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Question Number 156993 by amin96 last updated on 18/Oct/21

$$\text{Extra left or missing right}$$ $$0<x<\pi$$

Commented byMathSh last updated on 18/Oct/21

$$0<x<\pi \mathrm{or}0⩽x⩽\pi .?$$

Commented byamin96 last updated on 18/Oct/21

$$no0<x<\pi$$

Commented byMathSh last updated on 18/Oct/21

$$\text{Double subscripts: use braces to clarify}$$ $${\mathrm{f}}_{\mathbf{n}}\left(\mathrm{x}\right)\sim \alpha {\mathrm{n}}^{-\mathbf{r}}$$ $${\mathrm{f}}_{\mathbf{n}+1}=\sin {\mathrm{f}}_{\mathbf{n}}={\mathrm{f}}_{\mathbf{n}}-\left(\frac{1}{6}\right){\mathrm{f}}_{\mathbf{n}}^3+\mathrm{O}\left({\mathrm{f}}_{\mathbf{n}}^5\right)$$ $$\alpha {\mathrm{n}}^{-\mathbf{r}}{(1+\frac{1}{\mathrm{n}})}^{-\mathbf{r}}\sim \alpha {\mathrm{n}}^{-\mathbf{r}}[1-\left(\frac{1}{6}\right){\left(\alpha {\mathrm{n}}^{-\mathbf{r}}\right)}^2+\mathrm{O}\left({\mathrm{n}}^{-4\mathbf{r}}\right)]$$ $$-{\mathrm{rn}}^{-1}=-\frac{{\alpha }^2}{6}{\mathrm{n}}^{-2\mathbf{r}}$$ $$\alpha =\sqrt{3}\to \mathrm{r}=\frac{1}{2}$$ $$\Rightarrow {\underset{\mathbf{n}\to \mathrm{\infty }}{\lim n}}^{\mathbf{r}}{\mathrm{f}}_{\mathbf{n}}\left(\mathrm{x}\right)=\alpha =\sqrt{3}\bullet$$

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