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Question Number 23012 by A1B1C1D1 last updated on 25/Oct/17

Commented by ajfour last updated on 25/Oct/17

$$ihadsolvedyourprevious$$$$question{\int }_0^2{\int }_{y/2}^1{e}^{x^2}dxdy=?$$

Commented by A1B1C1D1 last updated on 25/Oct/17

$$\mathrm{Yes}!$$

Answered by ajfour last updated on 25/Oct/17

$$=\underset{x\to 0}{\lim }\left(\frac{[1+2x+\frac{4x^2}{2!}+\frac{8x^3}{3!}+]..-[1-2x+\frac{4x^2}{2!}-\frac{8x^3}{3!}]-4x}{x-x+\frac{x^3}{3!}-...}\right)$$$$=\underset{x\to 0}{\lim }\left(\frac{2\times \frac{8x^3}{3!}+...}{\frac{x^3}{3!}+....}\right)=16.$$

Commented by A1B1C1D1 last updated on 26/Oct/17

$$\mathrm{Thank}\mathrm{you}.$$

Commented by A1B1C1D1 last updated on 26/Oct/17

$$\mathrm{You}\mathrm{did}\mathrm{the}\mathrm{expansion}\mathrm{of}\mathrm{the}\mathrm{sine}\mathrm{function}\mathrm{in}\mathrm{infinite}\mathrm{series},\mathrm{right}?$$

Answered by FilupS last updated on 26/Oct/17

$$\underset{x\to 0}{\lim }\frac{e^{2x}-e^{-2x}-4x}{x-\sin \left(x\right)}$$$$\underset{x\to 0}{\lim }\frac{2\left(\frac{e^{2x}-e^{-2x}}{2}\right)-4x}{x-\sin \left(x\right)}$$$$u=2x$$$$\underset{u\to 0}{\lim }\frac{2\left(\frac{e^u-e^{-u}}{2}\right)-2u}{\frac{u}{2}-\sin \left(\frac{u}{2}\right)}$$$$\underset{u\to 0}{\lim }\frac{2\sin \left(u\right)-2u}{\frac{u}{2}-\sin \left(\frac{u}{2}\right)}=\frac{0}{0}$$$$L^{\prime }H\mathrm{opitals}\mathrm{rule}$$$$\underset{u\to 0}{\lim }\frac{2\cos \left(u\right)-2}{\frac{1}{2}+\frac{1}{2}\cos \left(\frac{u}{2}\right)}$$$$L^{\prime }H\mathrm{opitals}\mathrm{rule}$$$$\underset{u\to 0}{\lim }\frac{-2\sin \left(u\right)}{-\frac{1}{4}\sin \left(\frac{u}{2}\right)}$$$$=\frac{2}{1/4}$$$$=8???$$

Commented by A1B1C1D1 last updated on 26/Oct/17

$$16...$$

Commented by FilupS last updated on 26/Oct/17

$$\mathrm{I}\mathrm{cant}\mathrm{find}\mathrm{my}\mathrm{mistake}$$

Commented by A1B1C1D1 last updated on 26/Oct/17

$$\mathrm{I}\mathrm{also}\mathrm{can}\mathrm{not}\mathrm{see}.$$

Commented by ajfour last updated on 26/Oct/17

$$\sin uor\sinh u?$$

Commented by A1B1C1D1 last updated on 26/Oct/17

$$\mathrm{It}\mathrm{should}\mathrm{be}\sinh \mathrm{u}...$$

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