frac{1}{2}\Rightarrow {\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{U}}_{\mathrm{n}}={\mathrm{e}}^{\frac{1}{2}}=\sqrt{\mathrm{e}}$$

Commented by PRITHWISH SEN 2 last updated on 28/May/20

$$\mathrm{excellent}\mathrm{sir}.\mathrm{By}\mathrm{using}\mathrm{squeeze}\mathrm{theorem}.$$

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