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Question Number 69607 by malwaan last updated on 25/Sep/19

$$\mathit{w}\mathit{i}\mathit{t}\mathit{h}\mathit{o}\mathit{u}\mathit{t}\mathit{u}\mathit{s}\mathit{i}\mathit{n}\mathit{g}\mathit{l}\mathit{h}\mathit{o}\mathit{s}\mathit{p}\mathit{i}\mathit{t}\mathit{a}\mathit{l}\mathit{p}\mathit{l}\mathit{e}\mathit{a}\mathit{s}\mathit{e}$$$$\mathit{p}\mathit{r}\mathit{o}\mathit{v}\mathit{e}\mathit{t}\mathit{h}\mathit{a}\mathit{t}$$$$\underset{\mathit{x}\to 0}{\mathit{l}\mathit{i}\mathit{m}}\frac{\mathit{x}-\mathit{s}\mathit{i}\mathit{n}\mathit{x}}{{\mathit{x}}^3}=\frac{1}{6}$$$$\mathit{I}\mathit{w}\mathit{a}\mathit{n}\mathit{t}\mathit{e}\mathit{v}\mathit{e}\mathit{r}\mathit{y}\mathit{m}\mathit{e}\mathit{t}\mathit{h}\mathit{o}\mathit{d}$$$$\mathit{p}\mathit{o}\mathit{s}\mathit{s}\mathit{i}\mathit{b}\mathit{l}\mathit{e}\mathit{b}\mathit{e}\mathit{c}\mathit{a}\mathit{u}\mathit{s}\mathit{e}\mathit{s}\mathit{o}\mathit{m}\mathit{e}\mathit{o}\mathit{n}\mathit{e}$$$$\mathit{c}\mathit{h}\mathit{a}\mathit{l}\mathit{l}\mathit{e}\mathit{n}\mathit{g}\mathit{e}\mathit{m}\mathit{e}$$

Commented by mathmax by abdo last updated on 26/Sep/19

$$wehavesinx=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n{x}^{2n+1}}{(2n+1)!}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...\Rightarrow$$$$-sinx=-x+\frac{x^3}{3!}-\frac{x^5}{5!}+\frac{x^7}{7!}+o\left(x^7\right)\Rightarrow x-sinx=$$$$\frac{x^3}{3!}-\frac{x^5}{5!}+\frac{x^7}{7!}+o\left(x^7\right)\Rightarrow \frac{x-sinx}{x^3}=\frac{1}{3!}-\frac{x^2}{5!}+\frac{x^4}{7!}+o\left(x^4\right)\Rightarrow$$$${lim}_{x\to 0}\frac{x-sinx}{x^3}=\frac{1}{3!}=\frac{1}{6}$$

Commented by Prithwish sen last updated on 25/Sep/19

$$\mathrm{li}\underset{\mathrm{x}\to 0}{\mathrm{m}}\frac{\mathrm{x}-(\mathrm{x}-\frac{{\mathrm{x}}^3}{3!}+\frac{{\mathrm{x}}^5}{5!}-....)}{{\mathrm{x}}^3}=\mathrm{li}\underset{\mathrm{x}\to 0}{\mathrm{m}}\frac{1}{3!}-\frac{{\mathrm{x}}^2}{5!}+\mathrm{the}\mathrm{higher}\mathrm{power}\mathrm{of}\mathrm{x}$$$$=\frac{1}{3!}=\frac{1}{6}\mathbf{proved}$$

Commented by malwaan last updated on 26/Sep/19

$$thankyousir$$

Commented by malwaan last updated on 27/Sep/19

$$thankyousomuch$$

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